solution manual buku smith-corripio

(d) Automatic sprinkler system for fires.

The controller is On/Off and the control i feedback with respect to temperature.,

M, D, A: temperature element/controller TE/C. Usually a bi-metallic strip that pushes the latch on the mechanism holding the slice of bread. The bread is released and the heating element de-energized when the temperature reaches the value set by the set point. TE/CSP

(c) Toaster.

The controller is On/Off and the control is feedback on the temperature variable.

M: Temperature element TE, usually a gas-filled bulb

D: Temperature controller TC.

A: Solenoid S that operates the heating element in the oven

(b) Cooking oven.

TETC

S

SP

The controller is On/Off and the control is feedback.

M: Temperature element TE in thermostat TE/C

D: Mercury switch in thermostat TE/C

A: Solenoid S that turns unit (AC/H) on and off. TE/C S

AC/H

SP

(a) House air-conditioning/heating.

Identificaton of the M-D-A components, controller type, instrumentation diagram, and type of control.

Problem 1-1. Automation in daily life.

Smith & Corripio, 3rd edition

Watermain

TE/C

M, D, A: temperature element/controller TE/C, a rod that gives to the water pressure at a set temperature, allowing the water to spray over the fire.

The controller is On/Off with single action, and the control is feedback.

(e) Automatic cruise speed control.

ST

SCSP

S

Air

EngineTransmission

M: Speed sensor and transmitter ST on the transmission

D: Speed controller SC

A: Damper on air intake to the engine throttles the air varying the power delivered by the engine

Controller is regulating and control is feedback.

(f) Refrigerator.

TE/C SSP

M: temperature sensor TE, usually a gas-filled bulb

D: Temperature controller C, mechanically linked to the sensor

A: Solenoid S that turnsd the refrigeration compressor on and off

The controller is On/Off and the control is feedback.

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Problem 1-2. Automatic shower temperature diagram.

TETC

Hotwater

Coldwater

S

SPM: temperature sensor TE, a gas-filled bulb

D: temperature controller TC, mechanilly integrated to the sensor, but with a signa output

A: solenoid operated control valve on the hot water line.

The cold water valve is operated manually.


F s( )s

s2ω

2+

=

12

1s i ω⋅−

1s i ω⋅+

+= s i ω⋅− s+ i ω⋅+

2 s i ω⋅−( )⋅ s i ω⋅+( )= 2 s⋅

2 s2ω

2+( )⋅

= s

s2ω

2+

=

12

1−s i ω⋅−

e s i ω⋅−( )t−∞

0

⋅1−

s i ω⋅+e s i ω⋅+( )t−

∞

0

⋅+=

12 0

∞

te s i ω⋅−( )t−⌠

⌡d

0

∞

te s i ω⋅+( )t−⌠

⌡d+=

F s( )0

∞

tcos ωt⋅ e st−⋅

⌠

⌡d=

0

∞

tei ωt⋅ e i− ωt⋅

−

2e st−

⌠

⌡d=f t( ) cos ωt⋅=(c)

F s( )1

s a+=

F s( )0

∞

te at− e st−⌠

⌡d=

0

∞

te s a+( )t−⌠

⌡d= 1−

s a+e s a+( )t−

∞

0

⋅= 1s a+

=

where a is constantf t( ) e at−=(b)

F s( )1

s2=

F s( )t−

se st−

∞

0

⋅1s 0

∞

te st−⌠

⌡d⋅+= 0 0−

1

s2e st−

∞

0

⋅−= 1

s2=

v1−s

e st−=du dt=

dv e st− dt=u t=By parts:F s( )0

∞

tt e st−⋅

⌠

⌡d=f t( ) t=(a)

F s( )0

∞

tf t( ) e st−⌠

⌡d=

Problem 2-1. Derivation of Laplace transforms from its definitionSmith & Corripio, 3rd. edition

(d) f t( ) e at− coss ωt⋅=F s( )

0

∞

te at− cos ωt⋅ e st−⋅

⌠

⌡d=

0

∞

te at− ei ωt⋅ e i− ωt⋅+

2⋅ e st−

⌠

⌡d=

12 0

∞

te s a+ i ω⋅+( )t−⌠

⌡d

0

∞

te s a+ i ω⋅−( )− t⌠

⌡d+=

12

1−s a+ i ω⋅+

e s a+ i ω⋅+( )t−∞

0

⋅1−

s a+ i ω⋅−e s a+ i ω⋅−( )t−

∞

0

⋅+=

12

1s a+ i ω⋅+

1s a+ i ω⋅−

+= s a+ i ω⋅− s+ a+ i ω⋅+

2 s a+ i ω⋅+( ) s a+ i ω⋅−( )=

2 s a+( )

2 s a+( )2ω

2+⋅

= s a+

s a+( )2ω

2+

= F s( )s a+

s a+( )2ω

2+

=

All the results match results in Table 2-1.1


1s

1s 2+

+ 21

s 1+⋅−= 1

s1

s 2++

2s 1+

−=

F s( )1s

1s 2+

+2

s 1+−=

Used the linearity property.

(d) f t( ) u t( ) e t−− t e t−

⋅+= F s( ) L u t( )( ) L e t−( )− L t e t−⋅( )+= 1

s1

s 1+−

1

s 1+( )2+=

F s( )1s

1s 1+

−1

s 1+( )2+=


(e) f t( ) u t 2−( ) 1 e 2− t 2−( ) sin t 2−( )−= Let g t( ) u t( ) 1 e 2− tsin t⋅−( )= Then f t( ) g t 2−( )=

F s( ) e 2− s G s( )= e 2− s 1s

1

s 2+( )2 1+−=

Used the real translation theorem and linearity. F s( ) e 2− s 1s

1

s 2+( )2 1+−=


Smith & Corripio, 3rd editionProblem 2-2. Derive Laplace transforms from the properties and Table 2-1.1

(a) f t( ) u t( ) 2 t⋅+ 3 t2⋅+= F s( ) L u t( ) 2 t⋅+ 3 t2⋅+( )= L u t( )( ) 2 L t( )⋅+ 3 L t2( )⋅+=

1s

21

s2⋅+ 3

2!

s3⋅+= F s( )

1s

2

s2+

6

s3+=


(b) f t( ) e 2− t⋅ u t( ) 2 t⋅+ 3 t2⋅+( )= F s( ) L u t( ) 2 t⋅+ 3 t2⋅+( )s 2+

⋅= 1s

2

s2+

6

s3+

s 2+

⋅=

1s 2+

2

s 2+( )2+

6

s 2+( )3+=

F s( )1

s 2+

2

s 2+( )2+

6

s 2+( )3+=

Used the complex translation theorem.

(c) f t( ) u t( ) e 2− t+ 2e t−

−= F s( ) L u t( ) e 2− t+ 2 e t−

⋅−( )= L u t( )( ) L e 2− t( )+ 2 L e t−( )⋅−=

Must apply L'Hopital's rule:

∞s

11

22 s 2+( )

+6

3 s 2+( )2+ 1=lim

→Final value:

∞te 2− t u t( ) 2 t⋅+ 3t2+( ) 0 ∞⋅=lim

→0s

s1

s 2+

2

s 2+( )2+

6

s 3+( )2+ 0=lim

→L'Hopital's rule:

∞t

0

2e2t

2

2e2t+

6t

2e2t+ 0=lim

→Check!

(c) f t( ) u t( ) e 2− t+ 2e t−

−= F s( )1s

1s 2+

+2

s 1+−=

Initial value:

0tu t( ) e 2− t

+ 2e t−−( ) 1 1+ 2−( ) 0+=lim

→ ∞ss

1s

1s 2+

+2

s 1+−

∞

∞=lim

→

L'Hopital's rule:

∞s1

11

+21

− 0=lim→

Final value:

∞tu t( ) e 2− t

+ 2e t−−( ) 1 0+ 0+= 1=lim

→ 0ss

1s

1s 2+

+2

s 1+− 1 0+ 0+= 1=lim

→


Problem 2-3. Initial and final value check of solutions to Problem 2-2

(a) f t( ) u t( ) 2 t⋅+ 3t2+= F s( )1s

2

s2+

6

s3+=

Initial value:

0tu t( ) 2t+ 3t2+( ) 1=lim

→ ∞ss

1s

2

s2+

6

s3+⋅

∞s1

2s

+6

s2+ 1=lim

→=lim

→

Final value:

∞tu t( ) 2t+ 3t2+( ) ∞=lim

→ 0s1

2s

+6

s2+ ∞=lim

→Check!

(b) f t( ) e 2− t u t( ) 2t+ 3t2+( )= F s( )1

s 2+

2

s 2+( )2+

6

s 2+( )3+=

Initial value:

0te 2− t u t( ) 2t+ 3t2+( )lim

→ ∞ss

1s 2+

2

s 2+( )2+

6

s 2+( )3+

∞

∞=lim

→

1 1 0+ 0+( )= 1=


Check!

0ss

1s

1

s 1+( )2 1+− 1 0+= 1=lim

→∞t1 e 2− tsin t( )⋅− 1=lim

→

Final value:

∞ss

1s

1

s 1+( )2 1+− 1 0−= 1=lim

→0t1 e 2− tsin t⋅−( ) 1=lim

→

Initial value:

The test of the delayed fnction is not useful. Better to test the term in brackets, g(t):

F s( ) e 2− s 1s

1

s 1+( )2 1+−=f t( ) u t 2−( ) 1 e 2− t 2−( ) sin t 2−( )−=(e)

Check!∞t

1 0−1

1 et⋅

+ 1=lim→

L'Hopital's rule:

∞tu t( ) e t−

− t e t−⋅+( ) 1 0− ∞ 0⋅+=lim

→0s

1s

s 1+−

s

s 1+( )2+ 1 0− 0+= 1=lim

→

Final value: ∞s1

11

−1

2 s 1+( )+ 1 1− 0+= 0=lim

→

L'Hopital's rule:

∞ss

1s

1s 1+

−1

s 1+( )2+

∞

∞=lim

→0tu t( ) e t−

− t e t−⋅+( ) 1 1− 0 1⋅+= 0=lim

→

Initial value:

F s( )1s

1s 1+

−1

s 1+( )2+=f t( ) u t( ) e t−

− t e t−⋅+=(d)

Smith & Corripio, 3rd editionProblem 2-4. Laplace transform of a pulse by real translation theorem

f t( ) H u t( )⋅ H u t T−( )⋅−=

F s( ) H1s⋅ H e sT−

⋅1s⋅−= H

1 e sT−−

s⋅= F s( )

Hs

1 e sT−−( )=


0 2 40

2fd t( )

t0 2 4

0

2f t( )

t

f t( ) e

t0

τ e

t−

τ⋅:=

fd t( ) u t t0−( ) e

t t0−( )−

τ⋅:=

u t( ) 0 t 0<if

1 t 0≥if

:=τ 1:=t0 1:=Sketch the functions:F s( )

τ et0− s⋅

⋅

τ s⋅ 1+=

The result to part (b) agrees with the real translation theorem.

et0− s⋅ 1−

s1τ

+

⋅ es

1

τ+− λ⋅

⋅ ∞

0

⋅= et0− s⋅

s1τ

+

= τ et0− s⋅

⋅

τ s⋅ 1+=

F s( )t0−

∞

λu λ( ) e

λ−

τ es λ t0+( )−

⌠

⌡d= e

t0− s⋅

0

∞

λes

1

τ+ λ−

⌠

⌡d⋅=λ t t0−=Let

F s( )0

∞

tu t t0−( ) e

t t0−( )−

τ e st−

⌠

⌡d=f t( ) u t t0−( ) e

t t0−( )−

τ=

(b) Function is delayed and zero from t = 0 to t = t0:

F s( )τ e

t0

τ⋅

τ s⋅ 1+=F s( ) e

t0

τ 1

s1τ

+

= τ e

t0

τ⋅

τ s⋅ 1+=f t( ) e

t0

τ e

t−

τ=(from Table 2-1.1)

(a) Function is non-zero for all values of t > 0:

f t( ) e

t t0−( )−

τ=

Problem 2-5. Delayed versus non-delayed function

Y t( ) 2.5− e t− 2.5 u t( )+= (Table 2-1.1)

(b)9

d2 y t( )⋅

dt2⋅ 18

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Initial steady state: 4 y 0( )⋅ 8 x 0( ) 4−=

Subtract:9

d2 Y t( )⋅

dt2⋅ 18

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )=

Y t( ) y t( ) y 0( )−= Y 0( ) 0=

X t( ) x t( ) x 0( )−=Laplace transform:

9s2 Y s( ) 18s Y s( )⋅+ 4 Y s( )+ 8 X s( )= 81s⋅=

Solve for Y(s): Y s( )8

9s2 18s+ 4+

1s

= r118− 182 4 9⋅ 4⋅−+

2 9⋅:= r1 0.255−=

r218− 182 4 9⋅ 4⋅−−

2 9⋅:= r2 1.745−=

Expand in partial fractions:Y s( )

89 s 0.255+( ) s 1.745+( )s

=A1

s 0.255+

A2s 1.745+

+A3s

+=

A10.255−s

89 s 1.745+( )s

89 0.255− 1.745+( )⋅ 0.255−( )⋅

= 2.342−=lim→

=

Smith & Corripio, 3rd editionProblem 2-6. Solution of differential equations by Laplace transformsInput function: X t( ) u t( )= X s( )

1s

= (Table 2-1.1)

(a)d y t( )⋅

dt2 y t( )+ 5 x t( ) 3+=

Initial steady state: 2 y 0( ) 5 x 0( )= 3=

Subtract:d Y t( )⋅

dt2 Y t( )+ 5 X t( )= Y t( ) y t( ) y 0( )−= X t( ) x t( ) x 0( )−=

Laplace transform: sY s( ) Y 0( )− 2 Y s( )+ 5 X s( )= 51s⋅= Y 0( ) y 0( ) y 0( )−= 0=

Solve for Y(s):Y s( )

5s 2+

1s

=A1

s 2+

A2s

+=

Partial fractions:

A12−s

5s

2.5−=lim→

= A20s

5s 2+

2.5=lim→

=

Y s( )5−

s 1+

5s

+= Invert:

Y 0( ) 0=9d2 Y t( )⋅

dt2⋅ 12

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )=

Subtract initial steady state:

9d2 y t( )⋅

dt2⋅ 12

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=(d)

Y t( ) 1− 1.134i+( )e 0.5− 0.441i+( )t 1− 1.134i−( )e 0.5− 0.441i−( )t+ 2 u t( )+=

Invert using Table 2-1.1:

Y s( )1− 1.134i+

s 0.5+ 0.441i−

1− 1.134i−

s 0.5+ 0.441i++

2s

+=

A30s

8

9s2 9s+ 4+2=lim

→=A2 1− 1.134i−=

89 2 0.441i⋅( ) 0.5− 0.441i+( )

1− 1.134i+=A10.5− 0.441i+s

89 s 0.5+ 0.441i+( ) s

lim→

=

A1s 0.5+ 0.441i−

A2s 0.5+ 0.441i+

+A3s

+=

Y s( )8

9 s 0.5+ 0.441i−( ) s 0.5+ 0.441+( )s=Solve for Y(s), expand:

A21.745−s

89 s 0.255+( )s

89 1.745− 0.255+( ) 1.745−( )

= 0.342=lim→

=

A30s

89 s 0.255+( ) s 1.745+( )

89 0.255( ) 1.745( )

= 2.0=lim→

=

Y s( )2.342−

s 0.255+

0.342s 1.745+

+2s

+=

Invert with Table 2-1.1:Y t( ) 2.342− e 0.255− t 0.342e 1.745− t

+ 2 u t( )+=

(c) 9d2 y t( )⋅

dt2⋅ 9

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract initial steady state:9

d2 Y t( )⋅

dt2⋅ 9

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=

Laplace transform:9s2 9s+ 4+( )Y s( ) 8 X s( )= 8

1s⋅=

r19− 92 4 9⋅ 4⋅−+

2 9⋅:= r2

9− 92 4 9⋅ 4⋅−−

2 9⋅:= r1 0.5− 0.441i+=

Find roots:

r2 0.5− 0.441i−=

A2 0.027 0.022i−=3

2 2 2.598i⋅( ) 1− 2.598i+( ) 1.5− 2.598i+( )0.027 0.022i+=

A11.5− 2.598i+s

32 s 1.5+ 2.598i+( ) s 0.5+( )s

0.027 0.022i+=lim→

=

A1s 1.5+ 2.598i−

A2s 1.5+ 2.598i+

+A3

s 0.5++

A4s

+=

Y s( )3

2 s 1.5+ 2.598i−( ) s 1.5+ 2.598i+( ) s 0.5+( )s=Solve for Y(s) and expand:

polyroots

9

21

7

2

1.5− 2.598i−

1.5− 2.598i+

0.5−

=Find roots:

2s3 7s2+ 21s+ 9+( )Y s( ) 3 X s( )= 3

1s⋅=Laplace transform:

Y 0( ) 0=

2d3 Y t( )⋅

dt3⋅ 7

d2 Y t( )⋅

dt2⋅+ 21

d Y t( )⋅

dt⋅+ 9 Y t( )+ 3 X t( )=Subtract initial steady state:

2d3 y t( )⋅

dt3⋅ 7

d2 y t( )⋅

dt2⋅+ 21

d y t( )⋅

dt⋅+ 9 y t( )+ 3 x t( )=(e)

Y t( )4−

3t 2− e 0.667− t 2 u t( )+=Invert using Table 2-1.1:

A30s

8

9 s 0.667+( )22=lim

→=

A20.667−s

dds

89s 0.667−s

8−

9s22−=lim

→=lim

→=A1

0.667−s

89s

4−3

=lim→

=

Y s( )8

9 s 0.667+( )2s=

A1

s 0.667+( )2

A2s 0.667+

+A3s

+=Solve for Y(s) and expand:

r2 0.667−=

r1 0.667−=r212− 122 4 9⋅ 4⋅−−

2 9⋅:=r1

12− 122 4 9⋅ 4⋅−+

2 9⋅:=

Find roots:

9s2 12s+ 4+( )Y s( ) 8 X s( )= 81s⋅=Laplace transform:

A30.5−s

32 s 1.5+ 2.598i−( ) s 1.5+ 2.598i+( )s

0.387−=lim→

=

32 1 2.598i−( ) 1 2.598i+( ) 0.5−( )

0.387−= A40s

3

2s3 7s2+ 21s+ 9+

13

=lim→

=

Y s( )0.027 0.022i+

s 1.5+ 2.598i−

0.027 0.022i−

s 1.5+ 2.598i++

0.387−

s 0.5++

13

1s

+=


Y t( ) 0.027 0.022i+( )e 1.5− 2.598i+( )t 0.027 0.022i−( )e 1.5− 2.598i−( )t+ 0.387e 0.5− t

−13

u t( )+=



Y t( ) u t 1−( )8−

3t 1−( )⋅ 8− e 0.667− t 1−( )⋅

⋅ 8 e 0.333− t 1−( )⋅⋅+⋅=

Apply the real translation theorem in reverse to this solution:

Y s( )8−

31

s 0.667+( )2

8s 0.667+

−8

s 0.333++ e s−=

The partial fraction expansion of the undelayed signal is the same:

(Real translation theorem)

X s( )e s−

s13

+

=X t( ) u t 1−( ) e

t 1−( )−

3=(b) Forcing function:

Y t( )8−

3t 8− e 0.667− t 8e 0.333− t

+=Invert using Table 2-1.1:

Y s( )8−

31

s 0.667+( )2

8−s 0.667+

+8

s 0.333++=

A20.667−s

dds

89 s 0.333+( ) 0.667−s

8−

9 s 0.333+( )28−=lim

→=lim

→=

A30.333−s

8

9 s 0.667+( )28=lim

→=A1

0.667−s

89 s 0.333+( )

8−3

=lim→

=

8

9 s 0.667+( )2 s 0.333+( )=

A1

s 0.667+( )2

A2s 0.667+

+A3

s 0.333++=

Y s( )8

9s2 12s+ 4+( ) s13

+

=

X s( )1

s13

+

=From Table 2-1.1:X t( ) e

t−

3=(a) Forcing function:

Y 0( ) 0=9d2 Y t( )⋅

dt2⋅ 12

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )=

Problem 2-7. Solve Problem 2-6(d) with different forcing functions


(Final value theorem)

(b)9

d2 y t( )⋅

dt2⋅ 18

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract initial steady state: 9d2 Y t( )⋅

dt2⋅ 18

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=

Laplace transform and solve for Y(s): Y s( )8

9s2 18s+ 4+X s( )=

Find roots: r118− 182 4 9⋅ 4⋅−+

2 9⋅ min:= r2

18− 182 4 9⋅ 4⋅−−

2 9⋅ min:= r1 0.255− min 1−

=

r2 1.745− min 1−=

Invert using Table 2-1.1: Y t( ) A1 e 0.255− t⋅ A2 e 1.745− t

⋅+=

+ terms of X(s)

The response is stable and monotonic. The domnant root is: r1 0.255− min 1−=

Time for the response to decay to 0.67% of its initial value:5−

r119.6 min=

Final steady-state value for unit step input:0s

s8

9s2 18s+ 4+⋅

1s

lim→

2→


Smith & Corripio, 3rd editionProblem 2-8. Response characteristics of the equations of Problem 2-6

(a)d y t( )⋅

dt2 y t( )+ 5 x t( ) 3+=

Initial steady state: 2 y 0( ) 5 x 0( ) 3+=

Subtract:d Y t( )⋅

dt2 Y t( )+ 5 X t( )= Y t( ) y t( ) y 0( )−= X t( ) x t( ) x 0( )−=

Laplace transform: s Y s( )⋅ 2 Y s( )+ 5 X s( )= Y 0( ) y 0( ) y 0( )−= 0=


s 2+X s( )=

A1s 2+

= + terms of X(s)

Invert using Table 2-1.1: Y t( ) A1 e 2− t⋅= + terms of X(t)

The response is stable and monotonic.The dominant and only root is r 2− min 1−:=

Time for response to decay to within 0.67% of its initial value:5−r

2.5 min=

Final steady-state value for unit step input:

0ss

5s 2+⋅

1s

lim→

52

→ 2.5=

Time for oscillations to die:5−

0.5− min 1−10 min=

Final steady state value for a unit step imput:0s

s8

9s2 9s+ 4+⋅

1s

lim→

2→


(d) 9d2 y t( )⋅

dt2⋅ 12

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract initial steady state:9

d2 Y t( )⋅

dt2⋅ 12

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )=

Y 0( ) 0=


9s2 12s+ 4+X s( )=

Find roots: r112− 122 4 9⋅ 4⋅−+

2 9⋅ min:= r2

12− 122 4 9⋅ 4⋅−−

2 9⋅ min:= r1 0.667− min 1−

=

r2 0.667− min 1−=

Invert using Table 2-1.1: Y t( ) A1 t⋅ A2+( )e 0.667− t= + terms of X(t)

(c) 9d2 y t( )⋅

dt2⋅ 9

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract initial steady state: 9d2 Y t( )⋅

dt2⋅ 9

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=


9s2 9s+ 4+X s( )=

Find the roots: r19− 92 4 9⋅ 4⋅−+

2 9⋅ min:= r2

9− 92 4 9⋅ 4⋅−−

2 9⋅ min:= r1 0.5− 0.441i+ min 1−

=

r2 0.5− 0.441i− min 1−=

Invert using Table 2-3.1: Y t( ) D e 0.5− t⋅ sin 0.441t θ+( )= + terms of X(t)

The response is stable and oscillatory. The dominant roots are r1 and r2.

Period of the oscillations: T2π

0.441min 1−:= T 14.25 min=

Decay ratio: e 0.5− min 1− T 0.00081=


(Final value theorem)0s

s3

2s3 7s2+ 21s+ 9+

⋅1s

lim→

13

→Final steady state value for a unit step input:

5−r2

10 min=Time for response to die out:e 1.5− min 1− T 0.027=Decay ratio:

T 2.42 min=T2π

2.598min 1−:=The period of the oscillations is:

r2 0.5− min 1−=The response is stable and oscillatory. The dominant root is

r

1.5− 2.598i−

1.5− 2.598i+

0.5−

min 1−=r polyroots

9

21

7

2

min 1−:=

Find roots:

Y s( )3

2s3 7s2+ 21s+ 9+

X s( )=Laplace transform and solve for Y(s):

2d3 Y t( )⋅

dt3⋅ 7

d2 Y t( )⋅

dt2⋅+ 21

d Y t( )⋅

dt⋅+ 9 Y t( )+ 3 X t( )=Subtract initial steady state:

2d3 y t( )⋅

dt3⋅ 7

d2 y t( )⋅

dt2⋅+ 21

d y t( )⋅

dt⋅+ 9 y t( )+ 3 x t( )=(e)

(Final value theorem)0s

s8

9s2 12s+ 4+⋅

1s

lim→

2→Final steady state value for a unit step input:

5−r1

7.5 min=Time required for the response to decay within 0.67% of its initial value:

r1 0.667− min 1−=The response is stable and monotonic. The dominant root is

Value of k: kM− g⋅y0

:= k 1.816Nm

=

Laplace transform:M s2⋅ Y s( ) k Y s( )⋅+ F s( )=


M s2⋅ k+

F s( )=A1

s ikM

⋅−

A2

s ikM

⋅+

+=

+ terms of F(s)

θ 0:=

D 1:=Invert using Table 2-3.1: Y t( ) D sin

kM

t s⋅ θ+⋅:= + terms of f(t)

The mobile will oscillate forever with a period of T 2πMk

⋅:= T 1.043 s=


Problem 2-9. Second-Order Response: Bird Mobile

-Mgf(t)

y(t)

-ky(t)

y = 0

Problem data: M 50gm:= y0 27− cm:=

Solution:

Force balance:

Md v t( )⋅

dt⋅ M− g⋅ k y t( )⋅− f t( )+=

Velocity:d y t( )⋅

dtv t( )=

Initial steady state: 0 M− g⋅ k y0⋅−=

Subtract and substitute:

Md2 Y t( )⋅

dt2⋅ k− Y t( )⋅ f t( )+=

Y 0( ) 0=

0 2 41

0

1

Y t( )

t

To more accurately reflect the motion of the bird mobile, we must add the resistance of the air. If we assume it to be a force proportional to the velocity:

Md2 Y t( )⋅

dt2⋅ k− Y t( )⋅ b

d Y t( )⋅

dt⋅− f t( )+=

With this added term the roots will have a negative real part, causing the oscillations to decay, as they do in practice:

Y s( )1

M s2⋅ b s⋅+ k+

F s( )= r1b− b2 4M k⋅−+

2M= b−

2Mi

kM

b2

4M2−⋅+=

Invert:b2 4M k⋅<

Y t( ) D e

b−

2Mt⋅

⋅ sinkM

b2

4M2− t θ+= + terms of f(t)


H 1:=T 1:=τ 1:=KH 1:=Invert using Table 2-1.1, and the real translation theorem:

Y s( ) K H1s

1

s1τ

+

−⋅ 1 e sT−−( )=

A20s

K H⋅τ s⋅ 1+

K H⋅=lim→

=A11−

τs

K H⋅τ s⋅

K− H⋅=lim

→

=

Y s( )K

τ s⋅ 1+H⋅

1 e sT−−

s⋅=

A1

s1τ

+

A2s

+ 1 e sT−−( )=Substitute:

X s( ) H1 e sT−−

s⋅=

From Example 2-1.1b:

(b) Pulse of Fig. 2-1.1b

0 2 40

0.5

1

Y t( )

t

Y t( )Kτ

e

t−

τ:=


Y s( )K

τ s⋅ 1+=

X s( ) 1=From Table 2-1.1:X t( ) δ t( )=(a) Unit impulse:

Y s( )K

τ s⋅ 1+X s( )=Laplace transform and solve for Y(s):

Y 0( ) 0=τd Y t( )⋅

dt⋅ Y t( )+ K X t( )⋅=

Problem 2-10. Responses of general first-order differential equation


Y t( ) KH u t( ) e

t−

τ− u t T−( ) 1 e

t T−( )−

τ−⋅−⋅:=

X t( ) H u t( ) u t T−( )−( )⋅:=

0 2 40

0.5

1

Y t( )

X t( )

t



The tank is an integrating process because its ouput, the level, is the time integral of its input, the inlet flow.

0 5 100

5

10

h t( )

t

f(t)

h(t)

A 1:=

h t( )1A

t:=Invert using Table 2-1.1:H s( )1A

1

s2=Substitute:

(Table 2-1.1)F s( )1s

=f t( ) u t( )=Response to a unit step in flow:

H s( )F s( )

1A s⋅

=Transfer function of the tank:

H s( )1

A s⋅F s( )=Laplace transform and solve for H(s):

h 0( ) 0=Ad h t( )⋅

dt⋅ f t( )=

Problem 2-11. Response of an integrating process


r2 1.745− min 1−=

τe21−

r2:=

τe2 0.573 min=

5 τe1⋅ 19.64 min=Time for response to decay within 0.67% of its initial value:

(b) 9d2 y t( )⋅

dt2⋅ 9

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract initial steady stateand divide by the Y(t) coefficient:

94

d2 Y t( )⋅

dt2⋅

94

d Y t( )⋅

dt⋅+ Y t( )+ 2 X t( )= Y 0( ) 0=

Compare coefficients to standard form: τ94

min:= τ 1.5 min= ζ9min4 2⋅ τ⋅

:= ζ 0.75=

K 2:=

Underdamped.Find roots: r1

9− 92 4 9⋅ 4⋅−+

2 9⋅ min:= r1 0.5− 0.441i+ min 1−

=

Frequency of oscillations: ω 0.441radmin

:= Period of oscillations: T2πω

:= T 14.25 min=


Problem 2-12. Second-order differeential equations of Problem 2-6.

Standard form of the second-order equation: τ2 d2 Y t( )⋅

dt2⋅ 2 ζ⋅ τ⋅

d Y t( )⋅

dt⋅+ Y t( )+ K X t( )⋅=

(b) 9d2 y t( )⋅

dt2⋅ 18

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract the initial steady state:9

d2 Y t( )⋅

dt2⋅ 18

d Y t( )⋅

dt⋅+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=

Divide by Y(t) coefficient:94

d2 Y t( )⋅

dt2⋅

184

d Y t( )⋅

dt⋅+ Y t( )+ 2 X t( )=

Match coeffients to standard form:τ

94

min:= τ 1.5 min= ζ18min4 2⋅ τ⋅

:= ζ 1.5=Equivalent time constants:

K 2:= Overdamped.

Find roots: r118− 182 4 9⋅ 4⋅−+

2 9⋅ min:= r1 0.255− min 1−

= τe11−

r1:= τe1 3.927 min=

r218− 182 4 9⋅ 4⋅−−

2 9⋅ min:=

ζ 1=

K 2:= Critically damped.Equivalent time constants:

Find roots: r112− 122 4 9⋅ 4⋅−+

2 9⋅ min:= r1 0.667− min 1−

= τe11−

r1:= τe1 1.5 min=

r212− 122 4 9⋅ 4⋅−−

2 9⋅ min:= r2 0.667− min 1−

= τe21−

r2:= τe2 1.5 min=

Time for response to decay to within 0.67% of its initial value: 5 τe1⋅ 7.5 min=


Decay ratio: e 0.5− min 1− T 0.00081= Percent overshoot:e

0.5− min 1− T

2 2.8 %=

Rise time:T4

3.56 min= Settling time:5−

0.5− min 1−10 min=

(c) 9d2 y t( )⋅

dt2⋅ 12

d y t( )⋅

dt⋅+ 4 y t( )+ 8 x t( ) 4−=

Subtract initial steady state anddivide by the coefficient of Y(t):

94

d2 Y t( )⋅

dt2⋅ 3

d Y t( )⋅

dt⋅+ Y t( )+ 2 X t( )=

Y 0( ) 0=

Compare coefficients to standard form:τ

94

min:= τ 1.5 min= ζ3min2 τ⋅

:=

Y s( ) K ∆x1−τ

1

s1τ

+2

1

s1τ

+

−1s

+⋅=

A21−

τs

dds

K ∆x⋅

τ2s 1−

τs

K− ∆x⋅

τ2

s2K− ∆x⋅=lim

→

=lim

→

=

A30s

K ∆x⋅

τ s⋅ 1+( )2K ∆x⋅=lim

→=A1

1−

τs

K ∆x⋅

τ2s

K− ∆x⋅

τ=lim

→

=

Y s( )K

τ s⋅ 1+( )2∆xs

=A1

s1τ

+2

A2

s1τ

+

+A3s

+=

Step response for the critically damped case:

Y t( ) K ∆x u t( )τe1

τe1 τe2−e

t−

τe1−

τe2τe2 τe1−

e

t−

τe2−⋅=

(2-5.10)Invert using Table 2-1.1:

Y s( ) K ∆xτe1−

τe1 τe2−

1

s1τe1

+

τe2τe2 τe1−

1

s1τe2

+

−1s

+⋅=

A30s

K ∆x⋅

τe1 s⋅ 1+( ) τe2 s⋅ 1+( )K ∆x⋅=lim

→=

A2K− ∆x⋅ τe2⋅

τe2 τe1−=A1

1−

τe1s

K ∆x⋅

τe1 τe2⋅ s1τe2

+⋅ s

K− ∆x⋅ τe1⋅

τe1 τe2−=lim

→

=

Y s( )K

τe1 s⋅ 1+( ) τe2 s⋅ 1+( )∆xs

=A1

s1τe1

+

A2

s1τe2

+

+A3s

+=

X s( )∆xs

=Step response, over-damped second-order differential equation:

Problem 2-13. Partial fraction expansion coefficients for Eqs. 2-5.10 to 2-5.13


Y s( )K

τ s⋅ 1+( )2r

s2=

A1

s1τ

+2

A2

s1τ

+

+A3

s2+

A4s

+=

Ramp response for critically damped case:

Y t( ) K rτe1

2

τe1 τe2−e

t−

τe1 τe22

τe2 τe1−e

t−

τe2+ t+ τe1 τe2+( )−⋅=

(2-5.12)


Y s( ) K rτe1

2

τe1 τe2−

1

s1τe1

+

τe22

τe2 τe1−

1

s1τe2

+

+1

s2+

τe1 τe2+

s−⋅=

K r τe1− τe2−( )⋅=

A40s

dds

K r⋅τe1 s⋅ 1+( ) τe2 s⋅ 1+( )⋅

⋅0s

K r⋅τe1− τe2 s⋅ 1+( )⋅ τe2 τe1 s⋅ 1+( )⋅−

τe1 s⋅ 1+( )2 τe2 s⋅ 1+( )2⋅lim

→=lim

→=

A30s

K r⋅τe1 s⋅ 1+( ) τe2 s⋅ 1+( )⋅

K r⋅=lim→

=

A2K r⋅ τe2

2⋅

τe2 τe1−=A1

1−

τe1s

K r⋅

τe1 τe2⋅ s1τe2

+⋅ s2⋅

K r⋅ τe12

⋅

τe1 τe2−=lim

→

=

Y s( )K

τe1 s⋅ 1+( ) τe2 s⋅ 1+( )⋅

r

s2=

A1

s1τe1

+

A2

s1τe2

+

+A3

s2+

A4s

+=

X s( )r

s2=Ramp response for the over-damped case:

Y t( ) K ∆x u t( )tτ

1+ e

t−

τ−⋅=

(2-5.11)


A11−

τs

K r⋅

τ2

s2K r⋅=lim

→

= A30s

K r⋅

τ s⋅ 1+( )2K r⋅=lim

→=

A21−

τs

dds

K r⋅

τ2

s2 1−

τs

2−K r⋅

τ2

s3⋅ 2 K⋅ r⋅ τ⋅=lim

→

=lim

→

=

A40s

dds

K r⋅

τ s⋅ 1+( )2 0s2−

K r⋅ τ⋅

τ s⋅ 1+( )3⋅ 2− K⋅ r⋅ τ⋅=lim

→=lim

→=

Y s( ) K r1

s1τ

+2

2 τ⋅

s1τ

+

+1

s2+

2 τ⋅

s−⋅=

Invert using Table 2-1.1:Y t( ) K r⋅ t 2 τ⋅+( )e

t−

τ t+ 2 τ⋅−⋅= (2-5.13)



X s( )∆xs

=Problem 2-14. Derive step reponse of n lags in series

Y s( )K

1

n

k

τk s⋅ 1+( )∏=

∆xs

=A0s

1

n

k

Ak

s1τk

+∑=

+=

A00s

K ∆x⋅

1

n

k

τk s⋅ 1+( )∏=

K ∆x⋅=lim→

=


Y t( ) K ∆x⋅ u t( )⋅

1

n

k

Ak e

t−

τk⋅∑

=

+=

Ak1−

τks

K ∆x⋅

s

1 j k≠( )⋅

n

j

s1τ j

+∏=

⋅

1

n

j

τ j∏=

⋅

K ∆x⋅

1−τk 1 j k≠( )

n

j

1−τk

1τ j

+

1

n

j

τ j∏=

⋅∏=

⋅

=lim

→

=

K− ∆x⋅

1τk

1

τkn 1−

⋅ τk⋅

1 j k≠( )⋅

n

j

τk τ j−( )∏=

⋅

=K− ∆x⋅ τk

n 1−⋅

1 j k≠( )

n

j

τk τ j−( )∏=

=

Substitute:

Y t( ) K ∆x u t( )

1

n

k

τkn 1−

1 j k≠( )

n

j

τk τ j−( )∏=

e

t−

τk∑=

−⋅= (2-5.23)


r1τ1 τ2+( )− τ1 τ2+( )2 4τ1 τ2 1 k2−( )⋅−+

2 τ1⋅ τ2⋅=

(b) The response is stable if both roots are negative if 0 < k2 < 1.

This term is positive as long as τ1, τ2, and k2 are positive, so the response is overdamped.

τ1 τ2−( )2 4τ1 τ2⋅ k2⋅+=

τ12

2τ1 τ2⋅− τ22

+ 4τ1 τ2⋅ k2⋅+=

τ1 τ2+( )2 4τ1 τ2⋅ 1 k2−( )⋅− τ12

2τ1 τ2⋅+ τ22

+ 4τ1 τ2⋅− 4τ1 τ2⋅ k2⋅+=

(a) The response is overdamped if the term in the radical is positive:

r1τ1 τ2+( )− τ1 τ2+( )2 4τ1 τ2 1 k2−( )⋅−+

2 τ1⋅ τ2⋅=

τ1 τ2⋅ s2⋅ τ1 τ2+( )s+ 1+ k2− 0=

Find the roots of the denominator:

ζτ1 τ2+

2 τ⋅ 1 k2−( )⋅=

τ1 τ2+

2 τ1 τ2⋅ 1 k2−( )⋅⋅=Damping ratio:

ττ1 τ2⋅

1 k2−=Time constant:K

k11 k2−

=Gain:Comparing coefficients:

Y s( )

k1

1 k2−

τ1 τ2⋅

1 k2−s2 τ1 τ2+

1 k2−s+ 1+

X s( )=

Rerrange interacting equation:

Y s( )K

τ2

s2 2ζ τ⋅ s⋅+ 1+

X s( )=

Standard form of the second-order differential equaton, Eq. 2-5.4:

Y s( )k1

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ k2−X s( )=

k1

τ1 τ2⋅ s2⋅ τ1 τ2+( )s+ 1+ k2−

X s( )=

Problem 2-15. Transfer function of second-order interacting systems.Smith & Corripio, 3rd edition

If τ1, τ2, and k2 are positive, and if k2 < 1, then the positive term in the numerator is always less in magnitude than the negative term, and the root is negative. The other root has to be negative because both terms in the numerator are negative. So, the response is stable.

(c) Effective time constants

As the response is overdamped, we can derive the formulas for the two effective time constants. These are the negative reciprocals of the two real roots:

τe12 τ1⋅ τ2⋅

τ1 τ2+ τ1 τ2−( )2 4τ1 τ2⋅ k2⋅+−

= τe12 τ1⋅ τ2⋅

τ1 τ2+ τ1 τ2−( )2 4τ1 τ2⋅ k2⋅++

=

The first of these is the dominant time constant.



The response canot be unstable for positive Kc. The time constant and damping ratio are always real and positive for positive gain.

Cannot be undamped for finite Kc.ζ 0=(iii) Undamped:

ζ cannot be negative for positive Kc13

Kc< ∞<0 ζ< 1<(ii) Underdamped:

Kc13

<43

1 Kc+>2

3 1 Kc+( )1>ζ 1>(i) Overdamped:

Ranges of the controller gain for which the response is:

ζ4

2 τ⋅ 1 Kc+( )⋅= 2

3 1 Kc+( )⋅=Damping ratio:

τ3

1 Kc+=Time constant:K

Kc1 Kc+

=Gain:

C s( )

Kc

1 Kc+

31 Kc+

s2 41 Kc+

s+ 1+

R s( )=

Rearrange feedback loop transfer function and compare coefficients:

C s( )K

τ2

2ζ τ⋅ s⋅+ 1+

R s( )=Standard second-order transfer function, Eq. 2-5.4:

This is a second-order process with a proportional controller.

C s( )Kc

3s 1+( ) s 1+( )⋅ Kc+R s( )=

Kc

3s2 4s+ 1+ Kc+=

Problem 2-16. Transfer function of a second-order feedback control loop


Y X t( )( )α

1 α 1−( )xb+2

X t( )=

Y X t( )( ) y x t( )( ) y xb( )−=X t( ) x t( ) xb−=Let

y x t( ) y xb( )1 α 1−( ) xb⋅+ α⋅ α xb⋅ α 1−( )⋅−

1 α 1−( )xb+2

x t( ) xb−( )+=

y x t( )( )α x t( )⋅

1 α 1−( )x t( )+=

(c) Eqilibrium mole fraction by relative volatility, Eq. 2-6.3:

PoΓ t( )( )

B po⋅ Tb( )

Tb C+( )2Γ t( )=

PoΓ t( )( ) po T t( )( ) po Tb( )−=Γ t( ) T t( ) Tb−=Let

po T t( )( ) po Tb( ) B

Tb C+( )2e

AB

Tb C+−

T t( ) Tb−( )+=

po T t( )( ) eA

B

T t( ) C+−

=

(b) Antoine equation for vapor pressure, Eq. 2-6.2:

Hd Γ t( )( ) a1 2a2 Tb⋅+ 3a3 Tb2

⋅+ 4a4 Tb3

⋅+ Γ t( )=

Hd Γ t( )( ) H T t( )( ) H Tb( )−=Γ t( ) T t( ) Tb−=Let

H T t( )( ) H Tb( ) a1 2a2 Tb⋅+ 3a3 Tb2

⋅+ 4a4 Tb3

⋅+ T t( ) Tb−( )+=

H T t( )( ) H0 a1 T t( )⋅+ a2 T2⋅ t( )⋅+ a3 T3

⋅ t( )+ a4 T4⋅ t( )+=

(use subscript b for base value)(a) Enthalpy as a function of temperature, Eq. 2-6.1:

Problem 2-17. Linearization of common process model functions.


(d) Flow as a function of pressure drop, Eq. 2-6.4:

f ∆p t( )( ) k ∆p t( )⋅=

f ∆p t( )( ) f ∆pb( ) k

2 ∆pb⋅∆p t( ) ∆pb−( )+=

Let ∆P t( ) ∆p t( ) ∆pb−= F ∆P t( )( ) f ∆p t( )( ) f ∆pb( )−=

F ∆P t( )( ) k

2 ∆pb⋅∆P t( )=

(e) Radiation heat transfer rate as a function of temperature, Eq. 2-6.5:

q T t( )( ) ε σ⋅ A⋅ T4⋅ t( )=

q T t( )( ) q Tb( ) 4 ε⋅ σ⋅ A⋅ Tb3

⋅ T t( ) Tb−( )+=

Let Γ t( ) T t( ) Tb−= Q Γ t( )( ) q T t( )( ) q Tb( )−=

Q Γ t( )( ) 4 ε⋅ σ⋅ A⋅ Tb3

⋅ Γ t( )⋅=


Tmax 610 K= Tmin 590 K=

Temperature range for which the heat transfer rate is within 5% of the linear approximation:

error ε σ⋅ A⋅ T4⋅ ε σ⋅ A⋅ Tb

4⋅ 4ε σ⋅ A⋅ Tb

3⋅ T Tb−( )+−= 0.05 ε σ⋅ A T4

⋅⋅( )=

Simplify and rearrange: T4 4 Tb3

⋅ T⋅− 3Tb4

+ 0.05T4=

As the error is always positive, the absolute value brackets can be dropped. Rearrange into a polynomial and find its roots:

0.95TTb

44

TTb

− 3+ 0=

polyroots

3

4−

0

0

0.95

1.014− 1.438i−

1.014− 1.438i+

0.921

1.108

=

Ignore the complex roots. The other two roots are the lower and upper limits of the range:

0.921TTb

≤ 1.108≤

For Tb 400K:= Tmin 0.921 Tb⋅:= Tmax 1.108Tb:= Tmin 368 K= Tmax 443 K=


Problem 2-18. Linearization of radiation heat transfer--range of accuracy.

q T( ) 4ε σ⋅ A⋅ T4⋅= Use subscript "b" for base value for linearization.

From the solution to Problem 2-17(e), the slope is:d q T( )⋅

dT4 ε⋅ σ⋅ A⋅ T3

⋅=

Temperature range for which the slope is within 5% of the slope at the base valueK 1.8R:=

error 4 ε⋅ σ⋅ A⋅ T3⋅ 4 ε⋅ σ⋅ A⋅ Tb

3⋅−= 0.05 4 ε⋅ σ⋅ A⋅ Tb

3⋅⋅=

Tmax3 1.05 Tb= 1.0164Tb=T

Tb

31− 0.05=

Simplify and rearrange:

Tmin3 0.95 Tb= 0.983Tb=

For Tb 400K:= Tmax3 1.05 Tb:= Tmin

3 0.95 Tb:= Tmax 407 K= Tmin 393 K=

Tb 600K:= Tmax3 1.05 Tb:= Tmin

3 0.95 Tb:=

Tb 600K:= Tmin 0.921 Tb⋅:= Tmax 1.108Tb:= Tmin 553 K= Tmax 665 K=

So the range for which the linear approximation is within 5% of the heat rate is much wider than the range for which the value of the slope is within 5% of the actual slope. We must keep in mind that the parameters of the dynamic model are a function of the slope, not the heat rate.


0 x≤ 0.362≤

(b) xmin 1.1 0.9,( ) 0.637= xmax 1.1 0.9,( ) 1.183= (one) 0.637 x≤ 1≤

(c) xmin 5 0.1,( ) 0.092= xmax 5 0.1,( ) 0.109= 0.092 x≤ 0.109≤

(d) xmin 5 0.9,( ) 0.872= xmax 5 0.9,( ) 0.93= 0.872 x≤ 0.93≤

The range of accuracy is narrower the higher α and the higher xb.

For the vapor composition: y x( )α x⋅

1 α 1−( )x+=

error

α x⋅

1 α 1−( )x+

α xb⋅

1 α 1−( )xb+

α

1 α 1−( )xb+2

x xb−( )+

1−= 0.05=

α x⋅1 α 1−( )x+

1 α 1−( )xb+2

α xb 1 α 1−( )xb+⋅ α x⋅+ α xb⋅−1− 0.05=

The error is always negative, so we can change signs and drop the absolute value bars:

Smith & Corripio, 3rd editionProblem 2-19. Equilibrium vapor composition--range of accuracy

y x( )α x⋅

1 α 1−( )x+= Use subscript "b" for base value for linearization.

From the solution to Problem 2-17(c):d y x( )⋅

dxα

1 α 1−( )x+2

=

For the slope:

errorα

1 α 1−( )x+2

α

1 α 1−( )xb+2

−= 0.05α

1 α 1−( )xb+2

=

Simplify and rearrange: 1 α 1−( )xb+

1 α 1−( )x+

2

1− 0.05=

Lower limit:1 α 1−( )xb+

1 α 1−( )xmin+1.05= xmin α xb,( )

1 α 1−( )xb+ 1.05−

1.05 α 1−( ):=

Upper limit: 1 α 1−( )xb+

1 α 1−( )xmax+0.95= xmax α xb,( )

1 α 1−( )xb+ 0.95−

0.95 α 1−( ):=

(a) xmin 1.1 0.1,( ) 0.143−= (zero) xmax 1.1 0.1,( ) 0.362=

0.40 x≤ 1≤

(c) α 5:= xb 0.1:=

polyroots

0.95 α 1−( )⋅

0.05− α 1−( )2 xb0.05xb

− 2 α 1−( )−

0.95 α 1−( )

0.605

1.653=

xmin 0.605xb:= xmax 1.653xb:= xmin 0.061= xmax 0.165= 0.061 x≤ 0.165≤

(d) α 5:= xb 0.9:=

polyroots

0.95 α 1−( )⋅

0.05− α 1−( )2 xb0.05xb

− 2 α 1−( )−

0.95 α 1−( )

0.577

1.732=

xmin 0.577xb:= xmax 1.732xb:= xmin 0.519= xmax 1.559= 0.519 x≤ 1≤


1 α 1−( )xb+2α x⋅ 0.95 1 α 1−( )x+ α α 1−( ) xb

2α x⋅+=

0.95 α 1−( )⋅ x2⋅ 0.95 α 1−( )2⋅ xb

2⋅ 0.95+ 1− 2 α 1−( )⋅ xb⋅− α 1−( )2 xb

2⋅− x⋅+ 0.95 α 1−( )⋅ xb⋅+

0.95 α 1−( ) xxb

20.05− α 1−( )2⋅ xb

0.05xb

− 2 α 1−( )−xxb⋅+ 0.95 α 1−( )+ 0=

Find the roots, one is the lower limit and the other one the upper limit:

(a) α 1.1:= xb 0.1:=

polyroots

0.95 α 1−( )⋅

0.05− α 1−( )2 xb0.05xb

− 2 α 1−( )−

0.95 α 1−( )

0.138

7.231=

xmin 0.138xb:= xmax 7.231xb:= xmin 0.014= xmax 0.723= 0.014 x≤ 0.723≤

(b) α 1.1:= xb 0.9:=

polyroots

0.95 α 1−( )⋅

0.05− α 1−( )2 xb0.05xb

− 2 α 1−( )−

0.95 α 1−( )

0.444

2.25=

xmin 0.444xb:= xmax 2.25xb:= xmin 0.4= xmax 2.025=

2 k⋅ cAb⋅ cBb⋅ 2 hr 1−= k cAb

2⋅ 2 hr 1−

=

R CA t( ) CB t( ),( ) 2hr 1− CA t( ) 2hr 1− CB t( )+=

For cA 3kmole

m3:= 2 k⋅ cA⋅ cBb⋅ 2 k⋅ cAb⋅ cBb⋅− 1 hr 1−

=

(off by 50%)

k cA2

⋅ k cAb2

⋅− 2.5 hr 1−= (off by 125%)

For cB 2kmole

m3:= 2 k⋅ cAb⋅ cB⋅ 2 k⋅ cAb⋅ cBb⋅− 2 hr 1−

=

(off by 100%)

k cAb2

⋅ k cAb2

⋅− 0 hr 1−= (same as the base value)

These errors on the parameters of the linear approximation are significant, meaning that it is only valid for very small deviations of the reactant concentrations from their base values.



Problem 2-20. Linearization of chemical reaction rate. kmole 1000mole:=

r cA t( ) cB t( ),( ) k cA t( )2⋅ cB t( )=

Use subscript "b" for base value for linearization.

Problem parameters: k 0.5m6

kmole2hr:= cAb 2

kmole

m3:= cBb 1

kmole

m3:=

Linearize: r cA t( ) cB t( ),( ) r cAb cBb,( ) 2k cAb⋅ cBb cA t( ) cAb−( )⋅+ k cAb2

⋅ cB t( ) cBb−( )+=

Let R CA t( ) CB t( ),( ) r cA t( ) cB t( ),( ) r cAb cBb,( )−= CAb t( ) cA t( ) cAb−=

CB t( ) cB t( ) cBb−=

R CA t( ) CB t( ),( ) 2k cAb⋅ cBb⋅ CA t( )⋅ k cAb2

⋅ CB t( )⋅+=

At the given base conditions:

degC K:= mmHgatm760

:= mole% %:=

Numerical values for benzene at: pb 760mmHg:= Tb 95degC:= xb 50mole%:=

A 15.9008:= B 2788.51degC:= C 220.80degC:=

Let pob po Tb( )=

pob e

AB

Tb C+−

mmHg:= pob 1177 mmHg=

xb B⋅ pob⋅

pb Tb C+( )2⋅0.022

1degC

=pobpb

1.549=pob xb⋅

pb2

0.001021

mmHg=


Problem 2-21. Linearization of Raoult's Law for equilibrium vapor composition.

Raoult's Law: y T t( ) p t( ), x t( ),( )po T t( )( )

p t( )x t( )= po T t( )( ) e

AB

T t( ) C+−

=

Linearize: Use subscript "b" for base value for linearization.

y T t( ) p t( ), x t( ),( ) y Tb pb, xb,( )xbpb

δ

δT⋅ po T t( )( )⋅⋅ T t( ) Tb−( )⋅+

po Tb( )pb

x t( ) xb−( )+=

po− Tb( )xb

pb2

p t( ) pb−( )+

δ

δTeA

B

T t( ) C+−

⋅B

Tb C+( )2e

AB

Tb C+−

⋅=B po⋅ Tb( )⋅

Tb C+( )2=

Let Y Γ t( ) P t( ), X t( ),( ) y T t( ) p t( ), x t( ),( ) y Tb pb, xb,( )−= Γ t( ) T t( ) Tb−= P t( ) p t( ) pb−=

X t( ) x t( ) xb−=

Y Γ t( ) P t( ), X t( ),( )xb B⋅ po

⋅ Tb( )⋅

pb Tb C+( )2⋅Γ t( )

po Tb( )pb

X t( )+po Tb( ) xb⋅

pb2

P t( )−=

Y Γ t( ) P t( ), X t( ),( ) 0.022degC

Γ t( ) 1.549 X t( )+0.00102mmHg

P t( )−=

pob xb⋅

pb77.441 %= y Tb pb, xb,( ) 77.44mole%=


From the initial steady state: 0 fb cA.b cAb−( )⋅ k Tb( ) V⋅ cAb⋅−=

cAbfb cAib⋅

fb kb V⋅+:= cAb 9.231 10 5−

×kmole

m3=

Calculate parameters: τV

fb kb V⋅+:= K1

cAib cAb−

fb V kb⋅+:= K2

fbfb V kb⋅+

:= τ 0.01 s=

K1 0.046s kmole⋅

m6=

K3V− kb⋅ E⋅ cAb⋅

1.987kcal

kmole K⋅Tb

2⋅ fb V kb⋅+( )⋅

:=

K2 7.692 10 6−×=

fb V kb⋅+ 260.002m3

s= K3 3.113− 10 6−

×kmol

m3K=

Linearized equation:

0.01 sec⋅d CA t( )⋅

dt⋅ CA t( )+ 0.046

kmole

m3

s

m3F t( ) 7.692 10 6−

⋅ CAi t( )+ 3.113kmole

m3KΓ t( )−=



Problem 2-22. Linearization of reactor of Examples 2-6.4 and 2-6.1.

From the results of Example 2-6.4: τd CA t( )⋅

dt⋅ CA t( )+ K1 F t( )⋅ K2 CAi t( )⋅+ K3 Γ t( )⋅+=


τV

fb V k Tb( )⋅+= K1

cAib cAb−

fb V k Tb( )⋅+= K2

fbfb V k Tb( )⋅+

= K3V− k Tb( )⋅ E cAb⋅

R Tb2

⋅ fb V k Tb( )⋅+( )=

Problem parameters: V 2.6m3:= fb 0.002

m3

s:= cAib 12

kmole

m3:=

Let kb k Tb( )=Tb 573K:= kb 100s 1−

:= E 22000kcal

kmole:=

p t( ) ρ t( )v2 t( )

2⋅ po+= v t( ) 2

p t( ) po−( )ρ t( )

⋅=

Flow through the orifice caused by the bullet: wo t( ) ρ t( ) Ao⋅ v t( )⋅= Ao 2 ρ t( )⋅ p t( ) po−( )⋅⋅=

Ideal gas law: ρ t( )M p t( )⋅

Rg T 273K+( )⋅=

Substitute into mass balance:

V M⋅Rg T 273 K⋅+( )⋅

d p t( )⋅

dt⋅ wi t( ) Ao

2 M⋅Rg T 273K+( )⋅

p t( ) p t( ) po−( )⋅−=

Solve for the derivative:

d p t( )⋅

dtg wi t( ) p t( ),( )=

Rg T 273K+( )⋅

V M⋅wi t( ) Ao

2 M⋅Rg T 273K+( )⋅

p t( ) p t( ) po−( )⋅⋅−=

Linearize:d p t( )⋅

dtδ g⋅δ wi⋅

b

⋅ wi t( ) wb−( ) δ g⋅δ p⋅

b

⋅ p t( ) pb−( )+=

Let P t( ) p t( ) pb−= Wi t( ) wi t( ) wb−=

a1δ g⋅δ wi⋅

b

⋅= a1Rg T 273K+( )⋅

V M⋅:= a1 65.56

kPakg

=


Problem 2-23. Pressure in a compressed air tank when punctured.

Vp(t)

wi(t)wo(t)

po

Assumptions:Air obeys ideal gas law•Constant temperature•

Design conditions: kPa 1000Pa:=

pb 500 101.3+( )kPa:= M 29kg

kmole:=

Ao 0.785cm2:= T 70degC:= V 1.5m3

:=

Rg 8.314kPa m3

⋅

kmole K⋅⋅:= po 101.3kPa:=


Solution:

Mass balance on the tank: Vd ρ t( )⋅

dt⋅ wi t( ) wo t( )−=

Bernoulli's equation:


K 1.8R:=

If the compressor shuts down it will take approximately 5(42.8) = 214 sec (3.5 min) for the pressure transient to die out, according to the linear approximation. (See the results of the simulation, Problem 13-3, to see how long it actually takes.)

P s( )Wi s( )

Kτ s⋅ 1+

=Transfer function:

K 2.8 103×

kPa sec⋅

kg=τ 42.9 sec=

Ka1a2−

:=τ1a2−

:=Then

τd P t( )⋅

dt⋅ P t( )+ K Wi t( )⋅=Compare to standard form of first-order equation:

P 0( ) 0=1a2−

d P t( )⋅

dt⋅ P t( )+

a1a2−

Wi t( )=d P t( )⋅

dta1 Wi t( )⋅ a2 P t( )⋅+=Substitute:

a2 0.023− sec 1−=a2

Ao−

2 V⋅

2 Rg⋅ T 273 K⋅+( )⋅

M pb⋅ pb po−( )⋅

kPa1000Pa

⋅2 pb⋅ po−( )1000Pa

kPa⋅

m100cm

2:=

a2δ g⋅δ p⋅

b

⋅=Ao−

V

2 Rg⋅ T 273K+( )⋅

M⋅

12

pb pb p0−( )1−

2⋅ 2pb po−( )=

Γ t( ) T t( ) Tb−=

Substitute:d Γ t( )⋅

dta1 Γs t( )⋅ a2 Γ t( )⋅+= Γ 0( ) 0= (base is initial steady state)

Standard form of the first-order differential equation: τd Γ t( )⋅

dt⋅ Γ t( )+ K Γs t( )⋅=

Divide by -a2 and rearrange: 1a2−

d Γ t( )⋅

dt⋅ Γ t( )+

a1a2−

Γs t( )=

M cv⋅

4 ε⋅ σ⋅ A⋅ Tb3

⋅

d Γ t( )⋅

dt⋅ Γ t( )+

TsbTb

3

Γs t( )=

Compare coefficients: τM cv⋅

4 ε⋅ σ⋅ A⋅ Tb3

⋅= K

TsbTb

3

=

Laplace transform:Γ s( )Γs s( )

Kτ s⋅ 1+

=

The input variable is the temperature of the oven wall. See problem 13-4 for the simulation.



Problem 2-24. Temperature of a turkey in an oven.

T(t)Ts(t)

M

AssumptionsUniform turkey temperature•Negligible heat of cooking•Radiation heat transfer only•

Energy balance on the turkey:

M cv⋅d T t( )⋅

dt⋅ ε σ⋅ A⋅ Ts

4 t( ) T4 t( )−⋅=

Use subscript "b" for linearization base values.

Solve for the derivative:d T t( )⋅

dtg Ts t( ) T t( ),( )= ε σ⋅ A⋅

M cv⋅Ts

4 t( ) T4 t( )−=

Linearize: d T t( )⋅

dta1 Ts t( ) Tsb−( )⋅ a2 T t( ) Tb−( )⋅+=

where a1δ g⋅δTs b

⋅= 4 ε⋅ σ⋅ A⋅M cv⋅

Tsb3= a2

δ g⋅δT

b

⋅= 4− ε⋅ σ⋅ A⋅M cv⋅

Tb3=

Let Γs t( ) Ts t( ) Tsb−=

Q t( ) q t( ) qb−= a1δ g⋅δq

b

⋅= a2δ g⋅δT

b

⋅=

a11C

:= a24− α⋅ Tb

3⋅

C:= a1 5.556 10 3−

×R

BTU= a2 0.381− hr 1−

=

Substitute:d Γ t( )⋅

dta1 Q t( )⋅ a2 Γ t( )⋅+= Γ 0( ) 0= (base is initial value)

Standard form of first-order differential equation: τd Γ t( )⋅

dt⋅ Γ t( )+ K Q t( )⋅=

Divide by -a2 and rearrange:1a2−

d Γ t( )⋅

dt⋅ Γ t( )+

a1a2−

Q t( )=

C

4 α⋅ Tb3

⋅

d Γ t( )⋅

dt⋅ Γ t( )+

1

4α Tb3

⋅Q t( )=

Compare coefficients: τC

4α Tb3

⋅:= K

1

4α Tb3

⋅:= τ 2.62 hr= K 0.01458

R hr⋅

BTU=


Problem 2-25. Slab heated by an electric heater by radiation.

T(t)

Ts

q(t)

Assumptions:Uniform temperature of the slab•Heat transfer by radiation only•

Energy balance on the slab:

M cv⋅d T t( )⋅

dt⋅ q t( ) ε σ⋅ A⋅ T4 t( ) Ts

4−⋅−=

Let C M cv⋅= α ε σ⋅ A⋅=

Substitute Cd T t( )⋅

dt⋅ q t( ) α T4 t( ) Ts

4−−=

Problem parameters: Use subscript "b" to denote linearization base value.

C 180BTU

R:= α 5 10 8−

⋅BTU

hr R4⋅

:= Ts 540R:= Tb 700R:=

Solve for the derivative:d T t( )⋅

dtg q t( ) T t( ),( )= 1

Cq t( )

α

CT4 t( ) Ts

4−−=

Linearize:d T t( )⋅

dta1 q t( ) qb−( )⋅ a2 T t( ) Tb−( )⋅+=

Let Γ t( ) T t( ) Tb−=

Transfer function:Γ s( )Q s( )

Kτ s⋅ 1+

=


r1

1.8− 1.82

4 0.8⋅ 1 0.1Kc+( )⋅−+

2 0.8⋅= 1.8−

1.6

1.8

1.6

2 1 0.1Kc+

0.8−+=

Roots of the characteristic equation:

(b) Values of the controller gain for which the response is over-damped, critically damped, and under-damped

0.8s2

1.8s+ 1+ 0.1Kc+ 0=Characteristic equation:

C s( )

R s( )

0.1Kc

0.8s2

1.8s+ 1+ 0.1Kc+=Closed-loop transfer function:

τ1 τ2⋅ s2

⋅ τ1 τ2+( )s+ 1+ Kc K⋅+ 0=Characteristic equation:

C s( )

R s( )

KcK

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅⋅

1 KcK

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅⋅+

=Kc K⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ Kc K⋅+=

(a) Closed loop transfer function and characteristic equation of the loop.

τ2 0.8min:=τ1 1min:=K 0.10%TO

%CO:=Problem parameters:

Gc s( ) Kc=G1 s( )K

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Problem 6-1. Second-order loop with proportional controller.

Smith & Corripio, 3rd edition %CO %:=%TO %:=

τe11−

r1= τe1

2 τ1⋅ τ2⋅

τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−−

:= τe1 0.889 min=

τe21−

r2= τe2

2 τ1⋅ τ2⋅

τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−+

:= τe2 0.889 min=

Kc 0.2%CO

%TO:= (under-damped, time constant and damping ratio)

τ2

s2

2ζ τ⋅ s⋅+ 1+τ1 τ2⋅

1 Kc K⋅+s2 τ1 τ2+

1 Kc K⋅+s+ 1+=

Match coefficients: ττ1 τ2⋅

1 Kc K⋅+:= ζ

τ1 τ2+

2 τ⋅ 1 Kc K⋅+( )⋅:= τ 0.886 min= ζ 0.996=

(d) Steady-state offset for a unit step change in set point.

Final value theorem:∞t

Y t( )0s

s Y s( )⋅lim→

=lim→

R s( )1

s= (Table 2-1.1)

The response is critically damped when the term in the radical is zero:1.8

1.6

2 1 0.1Kc+

0.8− 0=

Kccd1

0.10.8

1.8

1.6

2

1−:= Kccd 0.125%CO

%TO=Critically damped:

Over-damped (real roots): Kc 0.125%CO

%TO< Under-damped: Kc 0.125

%CO

%TO>

The loop cannot be unstable for positive gain because,for real roots the radical cannot be greater than the negative term, so both roots are negative•for complex conjugate roots the real part is always negative, -1.8/1.6, or -(τ1+τ2)/2τ1τ2•

This is true for all positive values of the time constants and the product K. cK.

(c) Equivalent time constants for different values of the gain:

Kc 0.1%CO

%TO:= (over-damped, two equivalent time constans)

τe11−

r1= τe1

2 τ1⋅ τ2⋅

τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−−

:= τe1 0.935 min=

τe21−

r2= τe2

2 τ1⋅ τ2⋅

τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−+

:= τe2 0.847 min=

Kc 0.125%CO

%TO:= (critically damped, two equal real time constants)

Kc 0.1%CO

%TO:=

0ss

Kc K⋅

τ1 τ2⋅ s2

⋅ τ1 τ2+( ) s⋅+ 1+ Kc K⋅+⋅

1

slim→

9.9009900990099009901 10-3

⋅→

offset 1 0.0099−( )%TO:= offset 0.99 %TO=

Kc 0.125%CO

%TO:=

0ss

Kc K⋅

τ1 τ2⋅ s2

⋅ τ1 τ2+( ) s⋅+ 1+ Kc K⋅+⋅

1

slim→

1.2345679012345679012 10-2

⋅→


Kc 0.2%CO

%TO:=

0ss

Kc K⋅

τ1 τ2⋅ s2

⋅ τ1 τ2+( ) s⋅+ 1+ Kc K⋅+⋅

1

slim→

1.9607843137254901961 10-2

⋅→


These are very large offsets because the loop gains are so small.


0.00842%CO

%TOKc< 0.825

%CO

%TO<Under-damped (complex conjugate roots):

Kc 0.825%CO

%TO>andKc 0.00842

%CO

%TO<Over-damped (two real roots):

Kc 0.00842%CO

%TO=Kc

30 302

4 0.25⋅ 36⋅−−

2 36⋅:=

Kc 0.82491%CO

%TO=Kc

30 302

4 0.25⋅ 36⋅−+

2 36⋅:=0.25 30Kc− 36Kc

2+ 0=

The response is critically damped when the term in the radical is zero:

r1

1.5 6Kc−( )− 1.5 6Kc−( )2 4 0.5⋅ 1 6Kc+( )⋅−+

2 0.5⋅= 1.5− 6Kc+ 0.25 30Kc− 36Kc

2++=

Roots:

(b) Values of the gain for which the response is over-, critically, and under-damped

0.5 s2

⋅ 1.5 6Kc−( )s+ 1+ 6Kc+ 0=Characteristic equation:

C s( )

R s( )

Kc 6⋅ 1 s−( )

s 1+( ) 0.5s 1+( )⋅ Kc 6⋅ 1 s−( )+=Closed-loop transfer function:

(a) Closed-loop transfer function and characteristic equation of the loop.

Gc s( ) Kc%CO

%TO⋅=G1 s( )

6 1 s−( )

s 1+( ) 0.5 s⋅ 1+( )⋅

%TO

%CO=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Problem 6-2. Inverse-response second-order system with proportional controller.


ζ1.5 6Kc−( )min

2 τ⋅ 1 6Kc+( )⋅:= τ 0.423 min= ζ 0.127−=

(unstable)Try values that result in equivalent time constants:

Kc 0.005%CO

%TO:= τe1

1min

1.5 6 Kc⋅− 0.25 30Kc− 36Kc2

+−

:= τe1 0.868 min=

τe21min

1.5 6 Kc⋅− 0.25 30Kc− 36Kc2

++

:= τe2 0.559 min=

Kc 1%CO

%TO:= τe1

1min

1.5 6 Kc⋅− 0.25 30Kc− 36Kc2

+−

:= τe1 0.143− min=

(unstable)

τe21min

1.5 6 Kc⋅− 0.25 30Kc− 36Kc2

++

:= τe2 0.5− min=

(d) Offset for various values of the gain and a unit step change in set point.

Kc 0.10%CO

%TO:=

0ss

Kc 6⋅ 1 s−( )⋅

0.5s2

1.5 6 Kc⋅−( )s+ 1+ 6Kc+⋅

1

slim→

.37500000000000000000→

offset 1 0.375−:= offset 0.625%CO

%TO=

The response is unstable when Kc 0.25%CO

%TO> (one real root is positive or the real part of the

complex roots i positive)

(c) Effective time constants or time constant and damping ratio for various values othe gain:

0.5

1 6Kc+s2 1.5 6Kc−

1 6Kc+s+ 1+ τ

2s2

2ζ τ⋅ s⋅+ 1+=

Kc 0.1%CO

%TO:= τ

0.5min2

1 6Kc+:= ζ

1.5 6Kc−( )min

2 τ⋅ 1 6Kc+( )⋅:= τ 0.559 min= ζ 0.503=

Kc 0.125%CO

%TO:= τ

0.5min2

1 6Kc+:= ζ

1.5 6Kc−( )min

2 τ⋅ 1 6Kc+( )⋅:= τ 0.535 min= ζ 0.401=

Kc 0.2%CO

%TO:=

τ0.5min

2

1 6Kc+:= ζ

1.5 6Kc−( )min

2 τ⋅ 1 6Kc+( )⋅:= τ 0.477 min= ζ 0.143=

Kc 0.3%CO

%TO:=

τ0.5min

2

1 6Kc+:=

Kc 0.125%CO

%TO:=

0ss

Kc 6⋅ 1 s−( )⋅

0.5s2

1.5 6 Kc⋅−( )s+ 1+ 6Kc+⋅

1

slim→

.42857142857142857143→


%TO=

Kc 0.20%CO

%TO:=

0ss

Kc 6⋅ 1 s−( )⋅

0.5s2

1.5 6 Kc⋅−( )s+ 1+ 6Kc+⋅

1

slim→

.54545454545454545455→


%TO=

The offsets are high because the gains are small. Of course, since for gains greater than 0.25%CO/%TO the loop is unstable, offsets can only be high with a proportional controller.


K 1.8R:=

The real root cannot be negative for any positive value of the loop gain KK c because the radical is always smaller than the negative term. Also, for complex conjugate roots, the real part is always

r1

1 KKc+( )− τI⋅ 1 KKc+( )2 τI2

⋅ 4 τI⋅ τ⋅ KKc⋅−+

2τ I τ⋅=

No, there is no ultimate gain. This result just means that a negative loop gain will make the loop unstable. Another way to show it is to determine the roots of the characteristic equation:

KKcu 0:=ωu 0:=τ I− τ⋅ ωu2

⋅ KKcu+ i 1 KKcu+( ) ωu+ 0=Substitute s = iω:

(b) Is there an ultimate gain for this loop?

(no offset)

0s

KKc τ I s⋅ 1+( )⋅

τI τ⋅ s2

⋅ 1 KKc+( )τ I s⋅+ KKc+

KKc

KKc= 1=lim

→Offset: the steady state gain is:

τI τ⋅ s2

⋅ 1 KKc+( )τ I s⋅+ KKc+ 0=Characteristic equation:

C s( )

R s( )

KKc τI s⋅ 1+( )⋅

τI s⋅ τ s⋅ 1+( )⋅ KKc τ I s⋅ 1+( )⋅+=Closed-loop transfer functon:

(a) Closed-loop transfer function and characteristic equation of the loop. Offset.

τ 1:=To work in dimensionless units, t/τ, set:

Gc s( ) Kc 11

τ I s⋅+⋅=G1 s( )

K

τ s⋅ 1+=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Problem 6-3. First-order process and proportional-integral controller.


RK

1.8:=


As KKc increases τc decreases and the response is faster.

0 2 40

0.5

1

Y t( )

R t( )

t

Y t( ) u t( ) e

t−

τc−:=

Invert using Table 2-1.1:R t( ) u t( ):=

τc 1:=

u t( ) 0 t 0<if

1 t 0≥if

:=Y s( )

1

τc s⋅ 1+

1

s=

1

s

1

s1

τc+

−=τcτ

KKc=Let

Y s( )KKc τ s⋅ 1+( )⋅

τ s⋅ τ s⋅ 1+( )⋅ KKc τ s⋅ 1+( )⋅+

1

s=

KKc

τ s⋅ KKc+

1

s=

(Table 2-1.1)R s( )1

s=

(c) Response of the loop to a step change in set point for τI = τ as the gain varies from 0 to infinity.

Real1 KKc+( )−

2 τ⋅0<=

negative:

Tu 1.987 min=Tu 2

π

ωu:=

KIu 110%CO

%TO min⋅=ωu 3.162 min

1−=KIu

τ1 τ2+( ) ωu2

⋅

K:=

ωu1

τ1 τ2⋅:=

τ2 0.1min:=

(b) Ultimate gain and period for other values of the smaller time constant:

Tu 5.62 min=Tu

2π

ωu:=

KIu 22.5%CO

%TO min⋅=ωu 1.118 min

1−=KIu

τ1 τ2+( ) ωu2

⋅

K:=

ωu1

τ1 τ2⋅:=

τ1− τ2⋅ ωu3

ωu+ 0=τ1 τ2+( )− ωu2

KKIu+ 0=

τ1− τ2⋅ i⋅ ωu3

⋅ τ1 τ2+( )ωu2

− i ωu⋅+ KKIu+ 0 0i+=Substitute s = iωu:

τ1 τ2⋅ s3

τ1 τ2+( )s2+ s+ KKI+ 0=

τ1 τ2⋅ s2

⋅ τ1 τ2+( )s+ 1+K KI⋅

s+ 0=

Characteristic equation:

τ2 0.8min:=τ1 1min:=K 0.1%CO

%TO:=

(a) Ultimate gain and period with the parameters of Problem 6-1:

Gc s( )KI

s=G1 s( )

K

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Problem 6-4. Second-order process with pure integral controller.


τ2 2min:=ωu

1

τ1 τ2⋅:=

KIu

τ1 τ2+( ) ωu2

⋅

K:= ωu 0.707 min

1−= KIu 15

%CO

%TO min⋅=

Tu2π

ωu:=

Tu 8.886 min=

Reducing the non-dominat time constant increases the ultimate gain and reduces the ultimate period, as expected. When τ2 is increased to 2 min, it becomes the dominant time constant and the ultmate gain should be higher than for part (a). However, in this case K I has units of rate and, since the loop is slower, it results in a smaller ultimate gain.


τI0.8min

2

1.8min> 0.444min=or τI

τ1 τ2⋅

τ1 τ2+>

Notice that for some values of τI the ultimate frequency and period are complex. When this happens there is no ultimate gain and the loop is stable for all values of the gain. So, the loop is always stable as long as

Tu 2π 0.8min2

1.8min τ I⋅−⋅=KKcu

1.8min τI⋅

0.8min2

1.8min τ I⋅−=For the given numerical values:

Tu2π

ωu= 2π τ1 τ2⋅ τ1 τ2+( ) τI⋅−⋅=

KKcu

τ1 τ2+( ) τ I⋅

τ1 τ2⋅ τ1 τ2+( ) τ I⋅−=ωu

1

τ1 τ2⋅ τ1 τ2+( ) τI⋅−=

τ1− τ2⋅ τ I⋅ ωu2

⋅ τ I+ τ1 τ2+( )τI2ωu

2⋅+ 0=KKcu τ1 τ2+( )τ I ωu

2⋅=

τ1 τ2+( )− τ I ωu2

⋅ KKcu+ i τ1− τ2⋅ τI⋅ ωu3

⋅ τI 1 KKcu+( )ωu⋅++ 0 i0+=Substitute s = iωu:

τ1 τ2⋅ τI⋅ s3

⋅ τ1 τ2+( )τI s2

⋅+ τI 1 KKc+( )⋅ s⋅+ KKc+ 0=Characteristic equation of the loop:

(a) Ultimate gain and period as a function of integral time τI.

τ2 0.8min:=τ1 1min:=

Gc s( ) Kc 11

τ I s⋅+⋅=G1 s( )

K

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Problem 6-5. Second-order process with proportional-integral controller.



0 5 100

0.2

0.4

DR KKc( )

KKc

5 100

0.5

1

1.5

ζ KKc( )

KKc

DR e

2− π ζ⋅

1 ζ2−= e

2− π

2 0.8KKc 11

4 0.8⋅ KKc⋅−⋅⋅

= e

2− π⋅

3.2 KKc⋅ 1−= DR KKc( ) e

2− π

3.2KKc 1−:=

Under these conditions the decay ratio is, from Eq. 2-5.18:

KKc1

4 0.8⋅> 0.3125=These are complex for

r2

1− 1 4 0.8⋅ KKc⋅−−

2 0.8⋅=r1

1− 1 4 0.8⋅ KKc⋅−+

2 0.8⋅=

Roots:

ζ KKc( ) 1

2 0.8KKc⋅:=ζ

1min

2 τ⋅ KKc⋅=

1

2 0.8 KKc⋅⋅=τ

0.8min2

KKc=Damping ratio:

τ2

s2

2ζ τ⋅ s⋅+ 1+0.8

KKcs2 1

KKcs+ 1+=Standard second-order differential equation:

1

KKc 11

s+⋅

s 1+( ) 0.8s 1+( )⋅+ 1

KKc s 1+( )⋅

s s 1+( ) 0.8s 1+( )⋅+= 0.8s

2s+ KKc+= 0=

For these values there is no ultimate gain. THe characteristic equation becomes:

(b) Damping ratio and decay ratio with Kc equal to one half the ultimate and τI = 1 min

Block diagram of the control loop:

Fsset(s) E(s) M(s)

C(s)

Fs(s)Gc(s) Gv(s)

KT

Ksp+-

Size the flow transmitter for 150% of design flow: fsmax 1.5 fs⋅:=fsmax 225

kscf

hr=

Transmitter gain: KT100%TO

fsmax:= Ksp KT:= KT 0.444

%TO hr⋅

kscf=

PI Controller: Gc s( ) Kc 11

τ I s⋅+⋅=

Size the control valve for 100% overcapacity. From Eq. 5-2.3: Let Cf 0.9:=

GM lbmole⋅

29lb:= y

1.63

Cf

p1 p2−

p1 14.7psia+⋅:= y 1.181= fy y 0.148y

3−:= fy 0.937=

Smith & Corripio, 3rd edition degF R:= psia psi:=

kscf 1000ft3

:= psig psi:=Problem 6-6. Design of gas flow control loop.

FT

FC

fs(t)

m(t)

c(t)

fsset(t)

p1 p2

Design conditions: lbmole 453.59mole:=

fs 150kscf

hr:= p1 150psig:=

T1 60degF:= p2 80psig:=

M 29lb

lbmole:= α 50:=

τv 0.06min:= τI τv:=

Kc 0.9%CO

%TO:=

Assume the pressures and temperatures are constant and that the flow transmitter FT has a built-in square-root extractor so that the signal c(t) is proportional to the flow fs(t). The valve is equal-percentage and the controller is PI.


So the closed-loop responds faster than the valve, and has no offset.

τc 0.026 min=τc

τv

KT Kc⋅ Kv⋅:=Closed-loop time constant:

Fs s( )

Fsset

s( )

Ksp Kc⋅ Kv⋅

τv s⋅ KT Kc⋅ Kv⋅+= 1

τc s⋅ 1+=Ksp KT=τI τv=With

Fs s( )

Fsset

s( )⋅

Ksp Gc s( )⋅ Gv s( )⋅

1 KT Gc s( )⋅ Gv s( )⋅+=

Ksp Kc⋅ 11

τ I s⋅+⋅

Kv

τv s⋅ 1+⋅

1 KT Kc⋅ 11

τI s⋅+⋅

Kv

τv s⋅ 1++

=

Closed-loop transfer function and time constant of the loop:

Gv s( )Kv

τv s⋅ 1+=

Transfer function of the valve:

Kv 5.87kscf

hr %CO⋅=Kv

ln α( )100%CO

fs:=

Valve gain, equal-percentage, constant pressures, Eq. 5-2.24:

Cvmax 110gal

min psi⋅:=

From Fig. C-10.1, p. 532, a 3-in Masoneilan valve is the smallest for this service:

Cvmax 58.91gal

min psi⋅=

Cvmax

200 %⋅ fs⋅ G T1 460 R⋅+( )⋅⋅

Cf p1 14.7psia+( )⋅ fy⋅

gal hr⋅

0.836kscf min⋅

psi

R⋅:=

Size the flow transmitter for :


wmax:= Ksp KT:= KT 0.02

%TO hr⋅

lb=

PI Controller: Gc s( ) Kc 11

τ I s⋅+⋅=

Size the control valve for 100% overcapacity. From Eq. 5-2.3:

GM lbmole⋅

29lb:= y

1.63

Cf

p1 p2−

p1 14.7psia+⋅:= y 1.319= fy y 0.148y

3−:= fy 0.979=

From the steam table the saturated steam pressure at: p1 14.7psia+ 59.7 psia= Tsat 292degF:=


Problem 6-7. Steam flow control loop.

FT

FC

w(t)

m(t)

c(t)

wset(t)

p1 p2

Design conditions: lbmole 453.59mole:=

w 3500lb

hr:= p1 45psig:=

Tsh 50degF:= p2 20psig:=

M 18lb

lbmole:= wmax 5000

lb

hr:=

τI τv:=Linear valve.

Cf 0.8:= Kc 0.5%CO

%TO:=

Assume the pressures and temperatures are constant and that the flow transmitter FT has a built-in square-root extractor so that the signal c(t) is proportional to the flow w(t). The valve is linear and the controller is PI.

Block diagram of the control loop:

Wset(s) E(s) M(s)

C(s)

W(s)Gc(s) Gv(s)

KT

Ksp+-


So the closed-loop responds slightly slower than the valve, and has no offset. What can be adjusted to speed-up the response of the closed loop?

KT Kc⋅ Kv⋅ 0.913=τc

τv

KT Kc⋅ Kv⋅:=Closed-loop time constant:

Fs s( )

Fsset

s( )

Ksp Kc⋅ Kv⋅

τv s⋅ KT Kc⋅ Kv⋅+=

1

τc s⋅ 1+=Ksp KT=τI τv=With

W s( )

Wset

s( )⋅

Ksp Gc s( )⋅ Gv s( )⋅

1 KT Gc s( )⋅ Gv s( )⋅+=

Ksp Kc⋅ 11

τ I s⋅+⋅

Kv

τv s⋅ 1+⋅

1 KT Kc⋅ 11

τI s⋅+⋅

Kv

τv s⋅ 1++

=

Closed-loop transfer function and time constant of the loop:

Gv s( )Kv

τv s⋅ 1+=

Transfer function of the valve:

Kv 91.31lb

hr %CO⋅=Kv

wvmax

100%CO:=

Valve gain, linear, constant pressures, Eq. 5-2.23:

wvmax 9131lb

hr=

wvmax0.836kscf min⋅

gal hr⋅

R

psi⋅ Cvmax⋅ Cf⋅

p1 14.7psia+

G T1 460R+( )⋅⋅ fy

M lbmole⋅

0.380kscf⋅:=

Cvmax 110gal

min psi⋅:=

From Fig. C-10.1, p. 532, a 3-in Masoneilan valve is the smallest for this service:

Cvmax 84.3gal

min psi⋅=

Cvmax

200 %⋅ fs⋅ G T1 460 R⋅+( )⋅⋅

Cf p1 14.7psia+( )⋅ fy⋅

gal hr⋅

0.836kscf min⋅

psi

R⋅:=

T1 342 degF=T1 Tsat Tsh+:=fs 73.889kscf

hr=fs

w

M0.380⋅

kscf

lbmole:=

Real part: ωu4

6ωu2

− 1+ Kcu+ 0=Kcu ωu min⋅( )4− 6min

2ωu( )2⋅+ 1−:= Kcu 4

%CO

%TO=

(b) G1 s( )1

s 1+( )2

= s 1+( )2

Kc+ 0= s2

2s+ 1+ Kc+ 0=

Substitute s = iωu at Kc = Kcu: ωu2

− 2ωu i⋅+ 1+ Kcu+ 0 0i+=

Imaginary part: 2ωu 0= ωu 0:= (There is no ultimate gain)

Real part: ωu2

− 1+ Kcu+ 0= Kcu 1−%CO

%TO:=

The loop becomes monotonically unstable when the controller gain is less than -1%CO/%TO.


Problem 6-8. Ultimate gain and period of various process transfer functions.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Proportional controller: Gc s( ) Kc= Characteristic equation: 1 Kc G1 s( )⋅+ 0=

(a) G1 s( )1

s 1+( )4

= s 1+( )4

Kc+ 0= s4

4s3

+ 6s2

+ 4s+ 1+ Kc+ 0=

Substitute s = iωu at Kc = Kcu: ωu4

4ωu3

i− 6ωu2

− 4ωu i⋅+ 1+ Kcu+ 0 0i+=

Imaginary part: 4− ωu3

4ωu+ 0= ωu 1min1−

:= Tu2π

ωu:= Tu 6.28 min=


7 0.5Kcu+( )ωu+ 0=

8− ωu2

7+ 7ωu2

+ 0.5− 0= ωu 6.5 min1−

:= Tu2π

ωu:=

Kcu 14min2ωu

2⋅ 1−:= Kcu 90

%CO

%TO= Tu 2.46 min=

(e) G1 s( )1

4s 1+( ) 0.2s 1+( )⋅ 0.1s 1+( )⋅= 0.08s

31.22s

2+ 4.3s+ 1+ Kc+ 0=

Substitute s = iωu at Kc = Kcu: 0.08− ωu3

i 1.22ωu2

− 4.3ωu i⋅+ 1+ Kcu+ 0 0i+=

Imaginary part: 0.08− ωu.3

4.3ωu+ 0=ωu

4.3

0.08min

1−:= Tu

2π

ωu:=

Real part: 1.22− ωu2

1+ Kcu+ 0=Tu 0.857 min=

Kcu 1.22min2ωu

21−:= Kcu 64.6

%CO

%TO=

(c) G1 s( )1

4s 1+( ) 2 s⋅ 1+( )⋅ s 1+( )⋅= 8s

314s

2+ 7s+ 1+ Kc+ 0=

Substitute s = iωu at Kc = Kcu: 8− ωu3

i 14ωu2

− 7ωu i⋅+ 1+ Kcu+ 0 0i+=


7ωu+ 0=ωu

7

8min

1−:= Tu

2π

ωu:= Tu 6.72 min=

Real part: 14− ωu2

1+ Kcu+ 0= Kcu 14min2ωu

21−:= Kcu 11.25

%CO

%TO=

(d) G1 s( )0.5s 1+

4s 1+( ) 2s 1+( )⋅ s 1+( )⋅= 8s

314s

2+ 7 0.5Kc+( )s+ 1+ Kc+ 0=

Substitute s = iωu at Kc = Kcu: 8− ωu3

i 14ωu2

− 7 0.5Kcu+( )ωu i⋅+ 1+ Kcu+ 0 0i+=


1+ Kcu+ 0= Kcu 14ωu2

1−=


Tu 1.797 min=

Tu2π

ωu:=ωu

1 Kcu+

1.8min

1−:=1.8− ωu

21+ Kcu+ 0=Real part:

Kcu 21%CO

%TO=Kcu

6.3

0.3:=6.3 0.3Kcu−( )ωu 0=Imaginary part:

1.8− ωu2

6.3 0.3Kcu−( )ωu i⋅+ 1+ Kcu+ 0 0i+=Substitute s = iωu at Kc = Kcu:

1.8s2

6.3 0.3Kc−( )s+ 1+ Kc+ 0=

e0.6− s 1 0.3s−

1 0.3s+=Padé approximation:6s 1+ Kc e

0.6− s⋅+ 0=G1 s( )

e0.6− s

6s 1+=(f)

KIu 4− min3ωu

4⋅ 4min ωu

2⋅+:= Tu

2π

ωu:=

KIu 0.569%CO

%TO min⋅= Tu 15.17 min=

Must use the smaller ultimate frequency, as the ultimate gain for the other value is negative.

(b) G1 s( )1

s 1+( )2

= s3

2s2

+ s+ KI+ 0=

Substitute s = iωu at KI = KIu: ωu3

− i 2ωu2

− ωu i⋅+ KIu+ 0 0i+=

Imaginary part: ωu3

− ωu+ 0= ωu 1min1−

:= Tu2π

ωu:= Tu 6.28 min=


KIu+ 0= KIu 2minωu2

:= KIu 2%CO

%TO min⋅=


Problem 6-9. Ultimate gain and period with integral controller.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Integral controller: Gc s( )KI

s= Characteristic equation: 1

KI

sG1 s( )+ 0=

(a) G1 s( )1

s 1+( )4

= s5

4s4

+ 6s3

+ 4s2

+ s+ KI+ 0=

Substitute s = iωu at KI = KIu: ωu5i 4ωu

4+ 6ωu

3i− 4ωu

2− ωu i⋅+ KIu+ 0 0i+=

Imaginary part: ωu5

6ωu3

− ωu+ 0=

ωu6 6

24−+

2min2

:= ωu 2.414 min1−

= ωu6 6

24−−

2min2

:= ωu 0.414 min1−

=

Real part: 4 ωu4

⋅ 4ωu3

− KIu+ 0=

14− ωu3

1 4ωu4

− 3.5ωu2

+ ωu+ 0= 4− ωu4

10.5ωu2

− 1+ 0=

ωu10.5 10.5

24 4−( )⋅−−

2 4−( )⋅ min2

:= Tu2π

ωu:= Tu 20.71 min=

KIu 8− min3ωu

4⋅ 7min ωu

2⋅+:= KIu 0.576

%CO

%TO min⋅=

(e) G1 s( )1

4s 1+( ) 0.2s 1+( )⋅ 0.1s 1+( )⋅= 0.08s

41.22s

3+ 4.3s

2+ s+ KI+ 0=

Substitute s = iωu at KI = KIu: 0.08ωu4

1.22ωu3

i− 4.3ωu2

− ωu i⋅+ KIu+ 0 0i+=

Imaginary part: 1.22− ωu3

ωu+ 0= ωu1

1.22min

1−:= Tu

2π

ωu:= Tu 6.94 min=

Real part: 0.08ωu4

4.3ωu2

− KIu+ 0= KIu 0.08− min3ωu

4⋅ 4.3min ωu

2⋅+:=

KIu 3.47%CO

%TO min⋅=

(c) G1 s( )1

4s 1+( ) 2s 1+( )⋅ s 1+( )⋅= 8s

414s

3+ 7s

2+ s+ KIu+ 0=

Substitute s = iωu at KI = KIu: 8ωu4

14ωu3

i− 7ωu2

− ωu i⋅+ KIu+ 0 0i+=


ωu+ 0= ωu1

14min

1−:= Tu

2π

ωu:= Tu 23.51 min=

Real part: 8ωu4

7ωu2

− KIu+ 0= KIu 8− min3ωu

4⋅ 7min ωu

2⋅+:= KIu 0.459

%CO

%TO min⋅=

(d) G1 s( )0.5s 1+

4s 1+( ) 2s 1+( )⋅ s 1+( )⋅= 8s

414s

3+ 7s

2+ 1 0.5KI+( )s+ KI+ 0=

Substitute s = iωu at KI = KIu: 8ωu4

14ωu3

i− 7ωu2

− 1 0.5 KIu⋅+( )ωu i⋅+ KIu+ 0 0i+=

Real part: 8ωu4

7ωu2

− KIu+ 0= KIu 8− ωu4

7ωu2

+=

Imaginary part:


Tu 12.07 min=KIu 1.707%CO

%TO min⋅=KIu 6.3min ωu

2⋅:=

Tu2π

ωu:=ωu

1

3.69min

1−:=1.8− ωu

31 1.89ωu

2− ωu+ 0=Imaginary part:

KIu 6.3ωu2

=6.3− ωu2

KIu+ 0=Real part:

1.8− ωu3

i 6.3ωu2

− 1 0.3KIu−( )ωu i⋅+ KIu+ 0 0i+=Substitute s = iωu at KI = KIu:

1.8s3

6.3s2

+ 1 0.3KI−( )s+ KI+ 0=

e0.6− s 1 0.3s−

1 0.3s+=Padé approximation:6s

2s+ KI e

0.6− s⋅+ 0=G1 s( )

e0.6− s

6s 1+=(f)

τ τT+( )2 4τ τT⋅ KKc⋅− τ τT−<The dominant root is negative when:

τ τT+( )24τ τT⋅

1.8%CO%TO

=KKcτ τT+( )24 τ⋅ τT⋅

≤The roots are real as long as:

τ2

2τ τT⋅− τT2

+ 4τ τT⋅+ 4τ τT⋅ KKc⋅− τ τT+( )2 4τ τT⋅ KKc⋅−=The radical is:

r1τ τT−( )− τ τT−( )2 4τ τT⋅ 1− KKc+( )⋅−+

2 τ⋅ τT⋅=Dominant root:

τ τT⋅ s2⋅ τ τT−( )s+ 1− KKc+ 0=

τT 1min:=(b) Negligible valve time constant,

(negative real root)KKc 1%CO%TO

>The loop is stable for

r1 KKc−

τ=Root:τ s⋅ 1− KKc+ 0=

τT 0:=τv 0:=(a) Negligible valve and transmitter time constants:

τ s⋅ 1−( ) τv s⋅ 1+( )⋅ τT s⋅ 1+( )⋅ KKc+ 0=Charatcteristic equation of the loop:

τ 5min:=G1 s( )K

τs 1−( ) τv s⋅ 1+( )⋅ τT s⋅ 1+( )⋅=Gc s( ) Kc=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

D(s)

C(s)+ +

+-

Problem 6-10. Open-loop unstable process and proportional controller.


polyroots

1− KKc+

τ τv− τc−( )min 1−

τ τv⋅ τ τT⋅+( ) min 2−⋅

τ τv⋅ τT⋅( ) min 3−⋅

10.036−

0.966−

2.064 10 3−=

Positive root, unstable response

KKc 0.99:=

polyroots

1− KKc+


τ τv⋅ τ τT⋅+( ) min 2−⋅


10.037−

0.963−

0

=Root at zero, integrating response

KKc 1:=

Now, the response is unstable also for KKc < 1:

KKcu 43.1%CO%TO

=KKcu 1 τ τv⋅ τ τT⋅+ τv τT⋅−( )ωu2

+:=

Tu 2.25 min=τ τv⋅ τ τT⋅+ τv τT⋅−( )− ωu2

1− KKcu+ 0=Real part:

Tu2πωu

:=ωuτ τv− τT−

τ τv⋅ τT⋅:=τ− τv⋅ τT⋅ ωu

3⋅ τ τv− τT−( )ωu+ 0=Imaginary part:

τ− τv⋅ τT⋅ ωu3

⋅ i τ τv⋅ τ τT⋅+ τv τT⋅−( )ωu2

− τ τv− τT−( )ωu i⋅+ 1− KKcu+ 0 0i+=

Substitute s = ωi at KKc = KKcu:

τ τv⋅ τT⋅ s3⋅ τ τv⋅ τ τT⋅+ τv τT⋅−( )s2

+ τ τv− τT−( )s+ 1− KKc+ 0=Characteristic equation:

τT 1.0min:=τv 0.1min:=(c)

τ τT−( )−

2τ τT⋅0.4− min 1−

=Note: When the roots are complex, the real part is negative:

KKc 1%CO%TO

>So, the loop is stable for:

τ2

2τ τT⋅+ τT2

+ τ2

− 2τ τT⋅+ τT2

− 4τ τT⋅ KKc⋅<

τ τT+( )2 4τ τT⋅ KKc⋅− τ τT−( )2<

KKc 1.01:=

Negative real roots, stable response

polyroots

1− KKc+


τ τv⋅ τ τT⋅+( ) min 2−⋅


10.037−

0.961−

2.073− 10 3−=

So the range of the gain for which the response is stable is: 1%CO%TO

KKc< 43.1%CO%TO

<

Notice that for all three cases there is a lower limit on the loop gain for which the response is stable. This means the response is unstable when the feedback controller is on manual, KKc = 0.


τ 10 min=τV

f1b f2b+:=f2b 2.4

m3

min:=f1b 1.6

m3

min:=From the solution to Problem 3-18:

G2 s( )K2

τ s⋅ 1+=G1 s( )

K1τ s⋅ 1+

=H s( )KT

τT s⋅ 1+=Gv s( )

Kvτv s⋅ 1+

=Gc s( ) Kc=

Ksp Gc(s) Gv(s)Cset(s)

G1(s)

H(s)

C(s)M(s)E(s)G2(s)

F2(s)

F1(s)

+

- ++

From the solution top Problem 3-18, ignoring the inlet concentration disturbances for simplicity:

Use subscript "b" to denote base values for linearization.f1 0.1m3

min:=

Disturbance:

cmax 70kg

m3:=cmin 20

kg

m3:=

τT 3min:=Transmitter:

τv 0.1min:=pv 5psi:=

Linear control valve sized for 100% overcapacity.

cb 50kg

m3:=

c2b 30kg

m3:=fb 4.0

m3

min:=

c1b 80kg

m3:=V 40m3

:=

Problem parameters:

AC

AT

f(t)c(t)

f1(t)

f2(t)

c1(t)

c2(t)

V

m(t)

cset(t)

Problem 6-11. Analyzer control loop for blender of Problem 3-18.


C s( )Ksp Gc s( )⋅ Gv s( )⋅ G2 s( )⋅ Cset

⋅ s( )⋅ G1 s( ) F1 s( )⋅+

1 H s( ) Gc s( ) Gv s( ) G2 s( )+=

Characteristic equation: 1KT

τT s⋅ 1+Kc⋅

Kvτv. s⋅ 1+⋅

K2τ s⋅ 1+

+ 0=

τT τv⋅ τ⋅ s3⋅ τT τv⋅ τT τ⋅+ τv τ⋅+( ) s2

⋅+ τT τv+ τ+( )s+ 1+ KT Kc⋅ Kv⋅ K2⋅+ 0=

Let K KT Kv⋅ K2⋅:= K 0.542−=

Substitute s = ωui at KKc = KKcu:

τT− τv⋅ τ⋅ ωu3

⋅ i τT τv⋅ τT τ⋅+ τv τ⋅+( )ωu2

− τT τv+ τ+( )ωu i⋅+ 1+ KKcu+ 0 0i+=

Imaginary part: τT− τv⋅ τ⋅ ωu3

⋅ τT τv+ τ+( )ωu+ 0= ωuτT τv+ τ+

τT τv⋅ τ⋅:= Tu

2πωu

:=

Real part: τT τv⋅ τT τ⋅+ τv τ⋅+( )− ωu2

1+ KKcu+ 0= Tu 3.01 min=

KcuτT τv⋅ τT τ⋅+ τv τ⋅+( )ωu

21−

K:= Kcu 250−

%CO%TO

=

K1c1b cb−

f1b f2b+:= K2

c2b cb−

f1b f2b+:= K1 7.5

kg

m3

min

m3= K2 5−

kg

m3

min

m3=

Control valve: Cvmax200 %⋅ f2b⋅ gal⋅

3.785 10 3−⋅ m3

1pv

⋅:= Cvmax 567gal

min psi0.5⋅

=

From Fig. C-10.1, p. 532, an 8-in valve is the smallest with enough capacity:

Cvmax 640gal

min psi0.5⋅

:=

f2max Cvmax pv⋅3.785 10 3−

⋅ m3⋅

gal⋅:=

f2max 5.417m3

min= Kv

f2max100%CO

:= Kv 0.054m3

min %CO⋅=

Valve fails closed (air-to-open), to prevent overflowing the tank.

Transmitter: KT100 %TO⋅

cmax cmin−:= Ksp KT:= KT 2

%TO m3⋅

kg=

Closed-loop transfer function::


This is a highly oscillatory response, with 11 cycles before it settles. Students should be encouraged to study which controller gain actually gives quarter decay ratio.

Compare the results with the simulation of Problem 13-11.

5−

0.104− min 1−48.08 min=Settling time:

e 0.104− min 1− T 0.646=Decay ratio:

T 4.21 min=

T2π

1.494min 1−:=

polyroots

1 K Kc⋅+

τT τv+ τ+( )min 1−

τT τv⋅ τT τ⋅+ τv τ⋅+( )min 2−

τT τv⋅ τ⋅ min 3−⋅

10.226−

0.104− 1.494i+

0.104− 1.494i−

=

Response is stable and the dominant roots are complex conjugate. The period of oscillation is:

Although not asked in the problem, let us determine the roots of the characteristic equation:

KT− K1 f1⋅

1 KT Kc⋅ Kv⋅ K2⋅+0.022− %TO=

Offset0s

s

KT−

τT s⋅ 1+

K1τ s⋅ 1+⋅

f1s

⋅

1KT

τT s⋅ 1+Kc⋅

Kvτv s⋅ 1+⋅

K2τs 1+

+

⋅KT− K1⋅ f1⋅

1 KT Kc⋅ Kv⋅ K2⋅+=lim

→=

By the final-value theorem:

E s( ) Ksp Cset⋅ s( ) H s( ) C s( )⋅−= H s( )− C s( )⋅=

Kc 125−%CO%TO

=KcKcu

2:=Cset s( ) 0=F1 s( )

f1s

=Offset:

Direct-acting controller, because the dilute stream has a negative gain on the product composition.

(Eq. 4-2.39)(Eq. 4-2.41)G s( )35.77− 2.07s 1+( )

26.27s3 36.31s2+ 10.14s+ 1+

=

GF s( )31.79 0.976 s⋅ 1+( )⋅ 1 2.77 s⋅−( )⋅

26.27s3 36.31s2+ 10.14s+ 1+

=H s( ) KT=Gv s( ) Kv=Gc s( ) Kc=Ksp KT=

Ksp Gc(s) Gv(s)Tset(s)

GF(s)

H(s)

T(s)M(s)E(s)G(s)

Fc(s)

F(s)

+

- ++

Block diagram, from Section 4-2.3, ignoring disturbances for simplicity:

(use subscript "p" to denote base values for linearization)

fcp 0.8771ft3

min:=

Tmax 700R:=Tmin 640R:=

Transmitter: negligible time constant.

α 50:=

Control Valve: equal-percentage, negligible time constant. TC

TT

Tc(t)

cA(t)

f(t)

fc(t)

cAi(t)

Tci

V

m(t)

Tset(t)

Ti(t)

T(t)

Problem 6-12. Temperature control of non-isothermal reactor of Section 4-2.3 by manipulation of coolant flow to the jacket.



The ultimate frequency cannot be an imaginary number. This means that there is no ultimate gain and period for this loop. The reason is the G(s) has a net order of 2--one zero and three poles--and there are no additional lags for the valve and the transmitter. So, the loop cannot be made unstable with a proportional controller of positive gain.

ωu 0.406i min 1−=ωu

10.14 2.07−

26.27 2.07 36.31⋅−min 1−

:=

10.14 2.07− 26.27 2.07 36.31⋅−( )ωu2

=26.27− ωu3

10.14 2.07KKcu+( )ωu+ 0=Imaginary part:

KKcu 36.31ωu2

1−=36.31− ωu2

1+ KKcu+ 0=Real part:

26.27− ωu3

i 36.31ωu2

− 10.14 2.07 KKcu⋅+( )ωu i⋅+ 1+ KKcu+ 0 0i+=

Rearrange and substitute s = ωui at KKc = KKcu:

The positive K requires a positive Kc, that is, a reverse-acting controller.

K 2.046%TO%CO

=K 35.77−R min⋅

ft3Kv⋅ KT⋅:=Let

1 Kc Kv⋅ KT⋅35.77− 2.07 s⋅ 1+( )⋅

26.27s3 36.31s2+ 10.14s+ 1+

⋅+ 0=Characteristic equation of the loop:

KT 1.667%TO

R=KT

100%TOTmax Tmin−

:=Transmitter gain:

The control valve fails opened (air-to-close) to prevent overheating the reactor on loss of power. Thisis why its gain is negative.

Kv 0.034−ft3

min %CO⋅=Kv

ln α( )− fcp100%CO

:=Control valve gain, Eq. 5-2.24, p. 171:

(Eq. 4-2.39)(Eq. 4-2.41)G s( )35.77− 2.07s 1+( )

26.27s3 36.31s2+ 10.14s+ 1+

=

GF s( )31.79 0.976 s⋅ 1+( )⋅ 1 2.77 s⋅−( )⋅

26.27s3 36.31s2+ 10.14s+ 1+

=H s( ) KT=Gv s( ) Kv=Gc s( ) Kc=Ksp KT=

KspGc(s) Gv(s)

Tset(s)GF(s)

H(s)

T(s)

M(s)E(s)

G(s)Fc(s)

F(s)+

-

++

Block diagram, from Section 4-2.3, ignoring disturbances for simplicity:

(use subscript "p" to denote base values for linearization)

fp 1.3364ft3

min:=

Tmax 700R:=Tmin 640R:=

Transmitter: negligible time constant.

pv 5psi:=

Control Valve: linear, sized for 100% overcapacity, negligible time constant. Assume

TC

TT

Tc(t)

cA(t)

f(t)

fc(t)

cAi(t)

Tci

V

m(t)

Tset(t)

Ti(t)

T(t)

Problem 6-13. Temperature control of non-isothermal reactor of Section 4-2.3 by manipulation of the reactants flow.


The positive K requires a positive Kc, that is, a reverse-acting controller.

Rearrange and substitute s = ωui at KKc = KKcu:

26.27− ωu3

i 36.31 0.976 2.77−( ) KKcu+ ωu2

− 10.14 0.976 2.77−( ) KKcu⋅+ ωu i⋅+ 1+ KKcu+ 0=

Real part: 36.31 0.976 2.77−( ) KKcu+− ωu2

1+ KKcu+ 0= KKcu36.31ωu

21−

1 0.976 2.77−( ) ωu2

−

=

Imaginary part: 26.27− ωu3

10.14 0.976 2.77−( )KKcu+ ωu+ 0= 0.976 2.77− 1.794−=

0.976 2.77−( ) 2.704−=

26.27− ωu2

10.14+ 1.794−( )36.31ωu

21−

1 2.704−( )ωu2

−

+ 0=

26.27− 2.704( )⋅ ωu4

26.27− 10.14 2.704( )+ 1.794−( )36.31+[ ]ωu2

+ 10.14 1.794+( )+ 0=

26.27− 2.704( ) 71.034−= 26.27− 10.14 2.704( )+ 1.794−( )36.31+ 63.992−= 10.14 1.794+ 11.934=

71.034− ωu4

63.992ωu2

− 11.934+ 0=

Control valve size: Cvmax200% fp⋅

pv

7.48gal

ft3:= Cvmax 8.941

gal

min psi⋅=

From Fig. C-10.1, p. 532, a ½-in valve is required: Cvmax 11gal

min psi⋅:= fmax Cvmax pv⋅:=

Control valve gain, Eq. 5-2.23, p. 171: Kvfmax

100%CO:= Kv 0.033

ft3

min %CO⋅=

The control valve fails closed (air-to-open) to prevent overflowing the reactor on loss of power.


Tmax Tmin−:= KT 1.667

%TOR

=

Characteristic equation of the loop: 1 Kc Kv⋅ KT⋅31.79 0.976 s⋅ 1+( )⋅ 1 2.77s−( )⋅

26.27s3 36.31s2+ 10.14s+ 1+

⋅+ 0=

Let K 31.79R min⋅

ft3Kv⋅ KT⋅:= K 1.742

%TO%CO

=

ωu63.992 63.9922 4 71.034−( )⋅ 11.934( )−−

2 71.034−( ) min2:= Tu

2πωu

:= Tu 15.78 min=

Kcu1K

36.31min2ωu

21−

1 2.704min2ωu

2+

:= Kcu 1.91%CO%TO

=

The reason there is an ultimate gain in this case and not when the cooling water is manipulated (Problem 6-12) is the inverse response of the temperature to the reactants flow (negative zero in the transfer function).


K 1.8R:=

x3 16.667 %=x4 40.882 %=f4 3400galmin

=x4f3 x3⋅ f2 x2⋅+

f4:=f4 f3 f2+:=

f3 2400galmin

=x3f5 x5⋅

f3:=f3 f1 f5+:=Assume perfectly mixed tanks, constant density.

x7 90%:=f7 500galmin

:=x5 80%:=f5 500galmin

:=

x2 99%:=f2 1000galmin

:=f1 1900galmin

:=V 7000gal:=Problem parameters (Table P4-1):

Ksp Gc(s) Gv(s)X6

set(s)

G2(s)

H(s)

X6(s)M(s)E(s)G1(s)

F1(s)

F2(s)

+

- ++

Block diagram, considering only f1(t) anf f2(t) as input variables:

V V V

f5(t)

f1(t)

f2(t) f7(t)

f3(t) f4(t) f6(t)

x5(t) x2(t) x7(t)

x3(t) x4(t) x6(t)

AT

ACx6

set(t)m(t)

Problem 6-14. Analyzer control of three mixing tanks of Problem 4-3.


τ2Vf4

:= τ3Vf6

:= K6f4f6

:= K7x6 x4−

f6:= K3

f3f4

:= K5x4 x3−

f4:= K2

x3f3

:=

K8x2 x4−

f4:= τ1 2.917 min= τ2 2.059 min= τ3 1.795 min= K6 0.872= K8 0.017

% min⋅

gal=

K3 0.706= K5 7.122 10 3− % min⋅

gal= K2 6.944 10 3− % min⋅

gal= K7 1.615 10 3− % min⋅

gal=

Eliminate X3(s) and X4(s):

X4 s( )K2− K3⋅ K5 τ1 s⋅ 1+( )⋅−

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅F1 s( )

K8τ2 s⋅ 1+

F2 s( )+=

X6 s( ) G1 s( ) F1 s( )⋅ G2 s( ) F2 s( )⋅+=

G1 s( )K6 K2− K3⋅ K5 τ1 s⋅ 1+( )⋅−⋅ K7 τ1 s⋅ 1+( )⋅ τ2 s⋅ 1+( )⋅−

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ τ3 s⋅ 1+( )⋅=

G2 s( )K6 K8⋅ K7 τ2 s⋅ 1+( )⋅−

τ2 s⋅ 1+( ) τ3 s⋅ 1+( )⋅=

f6 f4 f7+:= x6f4 x4⋅ f7 x7⋅+

f6:= f6 3900

galmin

= x6 47.179 %=

Control valve: Equal-percentage valve, constant pressure drop, α = 50, negligible time constant.

Gv s( ) Kv= Kvln 50( )

100%COf1:= (Eq. 5-2.24, p. 171) Kv 74.3

galmin %CO⋅

=

The valve fails closed (air-to-open) to prevent overflowing the tanks on loss of power.

Analyzer Transmitter: negligible lag, 30 to 70% range:

H s( ) KT= KT100%TO70 30−( )%

:= KT 2.5%TO

%=

Proportional controller: Gc s( ) Kc=

Process Transfer Functions:

From the soution to Problem 4-3: X6 s( )1

τ3 s⋅ 1+K6 X4 s( )⋅ K7 F1 s( )⋅− K7 F2 s( )⋅−( )=

X4 s( )1

τ2 s⋅ 1+K3 X3 s( )⋅ K5 F1 s( )⋅− K8 F2 s( )⋅+( )=

(where we have added F2(s) as an input variable)X3 s( )

K2−

τ1 s⋅ 1+F1 s( )=

τ1Vf3

:=

Substitute s = ωui at Kc = K.cu:

τA− ωu3

⋅ i τB KA Kcu⋅−( )ωu2

− τC KB Kcu⋅−( )ωu i⋅+ 1+ KC Kcu⋅− 0 0i+=

Real part: τB KA Kcu⋅−( )− ωu2

1+ KC Kcu⋅− 0= Kcu−τB ωu

2⋅ 1−

KC KA ωu2

⋅−

=

Imaginary part: τA− ωu3

⋅ τC KBτB ωu

2⋅ 1−

KC KA ωu2

⋅−

⋅+ ωu+ 0=

KA τA⋅ ωu4

⋅ τA− KC⋅ KB τB⋅+ KA τC⋅−( )ωu2

+ KC τC⋅+ KB− 0=

Let a KA τA⋅:= b τA− KC⋅ KB τB⋅+ KA τC⋅−:= c KC τC⋅ KB−:=

a 19.419 min5= b 36.13 min3

= c 10.361 min=

ωub− b2 4 a⋅ c⋅−+

2 a⋅:= ωu 0.595i min 1−

=

The complex ultimate frequency mens that there is no ultimate gain. The process is stable for all negative Kc (direct-acting controller): increasing concentration increases controller output, opening the valve and increasing the flow of pure water. This dilutes the solution and brings the concentration down.

Characteristic equation of the loop: 1 H s( ) Gc s( )⋅ G s( )⋅ G1 s( )⋅+ 0=

1 KT Kc⋅ Kv⋅K6 K2− K3⋅ K5 τ1 s⋅ 1+( )⋅−⋅ K7 τ1 s⋅ 1+( )⋅ τ2 s⋅ 1+( )⋅−

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ τ3 s⋅ 1+( )⋅⋅+ 0=

τ1 τ2⋅ τ3⋅ s2⋅ τ1 τ2⋅ τ1 τ3⋅+ τ2 τ3⋅+ K7 KT⋅ Kc⋅ Kv⋅ τ1⋅ τ2⋅−( )s2

+ +

τ1 τ2+ τ3+ K6 K5⋅ KT⋅ Kc⋅ Kv⋅ τ1⋅− K7 KT⋅ Kc⋅ Kv⋅ τ1 τ2+( )⋅− s+ +

1+ K6 K2 K3⋅ K5+( )⋅ K7+ KT⋅ Kc⋅ Kv⋅− 0=

Let τA τ1 τ2⋅ τ3⋅:= τB τ1 τ2⋅ τ1 τ3⋅+ τ2 τ3⋅+:= τC τ1 τ2+ τ3+:= KA K7 KT⋅ Kv⋅ τ1⋅ τ2⋅:=

KB KT Kv⋅ K6 K5⋅ τ1⋅ K7 τ1 τ2+( )⋅+⋅:= KC K6 K2 K3⋅ K5+( )⋅ K7+ KT⋅ Kv⋅:=

τA 10.778 min3= τB 14.935 min2

= τC 6.77 min= KA 1.802 min2= KB 4.858 min=

KC 2.248=

KG2 0.013min %⋅

gal=KG2 K6 K8⋅ K7−:=G2 0( ) KG2=f2 10

galmin

:=

Offset Ksp x6set

⋅ KT x6⋅−= 0 KTG2 0( )

1 KT Kc⋅ Kv⋅ G1 0( )⋅+f2⋅−=

Offset for an increase of 10 gal/min in flow f2:

The reason there is no ultimate gain is that the transfer function of the process in the loop has a net order of one (three poles and two zeros). So, with negligible valve and transmitter lags, the loop cannot be made unstable with a proportional controller. In practice there will be some lags in the valve and transmitter, however small, and this will impose a limit on the controller gain. Note: In the second edition of this text the analyzer transmitter was specified to have a dead time of 2 min. This is a more realistic situation and did result in an ultimate gain and period.

polyroots

1 KC Kc⋅−

τC KB Kc⋅−( )min 1−

τB KA Kc⋅−( )min 2−

τA min 3−⋅

1.67− 103

2.105−

0.593−

=

Negative real roots. Loop is stable.

Roots of the characteristic equation:Kc 10000−%CO%TO

:=

polyroots

1 KC Kc⋅−



τA min 3−⋅

15.183−

2.326−

0.593−

=

Negative real roots. Loop is stable.


:=

polyroots

1 KC Kc⋅−



τA min 3−⋅

0.59−

0.481− 0.528i−

0.481− 0.528i+

=

Negative real root and complex conjugate roots with negative real parts. Loop is stable.


:=Test:

G1 0( ) KG1= KG1 K6 K2− K3⋅ K5−( )⋅ K7−:= KG1 0.012−min %⋅

gal=

Offset Kc( ) KT−KG2

1 KT Kc⋅ Kv⋅ KG1⋅+f2⋅:=

For Kc = -1 %CO/%TO: Offset 1−( ) 0.102− %TO=

Open-loop offset (Kc = 0): Offset 0( ) 0.332− %TO=

For a PI controller the offset is zero.

10 5 00.004

0.002

0

Offset X( )

X

Let X Kc=


f1 f0 fR+:=k2 k1:=

fR 0ft3

min:=f0 10

ft3

min:=k1 0.2min 1−

:=cA0 7lbmole

ft3:=V2 V1:=V1 125ft3:=

Problem data:

KT 20%TO ft3⋅

lbmole=KT

100 %TO⋅ ft3⋅

5lbmole:=τT 0.5min:=H s( )

KTτT s⋅ 1+

=Transmitter:

Gv s( ) Kv=Control valve:Gc s( ) Kc=Proportional controller:

Ksp Gc(s) Gv(s)

CA2set(s)

G2(s)

H(s)

CA2(s)M(s)E(s)G1(s)

F0(s)

CA0(s)

+

- ++

Block diagram:

V1

f0(t)

fR

f1(t)

cA0(t) cA1(t) cA2(t)

AT

ACcA2

set(t)m(t)

V2

Problem 6-15. Analyzer control of reactors in series of Problem 4-9.


τ1V1

f1 k1 V1⋅+:= τ2

V2f1 k2 V2⋅+

:= τ1 3.571 min=

τ2 3.571 min=

K1f0

f1 k1 V1⋅+:= K2

fRf1 k1 V1⋅+

:= K3f1

f1 k2 V2⋅+:= K1 0.286= K2 0=

K3 0.286=

CA1 s( )1

τ1 s⋅ 1+K1 CA0 s( )⋅ K4 F0 s( )+ K2 CA2 s( )⋅+( )= K4

cA0 cA1−

f1 k1 V1⋅+:= K4 0.143

lbmole min⋅

ft6=

CA2 s( )K3

τ2 s⋅ 1+CA1 s( )

K5τ2 s⋅ 1+

F0 s( )+= K5cA1 cA2−

f1 k2 V2⋅+:= K5 0.041

lbmole min⋅

ft6=

Substitute to eliminate CA1(s):

CA2 s( ) K3K1 CA0 s( )⋅ K4 F0 s( )⋅+ K2 CA2 s( )⋅+( )

τ2 s⋅ 1+( ) τ1 s⋅ 1+( )⋅

K5τ2 s⋅ 1+

F0 s( )+=

K1 K3⋅ CA0 s( )⋅ K3 K4⋅ K5 τ1 s⋅ 1+( )⋅+ F0 s( )⋅+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K3⋅−=

At the initial steady state: f1 k1 V1⋅+( ) cA1⋅ f0 cAo⋅ fR cA2⋅+= f1 k2 V2⋅+( )cA2 f1 cA1⋅=

cA1f0 cA0⋅

f1 k1 V1⋅+fR f1⋅

f1 k2 V2⋅+−

:= cA2f1 cA1⋅

f1 k2 V2⋅+:=

cA1 2lbmole

ft3= cA2 0.571

lbmole

ft3=

Control valve gain: linear sized for 100% overcapacity, assume pv 5psi:=

Cvmax200% f0⋅ 7.48⋅ gal

ft3 pv⋅:= Cvmax 66.903

gal

min psi⋅=

From Fig. C-10.1, p. 532, a 3-in valve is needed: Cvmax 110gal

min psi⋅:=

f0max Cvmax pv⋅ft3

7.48gal:= Kv

f0max100%CO

:= Kv 0.329ft3

min %CO⋅=

The valve fails closed (air-to-open) so as not to overflow the reactor on power failure.

Process transfer functions from the solution to Problem 4-9:

G2 s( )K1 K3⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K3⋅−=

OffsetKsp cA2

set⋅ KT cA2⋅−

1 KT Kc⋅ Kv⋅ G1 0( )⋅+=

0 KT G2 0( ) cA0−

1 KT Kc⋅ Kv⋅ G1 0( )⋅+=

Offset for an increase of 1 lbmole/ft3 in inlet reactant cncentration.

The imaginary value of the ultimate gain means that there is no ultimate gain. The loop is stable for all positive values of the controller gain. The reason is that the net order of the transfer function G1(s) is one (two poles and one zero). With an additional lag in the transmitter, the total order of the transfer function is two, not enough lags to produce instability with a proportional controller.

The controller gain is positive (reverse action): an increase in composition decreases the controller output. This decreases the flow of reactants and decreases the concentration.


KL τC⋅ τD 1 K2 K3⋅−( )⋅−

KL τA⋅ τD τB⋅−:=

τA− ωu3

⋅ τC τDτB ωu

2⋅ 1− K2 K3⋅+

KL⋅+ ωu+ 0=Imaginary part:

KcuτB ωu

2⋅ 1− K2 K3⋅+

KL=τB− ωu

2⋅ 1+ K2 K3⋅− KL Kcu⋅+ 0=Real part:

τA− ωu3

⋅ i τB ωu2

⋅− τC τD Kcu⋅+( )ωu i⋅+ 1+ K2 K3⋅− KL Kcu⋅+ 0 0i+=

Substitute s = ωui at Kc = Kcu:

K3 K4⋅ 0.041lbmole min⋅

ft6=KL 0.537=τD 0.959 min=τC 7.643 min=

τB 16.327 min2=τA 6.378 min3

=KL KT Kv⋅ K3 K4⋅ K5+( )⋅:=τD KT Kv⋅ K5⋅ τ1⋅:=

τC τT τ1+ τ2+ K2 K3⋅ τT⋅−:=τB τT τ1⋅ τT τ2⋅+ τ1 τ2⋅+:=τA τ1 τ2⋅ τT⋅:=Let

1+ K2 K3⋅− KT Kc⋅ Kv⋅ K3 K4⋅ K5+( )⋅+ 0=

τT τ1⋅ τ2⋅ s3⋅ τT τ1⋅ τT τ2⋅+ τ1 τ2⋅+( )s2

+ τT τ1+ τ2+ K2 K3⋅ τT⋅− KT Kc⋅ Kv⋅ K5⋅ τ1⋅+( )s+

1KT

τT s⋅ 1+Kc Kv⋅ G1 s( )⋅+ 0=Charactetristic equation of the loop:

G1 s( )CA2 s( )

F0 s( )=

K3 K4⋅ K5 τ1 s⋅ 1+( )+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K3⋅−=


X Kc=Let

0 5 100.02

0.01

0

Offset X( )

X

Compare these results with the simulation of this process in Problem 13-16.


Offset 0( ) 1.633− %TO=Open-loop offset (Kc = 0):

Offset 1( ) 1.062− %TO=For Kc = 1 %CO/%TO:

Offset Kc( ) KT−KG2

1 KT Kc⋅ Kv⋅ KG1⋅+cA0⋅:=

KG1 0.082lbmole min⋅

ft6=KG1

K3 K4⋅ K5+

1 K2 K3⋅−:=G1 0( ) KG1=

KG2 0.082=KG2K1 K3⋅

1 K2 K3⋅−:=G2 0( ) KG2=cA0 1

lbmole

ft3:=

Block diagram, ignoring the inlet temperature as a variable:

τT 10smin60s⋅:=Hs 1145.4

BTUlb

:=From the steam tables, the enthalpy of the steam is:

Tmax 100degF:=

Tmin 50degF:=Assumptions:

Perfectly mixed tank, constant volume, negligible heat losses•Constant density and physical properties•Steam is at atmospheric pressure•Transmitter has a range of 50 to 100ºF and a time constant of 10 s•

f1 2galmin

:=Disturbance:τv 4smin60s⋅:=

"p" denotes base value for linearization.

w2 t( ) 1.954lb

minvp t( )

pvpsi

⋅=

pv 10psi:=Control valve:

T3p 80degF:=

T1p 60degF:=f1p 25galmin

:=

Design conditions:

cp 0.8BTU

lb degF⋅:=

7lbgal

:=V 5gal:=

TT

TC

f1(t)

f3(t)

w2(t)

T1(t)

T3set(t)

Saturatedsteam

Liquid

T3(t)

m(t)

Problem parameters:

Problem 6-16. Temperature control of direct contact heater of Problem 4-5.


W2 t( ) w2 t( ) w2p−=F1 t( ) f1 t( ) f1p−=3 t( ) T3 t( ) T3p−=1 t( ) T1 t( ) T1p−=where

V⋅ cv⋅d 3 t( )⋅

dt⋅ a1 1 t( ) 3 t( )−( )⋅ a2 F1 t( )⋅+ a3 W2 t( )⋅+ a4 3 t( )⋅−=

Linearize:

Note: This value differs from the value given in the statement of Problem 4-5, probably because a different steam pressure was assumed.

w2p 2.529lb

min=w2p

f1p⋅ cp⋅ T3p T1p−( )⋅

Hs cp T3p 32degF−( )⋅−:=

At the initial steady state:

V⋅ cv⋅d T3 t( )⋅

dt⋅ f1 t( )⋅ cp⋅ T1 t( ) T3 t( )−( )⋅ w2 t( ) Hs cp T3 t( ) 32 degF⋅−( )⋅−⋅+=

Sustitute into enthalpy balance:

f3 t( )⋅ f1 t( )⋅ w2 t( )+=

f1 t( )⋅ w2 t( )+ f3 t( )⋅−d V⋅dt

⋅= 0=Mass balance, assuming constant volume:

f1 t( )⋅ cp⋅ T1 t( ) 32 degF⋅−( )⋅ w2 t( ) Hs⋅+ f3 t( )⋅ cp⋅ T3 t( ) 32 degF⋅−( )⋅− V⋅ cv⋅d T3 t( )⋅

dt⋅=

Enthalpy balance on the tank, neglecting heat losses:

KT 2%TOdegF

=Ksp KT:=KT100%TO

Tmax Tmin−:=H s( )

KTτT s⋅ 1+

=Temperature transmitter:

Kv 0.062lb

min %CO⋅=Kv 1.954

lbmin

pvpsi

⋅1

100%CO:=Gv s( )

Kvτv s⋅ 1+

=Linear control valve:

Gc s( ) Kc=Proportional controller:

Ksp Gc(s) Gv(s)

T3set(s)

G2(s)

H(s)

T3(s)M(s)E(s)G3(s)

W2(s)

F1(s)

+

- ++

Ultimate gain and period of the loop:

Characteristic equation: 1KT

τT s⋅ 1+Kc⋅

Kvτv s⋅ 1+⋅

K3τ s⋅ 1+⋅+ 0=

let KKc KT Kc⋅ Kv⋅ K3⋅=

τT τv⋅ τ⋅ s3⋅ τT τv⋅ τT τ⋅+ τv τ⋅+( ) s2

⋅+ τT τv+ τ+( )s+ 1+ KKc+ 0=

Substitue s = ωui at KKc = KKcu:

τT− τv⋅ τ⋅ ωu3

⋅ i τT τv⋅ τT τ⋅+ τv τ⋅+( ) ωu2

⋅− τT τv+ τ+( )ωu i⋅+ 1+ KKcu+ 0=

Imaginary part: τT− τv⋅ τ⋅ ωu3

⋅ τT τv+ τ+( ) ωu⋅+ 0= ωuτT τv+ τ+

τT τv⋅ τ⋅:= Tu

2πωu

:=

Real part: τT τv⋅ τT τ⋅+ τv τ⋅+( )− ωu2

1+ KKcu+ 0= Tu 0.448 min=

KcuτT τv⋅ τT τ⋅+ τv τ⋅+( )ωu

21−

KT Kv⋅ K3⋅:= Kcu 10.61

%CO%TO

=

Offset caused by change in inlet liquid flow at one half the ultimate gain:

OffsetKsp T3

set⋅ KT T3⋅−

1 KT Kc⋅ Kv⋅ K3⋅+=

0 KT K2⋅ f1⋅−

1 KT Kc⋅ Kv⋅ K3⋅+= Kc

Kcu2

:=

a1 f1p⋅ cp⋅:= a2 cp⋅ T1p T3p−( )⋅:= a3 Hs cp T3p 32degF−( )⋅−:= a4 w2p cp⋅:=

Rearrange: τd 3 t( )⋅

dt⋅ 3 t( )+ K1 1 t( )⋅ K2 F1 t( )⋅+ K3 W2 t( )⋅+= 3 0( ) 0=

where τ

V⋅ cp⋅

a1 a4+:= K1

a1a1 a4+

:= K2a2

a1 a4+:= K3

a3a1 a4+

:= τ 0.197 min=

K1 0.986= K2 0.789−degF min⋅

gal= K3 7.794

degF min⋅

lb=

Note: These values differ from the solution to Problem 4-5 because a different steam pressure is assumed here.

Laplace transform: 3 s( )K1

τ s⋅ 1+ 1 s( )K2

τ s⋅ 1+F1 s( )+

K3τ s⋅ 1+

W2 s( )+=

G2 s( )K2

τ s⋅ 1+= G3 s( )

K3τ s⋅ 1+

=

OffsetKT− K2⋅ f1⋅

1 KT Kc⋅ Kv⋅ K3⋅+:= Offset 0.516 %TO=

OffsetKT

0.258 degF=


Vd c3 t( )⋅

dt⋅ f t( ) c2 t( )⋅ f t( ) c3 t( )⋅− k V⋅ c3 t( )⋅−=

Vd c2 t( )⋅

dt⋅ f t( ) c1 t( )⋅ f t( ) c2 t( )⋅− k V⋅ c2 t( )⋅−=

Vd c1 t( )⋅

dt⋅ f t( ) ci t( )⋅ f t( ) c1 t( )⋅− k V⋅ c1 t( )⋅−=

Model of the reactors, from mass balances on each reactor, assumingPerfectly mixed, constant volume•Constant density and physical properties•

Kv 2.46gal

min %CO⋅=Kv

Cvmax pv⋅

100%CO:=

The valve fails closed (air-to-open) to prevent overflowing the reactors on power failure.

Cvmax 110gal

min psi⋅:=From Fig. C-10.1, page 532, a 3-in valve is required:

Cvmax 89.443gal

min psi⋅=Cvmax

200% fp⋅

pv:=pv 5psi:=

τv 0.1min:=Control valve: linear with constant pressure drop and sized for 100% overcapacity.

KT 100%TO gal⋅

lb=KT

100%TOcmax cmin−

:=cmax 1.0lbgal

:=cmin 0lbgal

:=Analyzer transmitter:

cip 4lbgal

:=fp 100galmin

:=k 0.1min 1−:=V 1000gal:=Problem data:

V V Vf(t) c1(t)

ci(t)

AT

ACc3

set(t)m(t)

c2(t) c3(t)

Problem 6-17. Composition control of three isothermal reactors in series.


τV

fp k V⋅+:= Ki

fpfp k V⋅+

:= K1cip c1p−

fp k V⋅+:= K2

c1p c2p−

fp k V⋅+:= K3

c2p c3p−

fp k V⋅+:=

τ 5 min= Ki 0.5= K1 0.01lb min⋅

gal2= K2 0.005

lb min⋅

gal2= K3 0.0025

lb min⋅

gal2=

Laplaxce transform:

C1 s( )Ki Ci s( )⋅ K1 F s( )⋅+

τ s⋅ 1+= C2 s( )

Ki C1 s( )⋅ K2 F s( )⋅+

τ s⋅ 1+= C3 s( )

Ki C2 s( )⋅ K3 F s( )⋅+

τ s⋅ 1+=

Combine to obtain the transfer functions:

G1 s( )C3 s( )

F s( )= 1

τ s⋅ 1+

Kiτs 1+

Ki K1⋅

τ s⋅ 1+K2+ K3+=

G1 s( )Ki Ki K1 K2 τ s⋅ 1+( )⋅+⋅ K3 τs 1+( )2⋅+⋅

τ s⋅ 1+( )3=

G1 s( )Ki

2 K1⋅ Ki K2⋅ τ s⋅ 1+( )⋅+ K3 τ s⋅ 1+( )2⋅+

τ s⋅ 1+( )3= G2 s( )

C3 s( )

Ci s( )=

Ki3

τ s⋅ 1+( )3=

Linearize:V

d C1 t( )⋅

dt⋅ fp Ci t( )⋅ fp k V⋅+( )C1 t( )− cip c1p−( )F t( )+=

Vd C2 t( )⋅

dt⋅ fp C1 t( )⋅ fp k V⋅+( )C2 t( )− c1p c2p−( )F t( )+=

Vd C3 t( )⋅

dt⋅ fp C2 t( )⋅ fp k V⋅+( )C3 t( )− c2p c3p−( )F t( )+=

where Cj t( ) Cj t( ) cjp−= F t( ) t( ) fp−=

At the initial steady-state:c1p

fp cip⋅

fp k V⋅+:= c2p

fp c1p⋅

fp k V⋅+:= c3p

fp c2p⋅

fp k V⋅+:=

Rearrange:

τd C1 t( )⋅

dt⋅ C1 t( )+ Ki Ci t( )⋅ K1 F t( )⋅+= C1 0( ) 0=

τd C2 t( )⋅

dt⋅ C2 t( )+ Ki C1 t( )⋅ K2 F t( )⋅+= C2 0( ) 0=

τd C2 t( )⋅

dt⋅ C2 t( )+ Ki Ci t( )⋅ K3 F t( )⋅+= C3 0( ) 0=

where

(c) Ultimate gain and period of the loop with a proportional controller.


Offset 12.5− %TO=Offset KT− Ki3 ci⋅:=For the open loop, Kc = 0:

OffsetKT

0.044−lbgal

=Offset 4.39− %TO=

OffsetKT Ki

3⋅ ci

1 KT Kc⋅ Kv⋅ KA⋅+−:=Offset

Ksp c3set

⋅ KT G2 0( )⋅ ci−

1 KT Kc⋅ Kv⋅ G1 0( )⋅+=

Kc 1.%CO%TO

:=ci 1lbgal

:=(b) Offset for a change in inlet concentration

KB 0.0075lb min⋅

gal2=K3 0.0025

lb min⋅

gal2=

KA 0.0075lb min⋅

gal2=τ 5 min=KB 2 K3⋅ Ki K2⋅+:=KA Ki

2K1 Ki K2⋅+ K3+:=

G1 s( )KA KB τ⋅ s⋅+ K3 τ

2⋅ s⋅+

τ s⋅ 1+( )3=G2 s( )

Ki3

τ s⋅ 1+( )3=

Gc s( ) Kc=Proportional controller:Gv s( )Kv

τv s⋅ 1+=H s( ) KT=Ksp KT:=

Ksp Gc(s) Gv(s)

C3set(s)

G2(s)

H(s)

C3(s)M(s)E(s)G1(s)

F(s)

Ci(s)

+

- ++

(a) Block diagram of the loop:


The imaginary value of the ultimate frequency shows that there is no ultimate gain for this loop. This is because the net order of the loop is one--three poles and two zeros--and it cannot be unstable for any positive value of the controller gain. The controller gain is positive, reverse acting: increases concentration decreases the signal to the valve. This decreases the reactants flow and the concentration decreases.


b− b2 4 a⋅ τB−( )⋅−−

2 a⋅:=

b 1.537− min3=a 1.962− 10 14− min5

=

b 3− τv⋅ τ2

⋅ KL⋅ τv τA⋅− τB 3⋅ τv⋅ τ⋅+:=a 3τv τ2

⋅ τA⋅ τB τv⋅ τ3

⋅−:=Let

3 τv⋅ τ2

⋅ τA⋅ τB τv⋅ τ3

⋅− ωu4

3− τv⋅ τ2

⋅ KL⋅ τv τA⋅− τB 3⋅ τv⋅ τ⋅+ ωu2

+ τB− 0=

3− τv⋅ τ2

⋅ ωu3

τv τB3 τv⋅ τ⋅ ωu

2⋅ τv τ

3⋅ ωu

4− 1−

KL τA ωu2

⋅−

⋅+ ωu+ 0=Imaginary part:

Kcu3 τv⋅ τ⋅ ωu

2⋅ τv τ

3⋅ ωu

4− 1−

KL τA ωu2

⋅−

=

τv τ3

⋅ ωu4

3 τv⋅ τ⋅ τA Kcu⋅+( )ωu2

− 1+ KL Kcu⋅+ 0=Real part:

τB 9.224 min=τA 15.373 min2=

KL 1.845%TO%CO

=KL KT Kv⋅ KA⋅:=τB KT Kv⋅ KB⋅ τ⋅:=τA KT Kv⋅ K3⋅ τ2

⋅:=where

τv τ3

⋅ ωu4

3τv τ2

⋅ ωu3

i− 3τv τ⋅ τA Kcu⋅+( )ωu2

− τv τB Kcu⋅+( )ωu i⋅+ 1+ KL Kcu⋅+ 0=

Rearrange and substitute s = ωui at Kc = Kcu:

1 KT Kc⋅Kv

τv s⋅ 1+⋅

KA KB τ⋅ s⋅+ K3 τ2

⋅ s2⋅+

τ s⋅ 1+( )3⋅+ 0=Characteristic equation of the loop:

Closed-loop transfer function:Gp s( )

Kpτp s⋅ 1+

=

Gsc s( )Ksc

τsc s⋅ 1+=KT 5

%TOpsi

=H s( )KT

τT s⋅ 1+=Ksp KT:=KT

100%TOPmax Pmin−

:=

Ksp Gc(s) G

sc(s)

Ps

set(s)

Gp(s)

H(s)

Ps(s)M(s)E(s)

Gp(s)

Fc(s)

Fi(s)

+

- -+

(a) Block diagram of the loop, closed-loop transfer function, and characteristic equation of the loop.

τT 1.2s:=Pmax 20psig:=Pmin 0psig:=Pressure transmitter, PC:

τsc 2.5s:=

Ksc 0.36kscf

min %CO⋅:=

Fc s( )Ksc

τsc s⋅ 1+M s( )=

τp 7.5s:=

Kp 0.5psi min⋅

kscf:=

Ps s( )Kp

τp s⋅ 1+Fi s( ) Fc s( )−( )=

Steam

SuctionDischarge

fi(t)

fc(t)

ps(t)

m(t)

PT

PC

SC

Problem Data:

Problem 6-18. Compressor suction pressure control. psig psi:=

kscf 1000ft3:=%TO %:=%CO %:=Smith & Corripio, 3rd edition

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OffsetKT

0.061− psi=

Offset 0.307− %TO=Offset0 KT Kp⋅ fi⋅−

1 KL Kc⋅+:=

OffsetKsp ps

set⋅ KT Kp⋅ fi⋅−

1 KL Kc⋅+=

Kc 7.9−%CO%TO

=KcKcu

2:=fi 1

kscfmin

:=(c) Offset caused by a change in inlet flow.

Kcu 15.9−%CO%TO

=

Tu 8.91 s=Tu2πωu

:=KcuτT τsc⋅ τT τp⋅+ τsc τp⋅+( )ωu

21−

KL:=Real part:

ωuτT τsc+ τp+

τT τsc⋅ τp⋅:=τT− τsc⋅ τp⋅ ωu

3⋅ τT τsc+ τp+( )ωu+ 0=Imaginary part:

KL 0.9−%TO%CO

=KL KT− Ksc⋅ Kp⋅:=where

τT− τsc⋅ τp⋅ ωu3

⋅ i τT τsc⋅ τT τp⋅+ τsc τp⋅+( )ωu2

− τT τsc+ τp+( )ωu i⋅+ 1+ KL Kcu⋅+ 0 0i+=

Rearrange the characteristic equation and sunbstiture s =ωui at Kc = Kcu:

Gc s( ) Kc=(b) Ultimate gain and period for a preportional controller.

The controller must be direct-acting (negative gain): increasing pressure increases the signal to te speed controller (SC). This increases the compressor speed and the flow through the compressor, decreasing the suctiion pressure.

1 H s( ) Gc s( )⋅ Gsc s( )⋅ Gp s( )⋅− 1KT

τT s⋅ 1+Gc s( )

Kscτsc s⋅ 1+

Kpτp s⋅ 1+

−= 0=

Characteristic equation of the loop:

Ps s( )Ksp− Gc s( )⋅ Gsc s( )⋅ Gp s( )⋅

1 H s( ) Gc s( )⋅ Gsc s( )⋅ Gp s( )−Ps

set s( )Gp s( )

1 H s( ) Gc s( )⋅ Gsc s( )⋅ Gp s( )⋅−( )Fi s( )+=

Ksp G

c(s) G

FC(s)

Tset(s)

G2(s)

H(s)

T(s)M(s)E(s)G1(s)

Fc(s)

Ti(s)

+

- -+

(a) Block diagram of the temperature control loop, valve fail position, controller action.

τFC 0.1min:=fmax 0.8m3

min:=fmin 0

m3

min:=Flow transmitter(FT):

τT 0.6min:=Tmax 70degC:=Tmin 20degC:=

Temperature transmitter (TT):

Tci 25degC:=T 45degC:=

Ti 70degC:=f 0.1m3

min:=

Design conditions:

cpc 4.2kJ

kg degC⋅:=c 1000

kg

m3:=

cp 3.8kJ

kg degC⋅:=800

kg

m3:=

Vc 1.1m3:=A 4m2

:=

U 200kJ

m2min degC⋅:=V 5m3

:=

Problem Data:

TC

TT

Tc(t)

f(t)

fc(t) Tci

V

m(t)

Tset(t)

Ti(t)

T(t)

FT

FCSP

Problem 6-19. Temperature control of stirred-tank cooler of Problem 4-7.

degC K:=kJ 1000joule:=Smith & Corripio, 3rd edition

Tc Tf ⋅ cp⋅ Ti T−( )⋅

U A⋅−:= Tc 35.5 degC=

f ⋅ cp⋅ Ti T−( )⋅ fc c⋅ cpc⋅ Tc Tci−( )⋅− 0= fcf ⋅ cp⋅ Ti T−( )⋅

c cpc⋅ Tc Tci−( )⋅:= fc 0.172

m3

min=

τ2Vc c⋅ cpc⋅

fc c⋅ cpc⋅ U A⋅+:= K3

c cpc⋅ Tc Tci−( )⋅

fc c⋅ cpc⋅ U A⋅+:= K4

U A⋅fc c⋅ cpc⋅ U A⋅+

:=

τ2 3.03 min= K3 28.94degC min⋅

m3= K4 0.525=

The coolant valve must fail opened (air-to-close) to prevent loss of coolant on power failure. This means that the flow controller must be direct acting: incease in flow increases the output to close the valve and reduce the flow.

The temperature controller must also be direct acting (negative gain): increasing temperature must increase the output to incrase the coolant flow and reduce the temperature.

(b) Ultimate gain and period of the loop with a proportional controller. Gc s( ) Kc=

Characteristic equation of the loop:1

KTτT s⋅ 1+

KcKFC

τFC s⋅ 1+⋅

K2 K3⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( ) K2 K4⋅−+ 0=

Rearrange and substitute s = ωui at Kc = Kcu:

τA ωu4

⋅ τB ωu3

⋅ i− τC ωu2

⋅− τD ωu⋅ i⋅+ 1+ K2 K4⋅− KL Kcu⋅+ 0 0i+=

Temperature transmitter: KT100%TO

Tmax Tmin−:= Ksp KT:= H s( )

KTτT s⋅ 1+

= KT 2%TOdegC

=

Flow contriol loop: KFCfmax fmin−

100%TO:= GFC s( )

KFCτFC s⋅ 1+

= KFC 8 10 3− m3

min %TO⋅=

From the results of Problem 4-7:

G1 s( )K2 K3⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K4⋅−= G2 s( )

K1 τ2 s⋅ 1+( )⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K4⋅−=

where τ1V⋅ cp⋅

f ⋅ cp⋅ U A⋅+:= K1

f ⋅ cp⋅

f ⋅ cp⋅ U A⋅+:= K2

U A⋅f ⋅ cp⋅ U A⋅+

:= τ1 13.77 min=

At the initial steady state (design) conditions: K1 0.275= K2 0.725=

f ⋅ cp⋅ Ti T−( )⋅ U A⋅ T Tc−( )⋅− 0=

KcuτA− ωu

4⋅ τC ωu

2⋅+ 1− K2 K4⋅+

KL:= Kcu 86.7−

%CO%TO

=

(c) Offset for change in inlet temperature with a proportional copntroller.

Ti 5degC:= KcKcu

2:= Kc 43.3−

%CO%TO

=

OffsetKsp T

set⋅ KT G2 0( )⋅ Ti−

1 KT Kc⋅ KFC⋅ G1 0( )⋅+= Offset

0 KT K1⋅ Ti⋅−

1 K2 K4⋅− KT Kc⋅ KFC⋅ K2⋅ K3⋅−:=

Offset 0.18− %TO=Offset

KT0.09− degC=

Open-loop, Kc = 0: Offset 0KT K1⋅ Ti⋅

1 K2 K4⋅−−:= Offset 4.44− %TO=

With a PI controller the offset is zero.

Students should verify these results with the simulation of this process in Problem 13-18.



is unlawful.

where τA τT τFC⋅ τ1⋅ τ2⋅:= τB τT τFC⋅ τ1⋅ τT τFC⋅ τ2⋅+ τFC τ1⋅ τ2⋅+ τT τ1⋅ τ2⋅+:=

τA 2.505 min4=

τC τT τFC⋅ 1 K2 K4⋅−( )⋅ τT τ1⋅+ τT τ2⋅+ τFC τ1⋅+ τFC τ2⋅+ τ1 τ2⋅+:=

τB 30.228 min3=

τD τT τFC+( ) 1 K2 K4⋅−( ) τ1+ τ2+:= KL KT− KFC⋅ K2⋅ K3⋅:=τC 53.54 min2

=

KL 0.336−%TO%CO

= τD 17.234 min=

Imaginary part: τB− ωu3

⋅ τD ωu⋅+ 0= ωuτDτB

:= Tu2πωu

:= Tu 8.32 min=

Real part: τA ωu4

⋅ τC ωu2

⋅− 1+ K2 K4⋅− KL Kcu⋅+ 0=

G 1.724=Use Eqs. 5-2.3 and 5-2.5, page 160.

Valve 1: y11.63Cf

p1 p2−

p1 14.7psia+⋅:= y1 1.187=

Cv1max200% f1⋅ G T 460R+( )⋅⋅

836 Cf⋅ p1 14.7psia+( )⋅ y1 0.148y13

−

gal psia⋅ hr⋅

min R0.5scf⋅

60minhr

:= Cv1max 90.9gal

min psi⋅=

From Fig. C-10.1, page 532, a 3-in valve is required: Cv1max 110gal

min psi⋅:=

Valve 3: y31.63Cf

p2 p3−

p2 14.7psia+⋅:= y3 0.908=

Cv3max200% f3⋅ G T 460R+( )⋅⋅

836 Cf⋅ p2 14.7psia+( )⋅ y3 0.148y33

−

gal psia⋅ hr⋅

min R0.5scf⋅

60minhr

:= Cv3max 62.6gal

min psi⋅=


min psi⋅:=

Valve 4: From initial steady state conditions: f4 f1 f3−:= f4 1000scfmin

=

y41.63Cf

p2 p4−

p2 14.7psia+⋅:= y4 1.284=

Smith & Corripio, 3rd edition degF R:= scf ft3:= psia psi:=

lbmole 453.59mole:=Problem 6-20. Pressure control og gas torage tank.

PT

PC

vp1(t)

vp3(t)

vp4(t)p2(t)

p1(t)

p3(t)

p4(t)

p2set(t)

Problem Data:MW 50

lblbmole

:=

V 550000ft3:= T 350degF:=

f1 1500scfmin

:= p1 90psi:=

f3 500scfmin

:= p3 30psig:=

p2 45psig:= p4 15psig:=

Pressure transmitter PT:

pmin 0psig:= pmax 100psig:=

(a) Size control valves for 100% overcapacity.Cf 0.9:= G

MW lbmole⋅

29lb:=

s 0.132lbscf

=

Valve 1: f1 t( )836 scf⋅ R⋅ min⋅

gal psia⋅ hr⋅

1 hr⋅

60 min⋅⋅

Cv1max Cf⋅ p1 t( ) 14.7 psia⋅+( )⋅

G T 460R+( )⋅⋅ y1 t( ) 0.148 y1 t( )3

− vp1 t( )=

4 eqns. 6 unks. (y1)y1 t( )

1.63Cf

p1 t( ) p2 t( )−

p1 t( ) 14.7psia+⋅=


gal psia⋅ hr⋅

1 hr⋅

60 min⋅⋅

Cv3max Cf⋅ p2 t( ) 14.7 psia⋅+( )⋅

G T 460R+( )⋅⋅ y3 t( ) 0.148 y3 t( )3

− vp3 t( )=


1.63Cf

p2 t( ) p3 t( )−

p2 t( ) 14.7psia+⋅=


gal psia⋅ hr⋅

1 hr⋅

60 min⋅⋅

Cv4max Cf⋅ p2 t( ) 14.7 psia⋅+( )⋅

G T 460R+( )⋅⋅ y4 t( ) 0.148 y4 t( )3

− vp4 t( )=


1.63Cf

p2 t( ) p4 t( )−

p2 t( ) 14.7psia+⋅=

Substitute ideal gas law: MW V⋅Rg T 460 R⋅+( ) s⋅

d p2 t( )⋅

dt⋅ f1 t( ) f3 t( )− f4 t( )−=

Linearize and substitute deviation variables:

MW V⋅Rg T 460+( )⋅ s⋅

d P2 t( )⋅

dt⋅ a1 VP1 t( )⋅ b1 P1 t( )⋅+ c1 P2 t( )⋅− a3 VP3 t( )⋅− b3 P2 t( )⋅− c3 P3 t( )⋅+=

Cv4max200% f4⋅ G T 460R+( )⋅⋅

836 Cf⋅ p2 14.7psia+( )⋅ y4 0.148y43

−

gal psia⋅ hr⋅

min R0.5scf⋅

60minhr

:= Cv4max 102.9gal

min psi⋅=


min psi⋅:=

(c) Block diagram of the loop considering all disturbances.

Transmitter PT: H s( ) KT= KT100%TO

pmax pmin−:= Ksp KT:= KT 1

%TOpsi

=

Process model:

Mass balance: Vd 2 t( )⋅

dt⋅ s f1 t( )⋅ s f3 t( )⋅− s f4 t( )⋅−=

1 eqn. 4 unks. ( 2, f1, f2, f3)

Ideal gas law: 2 t( )MW p2 t( ) 14.7psia+( )⋅

Rg T 460R+( )⋅= 2 eqns. 5 unks. (p2)

Rg 10.73psia ft3⋅

lbmole R⋅⋅:=

sMW 14.7⋅ psia

Rg 520⋅ R:=

ky111.63Cf

12⋅

p1 p2−

p1 14.7 psia⋅+

0.5−

⋅p1 14.7 psia⋅+( ) p1 p2−( )−

p1 14.7psia+( )2⋅:=

b1f1 t( )⋅

p1⋅= b1

kv Cv1max⋅ vp1⋅

100%COy1 0.148 y1

3⋅− p1 14.7psia+( ) 1 3 0.148⋅ y1

2− ky11+:=

ky11 0.00752 psia 1−= b1 18.819

scfmin psi⋅

=

ky12y1 t( )⋅

p2⋅= ky12

1.63Cf

12⋅

p1 p2−

p1 14.7 psia⋅+

0.5−

⋅1−

p1 14.7psia+( )⋅:= ky12 0.013− psi 1−

=

c1− f1 t( )⋅

p2⋅= c1

kv− Cv1max⋅ vp1⋅

100%COp1 14.7psia+( ) 1 3 0.148⋅ y1

2− ky12:= c1 7.878

scfmin psi⋅

=

ky32y3 t( )⋅

p2⋅= ky32

1.63Cf

12⋅

p2 p3−

p2 14.7 psia⋅+

0.5−

⋅p2 14.7 psia⋅+( ) p2 p3−( )−

p2 14.7psia+( )2⋅:=

b3f3 t( )⋅

p2⋅= b3

kv Cv3max⋅ vp3⋅

100%COy3 0.148 y3

3⋅− p2 14.7psia+( ) 1 3 0.148⋅ y3

2− ky32+:=

ky32 0.02266 psia 1−= b3 17.387

scfmin psi⋅

=

ky33y3 t( )⋅

p3⋅= ky33

1.63Cf

12⋅

p2 p3−

p2 14.7 psia⋅+

0.5−

⋅1−

p2 14.7psia+( )⋅:= ky33 0.03− psi 1−

=

a4− VP4 t( )⋅ b4 P2 t( )⋅− c4 P4 t( )⋅+

Let kv836scf R⋅ min⋅

gal psia⋅ hr⋅

hr60min

Cf

G T 460R+( )⋅:= kv 0.336

scf

gal psia⋅=

a1f1 t( )⋅

vp⋅= a1

kv100%CO

Cv1max⋅ p1 14.7psia+( )⋅ y1 0.148y13

−⋅:= a1 36.312scf

min %CO⋅=

a3f3 t( )⋅

vp⋅= a3

kv100%CO

Cv3max⋅ p2 14.7psia+( )⋅ y3 0.148y33

−⋅:= a3 17.565scf

min %CO⋅=

a4f4 t( )⋅

vp⋅= a4

kv100%CO

Cv4max⋅ p2 14.7psia+( )⋅ y4 0.148y43

−⋅:= a4 21.39scf

min %CO⋅=

Initial valve positions: vp1f1a1

:= vp3f3a3

:= vp4f4a4

:= vp1 41.3 %CO= vp3 28.5 %CO=

vp4 46.8 %CO=

ky11y1 t( )⋅

p1⋅=

K1a1

c1 b3+ b4+:= K3

a3c1 b3+ b4+

:=

K4a4

c1 b3+ b4+:= Kp1

b1c1 b3+ b4+

:= Kp3c3

c1 b3+ b4+:= Kp4

c4c1 b3+ b4+

:=

τ 534.3 min= K1 0.808psig%CO

= K3 0.391psi

%CO= K4 0.476

psi%CO

=

Kp1 0.419= Kp3 0.268= Kp4 0.131=

Process transfer function: Laplace transform:

P2 s( )1

τ s⋅ 1+K1 VP1 s( )⋅ K3 VP3 s( )⋅− K4 VP4 s( )⋅− Kp1 P1 s( )⋅+ Kp3 P3 s( )⋅+ Kp4 P4 s( )⋅+( )=

The very long time constant, approximately 9 hours, denotes that the pressure in the tank behaves as an integrating process. See discussion of controller tuning for integrating processes in Section 7-3.

Block diagram of the loop:

c3− f3 t( )⋅

p3⋅= c3


100%COp2 14.7psia+( ) 1 3 0.148⋅ y3

2− ky33:= c3 12.036

scfmin psi⋅

=

ky42y4 t( )⋅

p2⋅= ky42

1.63Cf

12⋅

p2 p4−

p2 14.7 psia⋅+

0.5−

⋅p2 14.7 psia⋅+( ) p2 p4−( )−

p2 14.7psia+( )2⋅:=

b4f4 t( )⋅

p2⋅= b4

kv Cv4max⋅ vp4⋅

100%COy4 0.148 y4

3⋅− p2 14.7psia+( ) 1 3 0.148⋅ y4

2− ky42+:=

ky42 0.01065 psia 1−= b4 19.691

scfmin psi⋅

=

ky44y4 t( )⋅

p4⋅= ky44

1.63Cf

12⋅

p2 p4−

p2 14.7 psia⋅+

0.5−

⋅1−

p2 14.7psia+( )⋅:= ky44 0.021− psi 1−

=

c4− f4 t( )⋅

p4⋅= c4


100%COp2 14.7psia+( ) 1 3 0.148⋅ y4

2− ky44:= c4 5.911

scfmin psi⋅

=

Rearrange equation into standard first-order form:

τd P2 t( )⋅

dt⋅ P2 t( )+ K1 VP1 t( )⋅ K3 VP3 t( )⋅− K4 VP4 t( )⋅− Kp1 P1 t( )⋅+ Kp3 P3 t( )⋅+ Kp4 P4 t( )⋅+=

where τMW V⋅

Rg T 460R+( )⋅ s⋅ c1 b3+ b4+( )⋅:=

Ksp Gc(s) K1

P2set(s)

H(s)

P2(s)VP1(s)E(s)1

F1(s)

P1(s)

+

-

-

+

Js + 1

K3

VP3(s)

K4

VP4(s)

Kp1

-

+

Kp3

Kp4

P4(s)

P3(s)

++

(c) Ultimate gain. The response of the loop cannot be unstable with a proportional controller with positive gain because the loop transfer function is first-order. There is no ultimate gain. The controller is reverse acting: increasing tank pressure decreases the controller output to close the control valve and decrease the inlet flow. This decreases the pressure in the tank.

The valve fails closed (air-to-open) to prevent over-pressuring the tank on instrument power failure.

(d) Offset for a proportinal controller and a change in set point. p2set 5psi:=

Kc 50%CO%TO

:=

OffsetKsp p2set⋅

1 KT Kc⋅ K1⋅+:= Offset 0.121 %TO= Offset

KT0.121 psi=



is unlawful.

XB 0( ) 0=WB t( ) w XB t( )⋅− xB WA t( ) WB t( )+( )⋅− Md XB t( )⋅

dt⋅=

Linearize and express in terms of deviation variables:

(b) Linearize the equations, transfer functions and block diagram:

2 eqns. 2 unks. (w, xB)wB t( ) w t( ) xB t( )⋅− M

d xB t( )⋅

dt⋅=Mass balance on component B:

w t( ) wA t( ) wB t( )+=wA t( ) wB t( )+ w t( )−d M⋅dt

= 0=Total mass balance:

Assumptions:Perfectly mixed•Constant mass and density•Negligible transportation lag•Inlet streams are pure A and B, respectively•

(a) Model of the composition control loop

Range is CL to CH.

C t( )xB t( )

mhom

=Analyzer AT measures solution conductivity:

q t( ) qmax m t( )⋅=wB t( ) Kv2 vp2 t( )⋅=wA t( ) Kv1 vp1 t( )⋅=Problem Data:

AT

AC

TT

TC

wA(t)

wB(t)

TA(t)

TB(t)

vp1(t)

vp2(t)

m2(t)

m(t)

xB(t)T(t)

b(t)

Problem 6-21. Temperature and analysis control of a heated mixer.Smith & Corripio, 3rd edition

Assumptions:perfectly mixed•constant mass and physical properties•

(c) Model of the temperature control loop

(negative set point scale)Ksp KT=

If the control valve is air-to-open, the controller must be direct acting, because of the negative gain in the transmitter: Increasing composition decreases the signal from the transmitter; the controller decreases the signal to the valve; this closes the valve and reduces the flow of component B, decreasing the composition. Notice also that

Ksp G

AC(s) K

v2KB

XB

set(s)

H(s)

XB(s)M2(s)E(s)1+

-

-

Js + 1

VP1(s)

++

Kv1KA

100

Block diagram of the composition loop:

WB s( ) Kv2 VP2 s( )⋅=Kv2

100%COM2 s( )=WA s( ) Kv1 VP1 s( )⋅=Control valves:

H s( ) KAT= 100− %TO⋅

CH CL−( )xB2

=B t( )100%TOCH CL−

−

xB2

XB t( )=Linearize:

b t( )C t( ) CL−

CH CL−100%TO= 100%TO

CH CL− xB t( )CL−=Transmitter:

XB s( )1

τ s⋅ 1+KB WB s( )⋅ KA WA s( )⋅−( )=Laplace transform and solve for output:

τd XB t( )⋅

dt⋅ XB+ KB WB t( )⋅ KA WA t( )⋅−=KB

1 xB−

w=KA

xBw

=τMw

=Let

Mw

d XB t( )⋅

dt⋅ XB t( )+

1 xB−

wWB t( )

xBw

WA t( )−=Rearrange in the standard first-order form:


The controller is reverse acting: increasing temperature decreases the controller output, decresing the rate of heat input. This decreases the temperature.

H s( ) KTT=100%TOTH .L−

=Temperature transmitter TT:Q s( )qmax

100%COM s( )=Electric heater:

K5wBw

=K4wAw

=K31

w cp⋅=K2

TB T−

w=K1

TA T−

w=τ

Mw

=

s( )1

τ s⋅ 1+K1 WA s( )⋅ K2 WB s( )⋅+ K3 Q s( )⋅+ K4 A s( )⋅+ K5 B s( )⋅+( )=

Laplace transform:0( ) 0=

Mw

d t( )⋅

dt⋅ t( )+

TA T−( )w

WA t( )TB T−( )

wWB t( )+

1w cp⋅

Q t( )+wAw A t( )+

wBw B t( )+=

Rearrange into the standard first-oder equation form:

wB cp⋅ B t( )⋅+ wA wB+( ) cp⋅ t( )⋅−

M cp⋅d t( )⋅

dt⋅ cp TA T−( )⋅ WA t( ) cp TB TA−( )⋅ WB t( )⋅+ Q t( )+ wA cp⋅ A t( )⋅+=


(d) Linearize the equation, transfer functions, block diagram.

M cp⋅d T t( )⋅

dt⋅ wA t( ) cp⋅ TA t( ) T t( )−( )⋅ wB t( ) cp⋅ TB t( ) T t( )−( )⋅+ q t( )+=

cp cpA= cpB= cv=Assume w t( ) wA t( ) wB t( )+=Substitute 1 eqn. 1 unk. (T)

M cv⋅d T t( )⋅

dt⋅ wA t( ) cpA⋅ TA t( ) Tref−( )⋅ wB t( ) cpB⋅ TB t( ) Tref−( )⋅+ q t( )+ w t( ) cp⋅ T t( ) Tref−( )⋅−=

Energy balance:

negligible heat losses and ransportantioon lag•

Ksp G

TC(s) q

maxK3

'set(s)

H(s)

M(s)E(s)1+

- Js + 1

VP2(s)

++

Kv2K2

100

'(s)

VP1(s)Kv1K1 K4

K5

'A(s)

'B(s)

++ +

(e) Characteristic equations of the control loops.

Analyzer control loop:

1 KAT GAC s( )⋅Kv2

100%CO⋅

KBτ s⋅ 1+⋅+ 1

100− %TO⋅

CH CL−( )xB2

GAC s( )Kv2

100%CO

1 xB−( )w τ s⋅ 1+( )⋅+= 0=

Temperature control loop:

1 KTT GTC s( )⋅qmax

100%CO⋅

K3τ s⋅ 1+⋅+ 1

100%TOTH TL−

GTC s( )qmax

100%CO⋅

1w cp⋅ τ s⋅ 1+( )⋅⋅+= 0=

For both loops: τMw

=

There is no ultimate gain for either loop because they are first-order when a proportional controller is used. The loops cannot be made unstable as long as the controller gains have the proper sign. In practice there will be lags on the transmitter and final control elements and there will be an ultimate gain and period.



is unlawful.

fi 50galmin

:= fA 50galmin

:=

(a) Size the control valve for 100% overcapacity.

GA MWA⋅ gal⋅

8.33lb:= Cvmax 200% fA⋅

Gpv

⋅:= Cvmax 77.475gal

min psi⋅=

From Fig. C-10.1, page 532, a 3-in valve is required. Cvmax 110gal

min psi⋅:=

Valve gain: KvCvmax

100%CO

pvG

⋅:= Kv 1.42gal

min %CO⋅=

The valve fails closed (air-to-open) to prevent overflowing the reactors on loss of instrument power.

(b) Model of the reactors.

Assumptions:Perfectly mixed reactors•Constant volume, temperatures, and physical properties.•Negligible transportation lags.•

A i= =

Total mass balance: f t( )⋅ A fA t( )⋅ i fi t( )⋅+= 1 eqn. 1 unk. (f)


Problem 6-22. Control of reactors in series.

LT LC

AT

AC

fi(t)

cAi(t)

cA1(t)

cA2set(t)

V

V

cA2(t)

DA

fA(t)

Problem Data:

V 500gal:= k1 0.25min 1−:=

A 2lbmole

gal:= k2 0.50min 1−

:=

MWA 25lb

lbmole:=

Control valve, linear, sized for 100% overcapacity.

pv 10psi:=

Analyzer transmitter:

cAL 0.05lbmole

gal:=

cAH 0.5lbmole

gal:= τT 0.5min:=

Design conditions: cAi 0.8lbmole

gal:=

τ2d CA2 t( )⋅

dt⋅ CA2 t( )+ K4 FA t( )⋅ K4 Fi t( )⋅+ K5 CA1 t( )⋅+= CA2 0( ) 0=

where τ1V

f V k1⋅+:= K1

A cA1−

f V k1⋅+:= K2

cAi cA1−

f V k1⋅+:= K3

fif V k1⋅+

:=

τ2V

f V k2⋅+:= K4

cA1 cA2−

f V k2⋅+:= K5

ff V k2⋅+

:=

τ1 2.222 min= K1 0.006123lbmole min⋅

gal2= K2 0.00079

lbmole min⋅

gal2= K3 0.222=

τ2 1.429 min= K4 0.00127lbmole min⋅

gal2= K5 0.286=

Laplace transform:

CA1 s( )K1 FA s( )⋅ K2 Fi s( )⋅+ K3 CAi s( )⋅+

τ1 s⋅ 1+=

Reactant balances:

Reactor 1: Vd cA1 t( )⋅

dt⋅ A fA t( )⋅ fi t( ) cAi t( )⋅+ V k1⋅ cA1 t( )⋅− f t( ) cA1 t( )⋅−=

2 eqns. 2 unks. (cA1)Reactor 2:

Vd cA2 t( )⋅

dt⋅ f t( ) cA1 t( )⋅ f t( ) cA2 t( )⋅− V k2⋅ cA2 t( )⋅−=

3 eqns. 3 unks. (cA2)

At the initial steady state: f fA fi+:= cA1A fA⋅ fi cAi⋅+

f V k1⋅+:=

f 100galmin

=

cA2f cA1⋅

f V k2⋅+:= cA1 0.622

lbmolegal

= cA2 0.178lbmole

gal=

(c) Linearize the model equations and obtain the block diagram.


Vd CA1 t( )⋅

dt⋅ A FA t( )⋅ cAi Fi t( )⋅+ fi CAi t( )⋅+ V k1⋅ f+( )CA1 t( )− cA1 FA t( ) Fi t( )+( )⋅−=

Vd CA2 t( )⋅

dt⋅ cA1 cA2−( ) FA t( ) Fi t( )+( )⋅ f CA1 t( )⋅+ f V k2⋅+( )CA2 t( )−=

Rearrange in the standard first-order form:

τ1d CA1 t( )⋅

dt⋅ CA1 t( )+ K1 FA t( )⋅ K2 Fi t( )⋅+ K3 CAi t( )⋅+= CA1 0( ) 0=

CA2 s( )

CAi s( )

G3 s( )

1 H s( ) Gc s( )⋅ Gv s( )⋅ G1 s( )⋅+=

CA2 s( )

Fi s( )

G2 s( )

1 H s( ) Gc s( )⋅ Gv s( )⋅ G1 s( )⋅+=

CA2 s( )

CA2set s( )⋅

Ksp Gc s( )⋅ Gv s( )⋅ G1 s( )⋅

1 H s( ) Gc s( )⋅ Gv s( )⋅ G1 s( )⋅+=

Ksp KT:=(d) Closed-loop transfer functions.

KT 222%TO gal⋅

lbmole=KT

100%TOcAH cAL−

:=H s( )KT

τT s⋅ 1+=Transmitter:Gv s( ) Kv=Control valve:

Ksp G

c(s) G

v(s)

CA2

set(s)

G2(s)

H(s)

CA2

(s)M(s)E(s)G1(s)

FA(s)

Fi(s)

+

- ++

G3(s)CAi(s)

+

Block diagram:

G3 s( )K5 K3⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=G2 s( )

K4 τ1 s⋅ 1+( )⋅ K5 K2⋅+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=G1 s( )

K4 τ1 s⋅ 1+( )⋅ K5 K1⋅+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

where

CA2 s( ) G1 s( ) FA s( )⋅ G2 s( ) Fi s( )⋅+ G3 s( ) CAi s( )⋅+=

CA2 s( )K4

τ2 s⋅ 1+FA s( ) Fi s( )+( )

K5τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅

K1 FA s( )⋅ K2 Fi s( )⋅+ K3 CAi s( )⋅+( )+=

Combine to obtain over-all transfer function:

CA2 s( )K4 FA s( )⋅ K4 Fi s( )⋅+ K5 CA1 s( )⋅+

τ2 s⋅ 1+=



is unlawful.

The imaginary value of the ultimate frequency means that there is no ultimate gain, that is, the loop is stable for all positive values of the gain. This is because the net order of the process is one--two poles and one zero--and, with the lag in the transmitter, the order is two. An order of at least three is needed to have an ultimate gain.

The controller is reverse acting: increasing reactants composition decreases the controller output, closing the reactant feed valve; this decreases the reactants flow and the composition of reactants.

ωu 1.021i min 1−=

ωuτD Kp τC⋅−

τD τB⋅ τA Kp⋅−:=τA− ωu

3⋅ τC τD

τB ωu2

⋅ 1−

Kp⋅+ ωu+ 0=

Imaginary part:

KcuτB ωu

2⋅ 1−

Kp=τB− ωu

2⋅ 1+ Kp Kcu⋅+ 0=Real part:

τA− ωu3

⋅ i τB ωu2

⋅− τC τD Kcu⋅+( )ωu i⋅+ 1+ Kp Kcu⋅+ 0 0 i⋅+=

Substitute s = ωui at Kc = Kcu

Kp 0.953=τD 0.89 min=

τC 4.151 min=τB 5 min2=Kp KT Kv⋅ K4 K5 K1⋅+( )⋅:=τD KT Kv⋅ K4⋅ τ1⋅:=

τA 1.587 min3=τC τ1 τ2+ τT+:=τB τT τ1⋅ τT τ2⋅+ τ1 τ2⋅+:=τA τT τ1⋅ τ2⋅:=

Let

1KT

τT s⋅ 1+Kc⋅ Kv⋅

K4 τ1 s⋅ 1+( )⋅ K5 K1⋅+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅⋅+ 0=Characteristic equation:

Gc s( ) Kc=Proportional controller:(e) Ultimate gain and period of the loop.

Substitute and simplify:

3 eqns. 3 unks.f f1 f2+=f1⋅ f2⋅+ f⋅=Total mass balance on tank 1:

2 eqns. 3 unks. (T4)

V2⋅ cp⋅d T4 t( )⋅

dt⋅ f ⋅ cp⋅ T3 t( ) Tref−( )⋅ ws t( )⋅+ f ⋅ cp⋅ T4 t( ) Tref−( )⋅−=

1 eqn. 2 unks. (f, T3)f− ⋅ cp⋅ T3 t( ) Tref−( )⋅Tank 2:

V1⋅ cv⋅d T3 t( )⋅

dt⋅ f1⋅ cp⋅ T1 t( ) Tref−( )⋅ f2⋅ cp⋅ T2 t( ) Tref−( )⋅+ U A⋅ Tc1 t( ) T3 t( )−( )⋅+=

Energy balances, Tank 1:

Assumptions:Constant volumes in the tanks•Each tank is perfectly mixed•Negligible heat losses•Heating coil temperature Tc1 is uniform (high flow of heating fluid)•Constant and uniform densities and specific heats•Negligible transportation lag.•

Model the process.

TT

TC

f2

T2(t)

T4set(t)

f1

T1(t)T

c1(t)

T3(t)

T4(t)

Steam

Cond.f

f

Problem 6-23. Temperature control of two heaters in series.


Gv s( )Kv

τv s⋅ 1+=Control valve:

G4 s( )K4

τ2 s⋅ 1+=

G3 s( )K3

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=G2 s( )

K2τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅

=G1 s( )K1

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

4 s( ) G1 s( ) 1 s( ) G2 s( ) 2 s( )+ G3 s( ) c1 s( )+ G4 s( ) W s( )⋅+=

4 s( )3 s( ) K4 W s( )⋅+

τ2 s⋅ 1+=3 s( )

K1 1 S( )⋅ K2 2 s( )⋅+ K3 c1 s( )⋅+

τ1 s⋅ 1+=

Laplace transform and solve for the outputs:

K4 f ⋅ cp⋅=τ2

V2⋅ cv⋅

f ⋅ cp⋅=

K2f2⋅ cp⋅

f ⋅ cp⋅ U A⋅+=K1

f1⋅ cp⋅

f ⋅ cp⋅ U A⋅+:=τ1

V1⋅ cv⋅

f ⋅ cp⋅ U A⋅+=where K3

U A⋅f ⋅ cp⋅ U A⋅+

:=

4 0( ) 0=τ2d 4 t( )⋅

dt⋅ 4 t( )+ 3 t( ) K4 W s( )⋅+=

3 0( ) 0=τ1d 3 t( )⋅

dt⋅ 3 t( )+ K1 1 t( )⋅ K2 2 t( )⋅+ K3 c1 t( )⋅+=

Rearrange into first-order standard form:

V2⋅ cv⋅d 4 t( )⋅

dt⋅ f ⋅ cp⋅ 3 t( )⋅ W s( )⋅+ f ⋅ cp⋅ 4 s( )⋅−=

V1⋅ cv⋅d 3 t( )⋅

dt⋅ f1⋅ cp⋅ 1 t( )⋅ f2⋅ cp⋅ 2 t( )⋅+ U A⋅ c1 t( )⋅+ f ⋅ cp⋅ U A⋅+( ) 3 t( )−=

Linearize and express in terms of the deviation variables:

V2⋅ cv⋅d T4 t( )⋅

dt⋅ f ⋅ cp⋅ T3 t( ) T4 t( )−( )⋅ ws t( )⋅+=

V1⋅ cv⋅d T3 t( )⋅

dt⋅ f1⋅ cp⋅ T1 t( ) T3 t( )−( )⋅ f2⋅ cp⋅ T2 t( ) T3 t( )−( )⋅+ U A⋅ Tc1 t( ) T3 t( )−( )⋅+=

Temperature transmitter TT: H s( )KT

τT s⋅ 1+= KT

100%TOTH TL−

= Ksp KT=

Block Diagram:

Ksp Gc(s) G

v(s)

'4set(s)

G1(s)

H(s)

M(s)E(s)G4(s)

W(s)+

- ++

G2(s)

+

G3(s)

'2(s)

'c1(s)

'1(s)

'4(s)+

Characteristic equation of the loop.

1 H s( ) Gc s( )⋅ Gv s( )⋅ G4 s( )⋅+ 1KT

τT s⋅ 1+Gc s( )

Kvτv s⋅ 1+

K4τ2 s⋅ 1+⋅+= 0=

τT τv⋅ τ2⋅ s3⋅ τT τv⋅ τT τ2⋅+ τv τ2⋅+( ) s2

⋅+ τT τv+ τ2+( ) s⋅+ 1+ KT Kv⋅ K4⋅ Gc s( )⋅+ 0=



is unlawful.

(20 BWG) L 974ft:= wm 0.178lbft

:= cpm 0.12BTU

lb degF⋅:=

Ao π Do⋅ L⋅:= Ao 127.5 ft2= CM wm L⋅ cpm⋅:= CM 20.805BTUdegF

=

Level transmitter: hL 7ft:= hH 10ft:= τLT 0.01min:=

Temperature transmitter: TL 100degF:= TH 300degF:= τTT 0.5min:=

(a) Size control valves for 50% overcapacity.Assume

Perfectly mixed tank•Constant and uniform densities, specific heats, and steam latent heat•The metal is at the same temperature as the condensing steam•Negligible heat losses and transpotation lags•Constant pressure drop across the steam valve.•


Problem 6-24. Temperature and level control of oil heater.

LTLC

vp2(t)

T3set(t)

fs(t) T1(t)

T3(t)

hset(t)

Steam

Condensate

3 ft

h(t)

TT

TC

p1(t)

p2 = 40 psia

5 ft

N2

AOAO

Tp3(t)

Problem Data: 53lb

ft3:= cp 0.45

BTUlb degF⋅

:= T1 70degF:= T3 200degF:= D 3ft:=

Steam, saturated at 115 psig (130 psia). from steam tables: Ts1 347degF:= s 873BTU

lb:=

p2 40psia:= p1 45psig:= p3 15psig:= U 136BTU

hr ft2degF⋅:= f1 100

galmin

:=

Heating coil: Do 0.5in:=

Cv2max 37.7gal

min psi⋅=

From Fig. C-10.1, page 532, a 2-in valve is required. Cv2max 46gal

min psi⋅:=

Size steam valve: ps1 130psia:= G1829

:= G 0.621= Assume Cf 0.9:=

y1.63Cf

ps1 ps−

ps1⋅:= y 0.42= fs

ws18lb

380scf60min

hr⋅:= fs 60138

scfhr

=

Cvsmax150 %⋅ fs⋅ G Ts1 460 R⋅+( )⋅⋅

Cf ps1⋅ y 0.148 y3⋅−( )⋅

gal hr⋅

836 min⋅ scf⋅

psiaR

⋅:= Cvsmax 50.4gal

min psi⋅=

From Fig. C-10.1, page 532, a 3-in valve is required. Cvsmax 110gal

min psi⋅:=

(b) Block diagram of the level control loop.

Mass balance: π D2⋅

47.48gal

ft3⋅ ⋅

d h t( )⋅

dt⋅ f1 t( )⋅ f3 t( )⋅−= 1 eqn. 3 unks. (h, f1, f3)

Valves: f1 t( ) Cv1max vp1 t( )⋅1

Gfp1 t( ) p2− g⋅ h t( ) 5 ft⋅−( )⋅

ft12 in⋅

2⋅−⋅⋅=

At the initial steady state: f3 f1:= f1− ⋅ cp⋅ T3 T1−( )⋅ ws s⋅+ 0=

wsf1⋅ cp⋅ T3 T1−( )⋅

s:= ws 47.5

lbmin

=

ws s⋅ U Ao⋅ Ts T3−( )⋅− 0= Ts T3ws s⋅

U Ao⋅+:= Ts 343.4 degF= ps 123psia:=

(steam tables)Size inlet oil valve:

Assume initially the level is at 50% of the level transmitter range: hhH hL+

2:=

pv1 p1 14.7psia+ p2− g⋅ h 5ft−( )⋅ft

12in

2⋅−:= h 8.5 ft= pv1 18.41 psi=

Gfft3⋅

62.4lb:= Cv1max 150% f1⋅

Gfpv1

⋅:= Cv1max 32.2gal

min psi⋅=

From Fig. C-10.1, page 532, a 3-in valve is required. Cv1max 46gal

min psi⋅:=

Size exit oil valve: pv2 p2 g⋅ h⋅ft

12in

2⋅+ p3 14.7psia+( )−:= pv2 13.43 psi=

Cv2max 150% f3⋅Gfpv2

⋅:=

a1 2.716gal

min psi⋅= a2 3.723

galmin psi⋅

= a3 1gal

min ft⋅= a4 1.37

galmin ft⋅

=

Rearrange model equation in standrd first-oder form:

τd H t( )⋅

dt⋅ H t( )+ K1 VP1 t( )⋅ K2 VP2 t( )⋅− K3 P1 t( )⋅+ K4 P3 t( )⋅+= H 0( ) 0=

where

τπD2

47.48gal

ft3⋅

1a3 a4+

:= K1f1max

a3 a4+:= K2

f3maxa3 a4+

:= K3a1

a3 a4+:= K4

a2a3 a4+

:=

τ 22.31 min= K1 90.4 ft= K2 77.2 ft= K3 1.146ftpsi

= K4 1.571ftpsi

=

Laplace transform:H s( )

1τ s⋅ 1+

K1 VP1 s( )⋅ K2 VP2 s( )⋅− K3 P1 s( )⋅+ K4 P3 s( )⋅+( )=

2 eqns. 3 unks.

f3 t( ) Cv2max vp2 t( )⋅1

Gfp2 p3 t( )− g⋅ h t( )⋅

ft12 in⋅

2⋅+⋅⋅=

3 eqns. 3 unks.

f1max Cv1maxpv1Gf

⋅:= f3max Cv2maxpv2Gf

⋅:= f1max 214.2galmin

= f3max 182.9galmin

=

Linearize the model equations and express in terms of deviation variables:

π D2⋅

47.48gal

ft3⋅

d H t( )⋅

dt⋅ f1max VP1 t( )⋅ f3max VP2 t( )⋅− a1 P1 t( )⋅+ a3 H t( )⋅− a2 P3 t( )⋅+ a4 H t( )⋅−=

where a1f1 t( )

p1= a2

f3 t( )−

p3= a3

f1 t( )−

h= a4

f3 t( )

h=

vp1f1

f1max:= vp2

f3f3max

:= vp1 0.467= vp2 0.547=

a1 Cv1max vp1⋅12⋅

1Gf pv1⋅

⋅:= a2 Cv2max− vp2⋅12⋅

1Gf pv2⋅

⋅ 1−( )⋅:=

a3 Cv1max− vp1⋅12⋅

1Gf pv1⋅

⋅ − g⋅ft2

144 in2⋅

⋅⋅:= a4 Cv2max vp2⋅12⋅

1Gf pv2⋅

⋅ g⋅ft2

144 in2⋅

⋅⋅:=

2 eqns. 3 nks. (ws)

2 eqns. 3 unks. (ws)

CMd Ts t( )⋅

dt⋅ s ws t( )⋅ U Ao⋅ Ts t( ) T3 t( )−( )⋅−=Energy balance on steam chest:

1 eqn. 2 unks. (T3, Ts)f3 t( )− ⋅ cp⋅ T3 t( ) Tref−( )⋅

πD2

47.48gal

ft3⋅ cv⋅

ddt⋅ h t( ) T3 t( )⋅( ) f1 t( )⋅ cp⋅ T1 t( ) Tref−( )⋅ U Ao⋅ Ts t( ) T3 t( )−( )⋅+=

Energy balance on tank:

(c) Block diagram and characteristic equation of the temperature control loop.

KLc

K1R(s)

K2

HLT(s)

M1(s)E(s)

Js + 1+

- +-

K3

+

K4

P1(s)

P3(s)

VP2(s)

H(s)+

1001

Block diagram of the level control loop:

GLc s( ) KLc=Proportional controller, LC: The controller is reverse acting: increasing level decreases the controller output closing the valve and decreasing the inlet flow.

KLT 33.33%TO

ft=KLT

100%TOhH hL−

:=HLT s( )KLT

τLT s⋅ 1+=Level transmitter LT:

VP1 s( )1

100%COM1 s( )=Level control valve:

τ1d 3 t( )⋅

dt⋅ 3 t( )+ K1− F1 t( )⋅ K2 1 t( )⋅+ K3 s t( )⋅+=

τ2d s t( )⋅

dt⋅ s t( )+ K4 Ws t( )⋅ 3 t( )+=

where τ1V⋅ cv⋅

f1⋅ cp⋅ U Ao⋅+:= K1

− cp⋅ T1 T3−( )⋅

f1⋅ cp⋅ U Ao⋅+:= K2

f ⋅ cp⋅

f ⋅ cp⋅ U Ao⋅+:= K3

U Ao⋅

f ⋅ cp⋅ U Ao⋅+:=

τ2CM

U Ao⋅:= K4

sU Ao⋅

:=τ1 2.357 min= τ2 0.072 min= K1 0.682

degF min⋅

gal=

K2 0.525= K3 0.475= K4 3.021degF min⋅

lb=

Laplace transform: 3 s( )K1− F1 s( )⋅ K2 1 s( )⋅+ K3 s s( )⋅+

τ1 s⋅ 1+= s s( )

K4 Ws s( )⋅ 3 s( )+

τ2 s⋅ 1+=

Steam valve:

ws t( )836 scf⋅ min⋅

hr gal⋅

Rpsia

⋅hr

60 min⋅⋅

18 lb⋅380 scf⋅⋅

Cvsmax vps t( )⋅ Cf⋅ ps1⋅ y 0.148 y3⋅−( )⋅

G Ts1 460R+( )⋅⋅=

3 eqns. 3 unks.

Let wsmax836 scf⋅ min⋅

hr gal⋅

Rpsia

⋅hr

60 min⋅⋅

18 lb⋅380 scf⋅⋅

Cvsmax Cf⋅ ps1⋅ y 0.148 y3⋅−( )⋅

G Ts1 460R+( )⋅⋅:=

wsmax 155.3lb

min=

Substitute mass balance into energy balance and simplify:

π D2⋅

47.48 gal⋅

ft3⋅ ⋅ cv⋅ h t( )⋅

d T3 t( )⋅

dt⋅ f1 t( ) cp⋅ T1 t( ) T3 t( )−( )⋅ U Ao⋅ Ts t( ) T3 t( )−( )⋅+=

ws t( ) wsmax vps t( )⋅=

Linearize the equations: Let VπD2

47.48gal

ft3⋅ h⋅:= V 449.4 gal= Assume cv cp:=

V⋅ cv⋅d 3 t( )⋅

dt⋅ cp⋅ T1 T3−( )⋅ F1 t( ) f1⋅ cp⋅ 1 t( ) 3 t( )−( )⋅+ U Ao⋅ s t( ) 3 t( )−( )⋅+=

CMd s t( )⋅

dt⋅ s Ws t( )⋅ U Ao⋅ s t( ) 3 t( )−(⋅−= 3 0( ) 0= s 0( ) 0=

Rearrange into standard first-order form:

Characteristic equation:

Ksp GTc

(s)

f1max

'3set(s)

G2(s)

HTT(s)

Ms(s)E(s)G3(s)

Ws(s)+

- ++

G1(s)

-a1

F1(s)P1(s)

'1(s)

'3(s)

+

100

M1(s)

a3

H(s)

100

wsmax

+

-

Block diagram of the temperature control loop:

a3 1gal

min ft⋅=a1 2.716

galmin psi⋅

=

f1max100%CO

2.142gal

min %CO⋅=F1 s( )

f1max100%CO

M1 s( ) a1 P1 s( )⋅+ a3 H s( )⋅−=Inlet flow:

GTc s( ) KTc 11

τ IT s⋅+⋅

τDT s⋅ 1+

αD τDT⋅ s⋅ 1+=PID controller:

KTT 0.5%TOdegF

=KTT100%TOTH TL−

:=HTT s( )KTT

τTT s⋅ 1+=

Temperature transmitter TT:

wsmax100%CO

1.553lb

min %CO⋅=Ws s( )

wsmax100%CO

Ms s( )=Steam control valve:

G3 s( )K3 K4⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K3−=

G2 s( )K2 τ2 s⋅ 1+( )⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K3−=G1 s( )

K1 τ2 s⋅ 1+( )⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K3−=where

3 s( ) G1 s( )− F1 s( )⋅ G2 s( ) 1 s( )+ G3 s( ) Ws s( )⋅+=Combine:



is unlawful.

Compare these results with the simulation of this process in Problem 13-24.

Kcu 38.9%CO%TO

=KcuK3 1− τB ωu

2⋅+

Kp:=

τB− ωu2

⋅ 1+ K3− Kp Kcu⋅+ 0=Real part:

Tu 1.069 min=Tu2πωu

:=ωuτCτA

:=τA− ωu3

⋅ τC ωu⋅+ 0=Imaginary part:

Kp 1.116%TO%CO

=

τC 2.692 min=τB 1.384 min2=τA 0.085 min3

=Kp KTTwsmax

100%CO⋅ K3 K4⋅:=

τC τTT 1 K3−( )⋅ τ1+ τ2+:=τB τTT τ1⋅ τTT τ2⋅+ τ1 τ2⋅+:=τA τTT τ1⋅ τ2⋅:=where

τA− ωu3

⋅ i⋅ τB ωu2

⋅− τC ωu⋅ i⋅+ 1+ K3− Kp Kcu⋅+ 0 0 i⋅+=Substitute s = ωui at KTc = Kcu:

GTc s( ) KTc=Proportional controller:

(d) Ultimate gain nad period of the temperature control loop.

1 HTT s( ) GTc s( )wsmax

100%CO⋅ G3 s( )+ 1

KTTτTT s⋅ 1+

GTc s( )wsmax

100%CO⋅

K3 K4⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K3−+= 0=

3 eqns. 4 unks.f1 t( ) Cv1 w g⋅ h1 t( )⋅ft

12in

2⋅⋅= kv1 h1 t( )⋅=Valves:

2 eqns. 4 unks. (h2, f2)π D2

2⋅

4⋅

d h2 t( )⋅

dt⋅ f1 t( )⋅ f2 t( )⋅−=tank 2:

1 eqn. 2 unks. (h1, f1)

π D12

⋅

4⋅

d h1 t( )⋅

dt⋅ fi t( )⋅ f3 t( )⋅+ fo t( )⋅− f1 t( )⋅−=Mass balance, tank 1:

Assume constant and uniform densities•constant valve positions and inlet valve pressures•

(a) Block diagram and transfer functions of level control loop.

HLT s( ) KT= 100%TOhmax

=Level transmitter LT has negligible time constant:

Kv Cv3pv3Gf

⋅=Gv s( )Kv

τv s⋅ 1+=

Linear control valve with constant pressure drop and time constant:

Fo s( )fmax

100%CO1

τp s⋅ 1+M1 s( )=Linear pump with a time constant:

LT

LC

f1(t)

.fi(t)

h2(t)

.f3(t)

fo(t)

h1(t)

f2(t).m1(t)

Problem 6-25. Level control of two tanks in series.


KLc

Gv(s)

R(s)

KT

M1(s)E(s)

(J1s+1)(J2s+1)+

- +-

.fmax

+

M1(s) Fo(s)

H2(s)K1K2

100(Jps+1)

Fi(s)

F3(s)


H2 s( )K2

τ2 s⋅ 1+H1 s( )=

K1 K2⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅Fi s( ) F3 s( )+ Fo s( )−( )=

H1 s( )K1

τ1 s⋅ 1+Fi s( ) F3 s( )+ Fo s( )−( )=Laplace transform:

K2kv1kv2

h2h1

⋅=τ2π D2

2⋅ h2⋅

2 kv2⋅=K1

2 h1⋅

kv1=τ1

π D12

⋅ h1⋅

2 kv1⋅=where

τ2d H2 t( )⋅

dt⋅ H2 t( )+ K2 H1 t( )⋅=

τ1d H1 t( )⋅

dt⋅ H1 t( )+ K1 Fi t( )⋅ K1 F3 t( )⋅+ K1 Fo t( )⋅−=

Rearrangle into the standard first-order form:

H2 0( ) 0=π D2

2⋅

4

d H2 t( )⋅

dt⋅

kv1

2 h1⋅H1 t( )

kv2

2 h2⋅H2 t( )−=

H1 0( ) 0=π D1

2⋅

4

d H1 t( )⋅

dt⋅ Fi t( ) F3 t( )+ Fo t( )−

kv1

2 h1⋅H1 t( )−=

Linearize the equations and express in terms of deviation variables::

4 eqns. 4 unks.f2 t( ) Cv2 w g⋅ h2 t( )⋅ft

12in

2⋅⋅= kv2 h2 t( )⋅=

(b) Characteristic equation and ultimate gain and period.

1 KT KLC⋅Kv

τv s⋅ 1+⋅

K1 K2⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅+ 0=

Substitute s = ωui at KLC = K.cu: Let Kp KT Kv⋅ K1⋅ K2⋅=100 %TO⋅

hmax

f3max100 CO⋅⋅

2 h2⋅

kv2⋅=

τv− τ1⋅ τ2⋅ ωu3

⋅ i τv τ1⋅ τv τ2⋅+ τ1 τ2⋅+( )ωu2

− τv τ1+ τ2+( ) ωu⋅ i⋅+ 1+ Kp Kcu⋅+ 0 0 i⋅+=

Imaginary part: τv− τ1⋅ τ2⋅ ωu3

⋅ τv τ1+ τ2+( )ωu+ 0= ωuτv τ1+ τ2+

τv τ1⋅ τ2⋅= Tu

2πωu

=

Real part: τv τ1⋅ τv τ2⋅+ τ1 τ2⋅+( )− ωu2

1+ Kp Kcu⋅+ 0= Kcuτv τ1⋅ τv τ2⋅+ τ1 τ2⋅+( )ωu

21−

Kp=



is unlawful.

Substitute, linearize the equations, and express in terms of linearized variables:

s 2.635 10 3− lbmolescf

=s14.7psiaRg 520⋅ R

:=

5 eqns. 5 unks.t( )p t( )Rg T⋅

lbmole

ft3=Ideal gas law:

4 eqns. 5 unks.y t( )1.63Cf

p t( ) p2−

p t( )⋅=

3 eqns. 5 unks. (y, p)

f3 t( )836scf min⋅

hr gal⋅

Rpsia

⋅hr

60minCv3 Cf⋅

p t( )

G t( ) T⋅⋅ y t( ) 0.148 y t( )3

⋅−( )=Exit valve:

2 eqns. 3 unks. (x3)

Vddt⋅ t( ) x3 t( )⋅( ) s f1 t( )⋅ x1 t( )⋅ s f2 t( )⋅ x2 t( )⋅+ s f3 t( )⋅ x3 t( )⋅−=

Methane mole balance:

1 eqn. 2 unks. ( , f3)Vd t( )⋅

dt⋅ s f1 t( )⋅ s f2 t( )⋅+ s f3 t( )⋅−=Total mole balance:

AssumeGas obeys ideal gas law•Temperature is constant•Perfectly mixed tank•

(a) Block diagram and transfer functions.

Gc s( ) Kc 11τ I s⋅

+⋅=

Controller is PI:

Sensor transmitter, lag τT, range hvL to hvH.

Fan driver, linear, lag τF, range 0 to f2max.

G t( ) a b x3 t( )⋅+=

hv t( ) c g x3 t( )⋅+=

Problem Data:

AT

AChvset(t)

Natural gas

Waste gas

f2(t) x2(t)

f1(t) x1(t)

f3(t) x3(t)

p2p(t)

hv

Problem 6-26. Control of heating value of fuel stream.


KFf2max

100%CO=F2 s( )

KFτF s⋅ 1+

M s( )=Variable speed fan:

HV s( ) g X3 s( )⋅=HV t( ) g X3 t( )⋅=Linearize the eating value equation:

P s( )1

τ2 s⋅ 1+

1a1

F1 s( )1a1

F2 s( )+a2a3

X3 s( )+=

X3 s( )1

τ1 s⋅ 1+K1 X1 s( )⋅ K2 X2 s( )⋅+ K3 F1 s( )⋅− K4 F2 s( )⋅+( )=Laplace transform:

τ2V

Rg T⋅ s⋅ a1⋅=K4

x2 x3−

f3=K3

x3 x1−

f3=K2

f2f3

=K1f1f3

=τ1V⋅f3 s⋅

=

τ2d P t( )⋅

dt⋅ P t( )+

1a1

F1 t( )⋅1a1

F2 t( )⋅+a2a1

X3 t( )+=

τ1d X3 t( )⋅

dt⋅ X3 t( )+ K1 X1 t( )⋅ K2 X2 t( )⋅+ K3 F1 t( )⋅− K4 F2 t( )⋅+=

Rearrange into standrad first-order form:

a2836− scf min⋅

hr gal⋅

Rpsia

⋅hr

60minCv3 Cf⋅

p

T⋅ y 0.148 y3

⋅−( ) b−

2 a b x3⋅+( )1.5=

a1836scf min⋅

hr gal⋅

Rpsia

⋅hr

60min

Cv3 Cf⋅

G T⋅y 0.148 y3

⋅− p 1 3 0.148⋅ y2⋅−( )⋅

1.63Cf 2⋅

p p2−

p

0.5− p2

p2+=

P 0( ) 0=V

Rg T⋅ s⋅

d P t( )⋅

dt⋅ F1 t( ) F2 t( )+ a1 P t( )⋅− a2 X3 t( )⋅+=

x3 0( ) 0=V⋅

s

d x3 t( )⋅

dt⋅ f1 X1 t( )⋅ f2 X2 t( )⋅+ f3 X3 t( )⋅− x1 x3−( )F1 t( )+ x2 x3−( ) F2 t( )⋅+=

Combine total and methane balances, and substitute valve equation:

a2− f3 t( )⋅

x3=a1

f3 t( )⋅

p=

where F3 t( ) a1 P t( )⋅ a2 X3 t( )⋅−=

V x3⋅

Rg T⋅d P t( )⋅

dt⋅

V P⋅Rg T⋅

d x3 t( )⋅

dt⋅+ s f1 X1 t( )⋅ f2 X2 t( )⋅+ f3 X3 t( )⋅− x1 F1 t( )⋅+ x2 F2 t( )⋅+ x3 F3 t( )⋅−( )⋅=

VRg T⋅

d P t( )⋅

dt⋅ s F1 t( )⋅ s F2 t( )⋅+ s F3 t( )⋅−=

Heating value transmitter: H s( )KT

τT s⋅ 1+= KT

100%TOhvH hvL−

= Ksp KT=

Block diagram of the heating value control

Ksp

KFK4

HVset(s)

K3

H(s)

M(s)E(s)

J1s + 1+

-+-

K2

+

K1

X2(s)

X1(s)

F1(s)

HV(s)+gG

c(s)

JFs + 1

The controller must be reverse acting (negative gain): increasing heating value decreases the controller output; this decreases the fan speed and the flow of natural gas, decreasing the heating value.

(b) Characteristic equation of the loop.

1KT

τT s⋅ 1+Gc s( )

KFτF s⋅ 1+⋅

K4τ1 s⋅ 1+

+ 0=

τT s⋅ 1+( ) τF s⋅ 1+( )⋅ τ1 s⋅ 1+( )⋅ KT Kc⋅ 11τ I s⋅

+⋅ KF⋅ K4⋅+ 0=



is unlawful.

τB 5.16 min2

= τC 6 min=

Imaginary part: τA− ωu3

⋅ τC ωu⋅+ 0= ωu

τC

τA:= Tu

2π

ωu:= Tu 2.29 min=

Real part: τB− ωu2

⋅ 1+ KKcu+ 0= Kcu1

KτB ωu

2⋅ 1−:= Kcu 15.1

%CO

%TO=

(a) Quarter decay ratio tuning parameters for a proportional controller.

Gc s( ) Kc= From Table 7-1.1: Kc

Kcu

2:= Kc 7.5

%CO

%TO=

(b) Quarter decay ratio tuning parameters for a proportional-integral controller.

Gc s( ) Kc 11

τ I s⋅+⋅=

From Table 7-1.1: Kc

Kcu

2.2:= τI

Tu

1.2:= Kc 6.9

%CO

%TO=

Smith & Corripio, 3rd edition %TO %:= %CO %:= Kd K:=

Problem 7-1. Feedback control of a third-order process.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

U(s)

C(s)+ +

+-

G1 s( )K

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ τ3 s⋅ 1+( )⋅=

K 2.5%TO

%CO:= τ1 5min:= τ2 0.8min:= τ3 0.2min:=

Characteristic equation of the loop: 1 Gc s( )K

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ τ3 s⋅ 1+( )⋅⋅+ 0=

Ultimate gain and period: Gc s( ) Kcu= s ωu i⋅=

τA− ωu3

⋅ i τB ωu2

⋅− τC ωu⋅ i⋅+ 1+ KKcu+ 0 0 i⋅+=

where τA τ1 τ2⋅ τ3⋅:= τB τ1 τ2⋅ τ1 τ3⋅+ τ2 τ3⋅+:= τC τ1 τ2+ τ3+:= τA 0.8 min3

=

τI 1.9 min=

(c) Quarter decay ratio tuning parameters for a series PID controller.

Gc s( ) Kc 11

τ I s⋅+⋅ τD s⋅ 1+( )= From Table 7-1.1: Kc

Kcu

1.7:= Kc 8.9

%CO

%TO=

τI

Tu

2:= τD

Tu

8:= τI 1.1 min=

τD 0.29 min=


T2π

2.0282min−

:=

T 3.098 min=

Damping ratio: ζ−

1 ζ2

−

0.2808−

2.0282= 2.0282 ζ⋅( )2 0.2808

21 ζ

2−( )= ζ

0.2808

0.28082

2.02822

+

:=

Decay ratio: e0.2808− min

1−T⋅

0.419= ζ 0.137=

The decay ratio is higher than one fourth.

b) PI Controller Gc s( ) Kc 11

τ I s⋅+⋅= Kc 6.9

%CO

%TO:= τI 1.9min:=

Roots of: τA s4

⋅ τB s3

⋅+ τC s2

⋅+ 1 K Kc⋅+( ) s⋅+K Kc⋅

τ I+ 0=


Problem 7-2. Feedback control loop of Problem 7-1.From the solution to Problem 7-1, the characteristic equation is:

τA s3

⋅ τB s2

⋅+ τC s⋅+ 1+ Gc s( ) K⋅+ 0=

with τA 0.8 min3

= τB 5.16 min2

= τC 6 min= K 2.5%TO

%CO=

(a) Roots of the characteritic equation, dominant roots, damping ratio and decay ratio.

a) Proportional controller Gc s( ) Kc=Kc 7.5

%CO

%TO:=

Find the roots: Dominant roots are the complex conjugate pair.

polyroots

1 K Kc⋅+

τC min1−

⋅

τB min2−

⋅

τA min3−

⋅

5.8883−

0.2808− 2.0282i−

0.2808− 2.0282i+

=Period:

The linear loop is simulated with one Simulink transfer function block to simulate the process and another block to simulate the controller. The controller block, Gc(s), is obtained from

(b) Simulate tye loop and plot responses to a unit step change in set point.

This close to the desired decay ratio of one fourth (0.25).

e0.483− min

1−T⋅

0.278=Decay ratio:0.483

0.4832

2.3722

+

0.2=Damping ratio:

T 2.649 min=

polyroots

K Kc⋅

τ Imin

1 K Kc⋅ 1τD

τI+⋅+

τC K Kc⋅ τD⋅+( )min1−

τB min2−

⋅

τA min3−

⋅

4.531−

0.952−

0.483− 2.372i−

0.483− 2.372i+

=

T2 π⋅

2.372min1−

:=Period:

The dominant roots are the complex conjugate roots.

τA s4

⋅ τB s3

⋅+ τC K Kc⋅ τD⋅+( ) s2

⋅+ 1 K Kc⋅ 1τD

τ I+⋅+ s⋅+

K Kc⋅

τI+ 0=Roots of:

τD 0.29 min=τI 1.1min:=

Gc s( ) Kc 11

τ I s⋅+⋅ τD s⋅ 1+( )=c) Series PID controller: Kc 8.9

%CO

%TO:=

This is a very undamped response; extremely high decay ratio.

e0.06− min

1−T⋅

0.818=Decay ratio:0.06

0.062

1.8812

+

0.032=Damping ratio:

T 3.34 min=

polyroots

K Kc⋅

τImin⋅

1 K Kc⋅+

τC min1−

⋅

τB min2−

⋅

τA min3−

⋅

5.776−

0.555−

0.06− 1.881i+

0.06− 1.881i−

=

T2 π⋅

1.8814min1−

:=Period:

The dominant roots are the complex conjugate roots:

P controller: a simple proportional gain•PI controller: from the Public Model Library, f0403PI (Fig. 13-4.3)•Series PID controller: Public Model Library, f0405PIDs (Fig. 13-4.5)•

All the initial conditions in the controller models are zero. The set point input, R(s), is a step input that changes from 0 to 1 at time = 1 min. The limits on the controller output must be changed to -100%CO to 100%CO for this linear system, so that it can be negative.

The Simulink block diagram for the loop is:

The plots for the three controllers, using the tuning parameters determined in Problem 7-2, are:

The responses are for the proportional (gold), PI (purple), and series PID (green). Notice how the periods of oscillation and decay ratios closely match the analytical results of part (a) of this problem. The proportional controller shows a very small offset:

1%TO

1 K Kc⋅+0.043 %TO=


Kc0.9

KPu

1−:= τI 3.33 t0e⋅:= Kc 1.6

%CO

%TO= τI 2 min=

(b) PI controller tuned for minimum IAE on disturbance inputs.

From Table 7-2.2: Kc0.984

KPu

0.986−:= τI

τ´

0.608Pu

0.707:= Kc 1.7

%CO

%TO= τI 1.2 min=

(c) PI controller tuned for minimum IAE on set-point inputs.


KPu

0.861−:= τI

τ´

1.02 0.323 Pu⋅−:= Kc 1.2

%CO

%TO= τI 1.5 min=

(d) PI controller tuned by controller syntesis for 5% overshoot on a set-point chang


KPu

1−:= τI τ´:= Kc 0.89

%CO

%TO= τI 1.3 min=



Problem 7-3. Feedback control of a second-order plus dead-time process.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

U(s)

C(s)+ +

+-

G1 s( )K e

t0− s⋅⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅= K 1.25

%TO

%CO:= τ1 1min:= τ2 0.6min:= t0 0.20min:=

τ2

τ10.6= τ´ 1.32 τ1⋅:= τ´ 1.32 min=First -order plus dead-time parameters from Fig. 7-2.8:

t´0 0.39 τ1⋅:= t´0 0.39 min=(the dead-time equivalent is added to the actual dead time)

PI controller: Gc s( ) Kc 11

τ I s⋅+⋅= t0e t´0 t0+:= Pu

t0e

τ´:= Pu 0.447=

(a) PI controller tuned for quarter-decay ratio response

From Table 7-2.1:

τI 0.82 min=

For parallel PID.

τD 0.482 τ´⋅ Pu1.137

:= τD 0.25 min=



KPu

0.869−:= τI

τ´

0.740 0.130 Pu⋅−:= Kc 1.7

%CO

%TO= τI 1.9 min=

For parallel PID.τD 0.348 τ´⋅ Pu

0.914:= τD 0.22 min=

(d) PID controller tuned by controller syntesis for 5% overshoot on a set-point change.

From Table 7-4.1, series: Kć

0.5

KPu

1−:= τÍ τ´:= τ´D

t0e

2:= Kć 0.89

%CO

%TO= τÍ 1.3 min=

τ´D 0.3 min=

Parallel PID: Kc Kć 1τ´D

τÍ+⋅:= τI τÍ τ´D+:= τD

τÍ τ´D⋅

τÍ τ´D+:=

Kc 1.1%CO

%TO= τI 1.6 min=

τD 0.24 min=



Problem 7-4. Process of Problem 7-3 with PID controller.

From the solution to Problem 7-3: τ´ 1.32min:= t0e 0.39 0.2+( )min:= Pu

t0e

τ´:= Pu 0.447=

Series PID: Gc s( ) Kć 11

τÍ s⋅+⋅ τ´D s⋅ 1+( )⋅= Parallel PID: Gc s( ) Kc 1

1

τ I s⋅+ τD s⋅+⋅=

(a) PID controller tuned for quarter-decay ratio response

From Table 7-2.1, series: Kć1.2

KPu

1−:= τÍ 2 t0e⋅:= Kć 2.1

%CO

%TO= τÍ 1.2 min=

τ´D 0.5 t0e⋅:= τ´D 0.3 min=

Parallel PID: Kc Kć 1τ´D

τÍ+⋅:= τI τÍ τ´D+:= τD

τÍ τ´D⋅

τÍ τ´D+:=

Kc 2.7%CO

%TO= τI 1.5 min=

τD 0.24 min=(b) PID controller tuned for minimum IAE on disturbance inputs.


KPu

0.921−:= τI

τ´

0.878Pu

0.749:= Kc 2.4

%CO

%TO=


KPu

0.986−:= τI

τ´

0.608Pu

0.707:= Kc 1.6

%CO

%TO= τI 1.3 min=



KPu

0.861−:= τI

τ´

1.02 0.323 Pu⋅−:= Kc 1.1

%CO

%TO= τI 1.5 min=

(d) PI controller tuned by controller syntesis for 5% overshoot on a set-point chang


KPu

1−:= τI τ´:= Kc 0.82

%CO

%TO= τI 1.3 min=



Problem 7-5. Process of Problem 7-3 with sampled-data PI controller.Sample time: T 0.1min:= From the solution to Problem 7-3: τ´ 1.32min:=

Use Eq. 7-2.18: t0e 0.39 0.2+( )minT

2+:= Pu

t0e

τ´:= Pu 0.485=

PI controller:Gc s( ) Kc 1

1

τ I s⋅+⋅=

(a) PI controller tuned for quarter-decay ratio response


KPu

1−:= τI 3.33 t0e⋅:= Kc 1.5

%CO

%TO= τI 2.1 min=

(b) PI controller tuned for minimum IAE on disturbance inputs.

Substitute:

1 et0− s⋅

−

1t0

2s⋅+ 1−

t0

2s+

1t0

2s+

=t0 s⋅

1t0

2s⋅+

=et0− s⋅ 1

t0

2s−

1t0

2s+

=

This is a PID controller with dead-time compensation. To eliminate the dead-time compensation term use the Padé approximation:

Gc s( )τ1 s⋅ 1+( ) τ2 s⋅ 1+( )

K et0− s⋅

⋅

et0− s⋅

τc s⋅ 1+ et0− s⋅

−

=Substitute:

C s( )

R s( )

et0− s⋅

τc s⋅ 1+=Gc s( )

1

G s( )

C s( )

R s( )

1C s( )

R s( )−

=Dahlin synthesis formula:

G s( )K e

t0− s⋅⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

(b) Second-order plus dead time.

τ´D τ2=τÍ τ1=Kć

τ1

K τc⋅=

Gc s( ) Kć 11

τÍ s⋅+⋅ τ´D s⋅ 1+( )=Compare with the series PID controller:

Gc s( )1

G s( )

1

τc s⋅=

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )K

1

τc s⋅⋅=

τ1

K τc⋅1

1

τ1 s⋅+ τ2 s⋅ 1+( )=

Dahlin syntheis formula:

G s( )K

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

(a) Assuming no dead time.Problem 7-6. Controller Synthesis for the process of Problem 7-3.


Gc s( )τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅

K

1t0

2s⋅+

τc s⋅ 1t0

2s+⋅ t0 s⋅+

⋅=τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅

t0

2s 1+⋅

K s⋅ τc t0+τc t0⋅

2s+⋅

=

τe

τc t0⋅

2 τc t0+( )⋅=

Gc s( )τ1

K τc t0+( )⋅1

1

τ1 s⋅+⋅ τ2 s⋅ 1+( )⋅

t0

2s⋅ 1+

τe s⋅ 1+⋅=

This is a series PID controller with a lead-lag unit attached. The corresponding tuning parameetrs are:

K´c

τ1

K τc⋅= τ´I τ1= τ´D τ2= and a second derivative with τ´D2

t0

2=

In practice astandard PID controller is used with the tuning parameters of Problem 7-4.



Problem 7-7. Simulation of the control loop of Problem 7-3.

To simulate the loop usea Simulink transfer function block•a Simulnk time delay block•a parallel PID controller from the Public Model Library, f406PIDp (Fig. 13-4.6).•

For this linear system all the initial conditions are zero, and the limits on the controller output are set to -100 to 100%CO to allow the output to go negative.

The Simulink diagram is:τ1 s⋅ 1+( ) τ2 s⋅ 1+( ) 0.6s2 1.6s+ 1+=

Two additional blocks have been added to calculate the integral of the absolute value of the error. Students are encouraged to adjust the controller parameters to minimize the IAE. However, they should also observe the time response of the controller output and the controlled variable.

A sample plot to a unit step change in set point at 1 minute is:

The PID tuning parameters for minimum IAE on set point changes (Problem 7-4(c)) were used:

Kc 1.8%CO%TO

:=

τI 1.9min:=

τD 0.22min:=



Problem 7-8. Quarter decay tuning of PI controller for the blender of Problem6-11.

From the solution to Problem 6-11: Kcu 250−%CO%TO

:= Tu 3.01min:=

PI controller quarter-decay tuning from Table 7-1.1: KcKcu2.2

:= τITu1.2

:= Kc 114−%CO%TO

=

τI 2.5 min=The negative gain means the controller is direct acting.



τD 1.5 min=τI 6 min=Kc 1.2%CO%TO

=τDt02

:=τI 2 t0⋅:=Kc1.2K

t0τ

1−

:=

From Table 7-2.1, the quarter-decay ratio tuning parameters for a series PID controller are:

t0 3 min=t0 t2 τ−:=

τ 6 min=τ 1.5 t2 t1−( ):=

t2 10 1−( )min:=

0.632 10.1⋅ %TO 6.383 %TO=

t1 6 1−( )min:=

0.283 10.1⋅ %TO 2.858 %TO=

By fit 3 (two-point method):

K 2.02%TO%CO

=

K10.1%TO

5%CO:=Gain:

In Problem 6-12 we found that there is no ultimate gain for reactor temperature control loop when the cooling water is the manipulated variable. By simulation of the linear loop, the open-loop response to a 5%CO step change at 1 minute is:

Problem 7-9. Quarter decay tuning of PID controller for the reactor of Problem 6-12.



τI 2 min=Kc 2.8−%CO%TO

=τI 3.33t0:=Kc0.9K

t0τ

1−

:=From Table 7-2.1:

t0 0.6 min=t0 t2 τ−:=

τ 4.2 min=τ 1.5 t2 t1−( ):=

t2 5.8 1−( )min:=

0.632 11.2− %TO( ) 7.078− %TO=

t1 3 1−( )min:=

0.283 11.2− %TO( ) 3.17− %TO=

Fit 3 (two-point method):

K 2.24−%TO%CO

=

K11.2− %TO5%CO

:=Gain:

In Problem 6-14 we found that there is no ultimate gain for the composition control loop. By simulation of the linear approximation, the response to a step change of 5%CO at 1 min is:

Problem 7-10. Quarter decay tuning of a PI controller for the three-tank process of Problem 6-14.


Kv 2.46gal

min %CO⋅:=KT 100

%TO gal⋅

lb:=τv 0.1min:=τ 5min:=where

1 KT Kc⋅ 11τ I s⋅

+⋅Kv

τv s⋅ 1+⋅

KA KB τ⋅ s⋅+ K3 τ2

⋅ s2⋅+

τ s⋅ 1+( )3⋅+ 0=

From the solution to Problem 3-17, the characteristic equation is:

Roots of the characteristic equation, damping ratio, and decay ratio.

τI 9.2 min=Kc 0.77%CO%TO

=τI 3.33 t0⋅:=Kc0.9K

t0τ

1−

:=From Table 7-2.1:

t0 2.75 mi=t0 t2 τ−:=

τ 11.25 m=τ 1.5 t2 t1−( )⋅:=

t2 15 1−( )min:=

0.632 ∆c⋅ 15.168 %TO=

t1 7.5 1−( )min:=

0.283 ∆c⋅ 6.792 %TO=

Two-point method:

K 4.8%TO%CO

=K∆c

5%CO:=

∆c 74 50−( )%TO:=

From this response we get:

In Problem 6-17 we found that there is no ultimate gain for the composition control loop. Because of the complex combination of poles and zeros, the open-loop parameters cannot be easily determined analytically. This problem is solved by simulation in Problem 13-23, where the following open-loop response to a step increase of 5%CO at 1 minute is obtained:

Problem 7-11. Quarter decay tuning of a PI controller for the reactors of Problem 6-17.



There is essentially no oscillation in the response. The response is complete in less than one complete oscillation. Students should verify this with the simulation of Problem 13-23 and experiment with other tuning parameters. A higher controller gain is indicated by these results.

e 0.112− min 1− T 0.014=Decay ratio:

0.112

0.1122 0.1642+

0.564=

Damping ratio:

T 38.3 min=T2π

0.164min 1−:=

polyroots

KLτ I

min⋅

τE

τC min 1−

τB min 2−

τAmin 3−

τv τ3

⋅ min 4−

9.905−

0.39−

0.112− 0.164i+

0.112− 0.164i−

0.082−

=

There is also a comlex conjugate root with period:

1−

0.082− min 1−12.2 min=

The dominant root is a real root with time constant:

Roots of the characteristic equaton:

τE 3.188=τC 22.176 min=τB 88.293 min2=τA 132.5 min3

=τv τ3

⋅ 12.5 min4=

KL 1.415=τE 1 KL+KT Kc⋅ Kv⋅ KB⋅ τ⋅

τ I+:=τC τv 3τ+ KT Kc⋅ Kv⋅ KB⋅ τ⋅+:=

KL KT Kc⋅ Kv⋅ KA⋅:=τB 3 τv⋅ τ⋅ 3τ2

+ KT Kc⋅ Kv⋅ K3⋅ τ2

⋅+:=τA 3 τv⋅ τ2

⋅ τ3

+:=where

τv τ3

⋅ s5⋅ τA s4

⋅+ τB s3⋅+ τC s2

⋅+ τE s⋅+KLτI

+ 0=

s τv s⋅ 1+( )⋅ τ3

s3⋅ 3 τ

2⋅ s2

⋅+ 3 τ⋅ s⋅+ 1+( )⋅ KT Kc⋅ Kv⋅ s1τI

+⋅ KA KB τ⋅ s⋅+ K3 τ2

⋅ s2⋅+⋅+ 0=

K3 0.0025lb min⋅

gal2:=KB 0.0075

lb min⋅

gal2:=KA 0.0075

lb min⋅

gal2:=

Tu 8.91s:=

Quarter-decay ratio tuning parametes for a PI controller:

From Table 7-1.1: KcKcu2.2

:= τITu1.2

:= Kc 7.2−%CO%TO

= τI 7.4 s=


From the solution to Problem 6-18: 1KT

τT s⋅ 1+Kc 1

1τ I s⋅

+⋅Ksc

τsc s⋅ 1+

Kpτp s⋅ 1+

− 0=

τT τsc⋅ τp⋅ s4⋅ τT τsc⋅ τT τp⋅+ τsc τP⋅+( ) s3

⋅+ τT τsc+ τp+( )s2+ 1 KL−( )s+

KLτ I

− 0=

where KL KT Kc⋅ Ksc⋅ Kp⋅:= KL 6.505−=

Find roots:

Smith & Corripio, 3rd edition kscf 1000ft3:=

Problem 7-12. Control of suction pressure for compressor of Problem 6-18.

Steam

SuctionDischarge

fi(t)

fc(t)

ps(t)

m(t)

PT

PC

SC

From the solution to Problem 6-18:

H s( )KT

τT s⋅ 1+= KT 5

%TOpsi

:=

τT 1.2s:=

Gsc s( )Ksc

τsc s⋅ 1+= τsc 2.5s:=

Ksc 0.36kscf

min %CO⋅:=

Gp s( )Kp

τp s⋅ 1+= τp 7.5s:=

Kp 0.5psi min⋅

kscf:=

Ultimate gain and period: Kcu 15.9−%CO%TO

:=

The dominant roots are the complex conjugate pair. The period of the oscillations is:

polyroots

KL−

τIsec

1 KL−

τT τsc+ τp+( )sec 1−

τT τsc⋅ τT τp⋅+ τsc τp⋅+( )sec 2−

τT τsc⋅ τp⋅ sec 3−⋅

1.16−

0.135−

0.036− 0.498i−

0.036− 0.498i+

= T2π

0.498sec 1−:=

T 12.62 s=

Damping ratio: 0.036

0.0362 0.4982+

0.072= Decay ratio: e 0.036− sec 1− T⋅ 0.635=

The damping ratio is too low and the decay ratio is too high. To reduce the oscillations a smaller gain is required.


K4 0.525:=

Ultimate gain and period: Kcu 86.7−%CO%TO

:= Tu 8.32min:=

Quarter-decay ratio tuning of a PI controller.

KcKcu2.2

:= τITu1.2

:= Kc 39−%CO%TO

= τI 6.9 min=



1KT

τT s⋅ 1+Kc 1

1τ I s⋅

+⋅KFC

τFC s⋅ 1+

K2 K3⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K4⋅−− 0=

τA s5⋅ τB s4

⋅+ τC s3⋅+ τD s2

⋅+ τE s+KLτ I

− 0=

where τA τT τFC⋅ τ1⋅ τ2⋅:= τB τT τFC⋅ τ1⋅ τT τFC⋅ τ2⋅+ τFC τ1⋅ τ2⋅+ τT τ1⋅ τ2⋅+:=

degC Kd:=Smith & Corripio, 3r edition

Problem 7-13. Temperature control of stirred-tank cooler of Problem 6-19.From the solution to Problem 6-19:

TC

TT

Tc(t)

f(t)

fc(t) Tci

V

m(t)

Tset(t)

Ti(t)

T(t)

FT

FCSP

GFC s( )KFC

τFC s⋅ 1+=

τFC 0.1min:=

KFC 0.008m3

min %CO⋅:=

H s( )KT

τT s⋅ 1+= τT 0.6min:=

KT 2%TOdegC

:=

G1 s( )K2 K3⋅

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ K2 K4⋅−=

τ1 13.77min:= τ2 3.03min:=

K2 0.725:= K3 28.94degC min⋅

m3:=


The damping ratio is too low and the decay ratio is too high. The controller gain should be decresaed to reduced the very oscillatory behavior.

e 0.02− min 1− T⋅ 0.782=Decay ratio:

0.02

0.022 0.5122+

0.039=Damping ratio:polyroots

KL−

τImin

τE

τD min 1−⋅

τC min 2−⋅

τB min 3−⋅

τA min 4−⋅

9.993−

1.882−

0.154−

0.02− 0.512i−

0.02− 0.512i+

=

Tu 8.32 min=T2π

0.512min 1−:=

The dominant roots are the complex conjugate pair. The period of oscillation is:

Find the roots:

τD 17.234 min=KL 13.23−=τE 13.849=τE 1 K2 K4⋅− KL−:=

τC 53.52 min2=

KL KT Kc⋅ KFC⋅ K2⋅ K3⋅:=τD τT τFC+( ) 1 K2 K4⋅−( ) τ1+ τ2+:=

τB 30.214 min3=

τC τT τFC⋅ 1 K2 K4⋅−( )⋅ τT τ1⋅+ τT τ2⋅+ τFC τ1⋅+ τFC τ2⋅+ τ1 τ2⋅+:=τA 2.503 min4

=

In the solution to Problem 6-22 we determined that there is no ultimate gain for the analyzer control loop. By simulation, the open-loop response to a 5% increase in the controller output is:

Quarter-decay tuning parameters for a PI controller.

K5 0.286:=K4 0.00127lbmole min⋅

gal2:=

K1 0.006123lbmole min⋅

gal2:=

τ2 1.429min:=τ1 2.222min:=

G1 s( )K4 τ1 s⋅ 1+( )⋅ K5 K1⋅+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=

KT 222%TO gal⋅

lbmole:=

τT 0.5min:=H s( )KT

τT s⋅ 1+=

Kv 1.42gal

min %CO⋅:=


LT LC

AT

AC

fi(t)

cAi(t)

cA1(t)

cA2set(t)

V

V

cA2(t)

A

fA(t)

Problem 7-14. Composition control of reactors in series of Problem 6-22.

lbmole 433.59mole:=Smith & Corripio, 3rd edition

Find the roots:

KL 2.736=τA 2.555 min=KL KT Kc⋅ Kv⋅ K4 K5 K1⋅+( )⋅:=τA KT Kc⋅ Kv⋅ K4⋅ τ1⋅:=where

τT τ1⋅ τ2⋅ s4⋅ τT τ1⋅ τT τ2⋅+ τ1 τ2⋅+( )s3

+ τT τ1+ τ2+ τA+( )s2+ 1 KL+( )s+

KLτ I

+ 0=

1KT

τT s⋅ 1+Kc 1

1τ I s⋅

+⋅ KvK4 τ1 s⋅ 1+( )⋅ K5 K1⋅+

τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅⋅+ 0=From the solution to Problem 2-22:


τI 2.5 min=τI 3.33 t0⋅:=

Kc 2.9%CO%TO

=Kc0.9K

t0τ

1−

:=

From Table 7-2.1:

t0 0.8 min=t0 t2 τ−:=

τ 2.3 min=τ 1.5 t2 t1−( )⋅:=

t2 4 1−( )min:=

0.632 4.7⋅ %TO⋅ 2.97 %TO=

t1 2.5 1−( )min:=

0.283 4.7⋅ %TO 1.33 %TO=

Two-point method:

K 0.94%TO%CO

=K4.75

:=Gain:

The dominant roots are the second pair of complex conjugate roots. The period of ocillation is:

polyroots

KLτI

min

1 KL+

τT τ1+ τ2+ τA+( )min 1−

τT τ1⋅ τT τ2⋅+ τ1 τ2⋅+( )min 2−

τT τ1⋅ τ2⋅ min 3−⋅

1.204− 0.813i−

1.204− 0.813i+

0.371− 0.435i−

0.371− 0.435i+

= T2π

0.435min 1−:=

T 14.44 min=

Damping ratio: 0.371

0.3712 0.4352+

0.649= Decay ratio: e 0.371− min 1− T 0.00471=

The damping ratio is high and the decay ratio is small, with practically no oscillations. The controller gain should be higher. The reason is that the quarter-decay ratio formulas are based on fit 1, not fit 3 (the two-point method), to determine the open-loop time constant and dead time.


KT

2.5%TO

%=K

T

100%TO

xmax

xmin

−:=

xmax

95%:=xmin

55%:=

Transmitter (AT):

xo

75%:=xin

95%:=

txi

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

17

19

21

25

29

33

75.0

75.0

75.0

75.0

75.0

75.0

75.1

75.3

75.4

75.6

75.7

75.9

76.1

76.2

76.3

76.4

76.6

76.7

76.8

76.9

77.0

77.0

:=Problem data: txm

0

1

1.5

2.5

3.5

4.5

5.5

6.5

7.5

8.5

9.5

10.5

11.5

13.5

15.5

17.5

19.5

21.5

25.5

29.5

33.5

75.0

75.0

75.0

75.0

74.9

74.6

74.3

73.6

73.0

72.3

71.6

70.9

70.3

69.3

6.6

68.0

67.6

67.4

67.1

67.0

67.0

:=

First column is time in •minutesSecond column is outlet % •moisture in the solids

∆xin

0.5%:=∆m 12.5%CO:=

Response to a step change of:

AT

ACM

Sludge

Ferric

Chloride

To

incinerator

Filtrate

Problem 7-15. Solid moisture control of a vacuum filter.%CO %:=%TO %:=Smith & Corripio, 3rd edition

KT

G1

s( )⋅1.6− e

5.71− s

6.5 s⋅ 1+

%TO

%CO=

∆xi

txi21 1,

txi0 1,

− %:= ∆xi

2 %= K2

∆xi

∆xin

:= KT

K2

⋅ 10%TO

%=

txi0 1,

% 0.283 ∆xi

⋅+ 75.566 %= t1

txi9 0,

txi10 0,

txi9 0,

−

75.566 txi9 1,

−

txi10 1,

txi9 1,

−⋅+:=

txi0 1,

% 0.632 ∆xi

⋅+ 76.264 %= t2

txi14 0,

txi15 0,

txi14 0,

−

76.264 txi14 1,

−

txi15 1,

txi14 1,

−⋅+:=

t1

8.66= t2

13.64= τ2

1.5 t2

t1

−( )min:= t02

t2

min⋅ τ2

−:= τ2

7.47 min= t02

6.17 min=

KT

G2

s( )⋅10e

6.17− s

7.47 s⋅ 1+

%TO

%=

(a) Block diagram of the moisture control loop.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

C(s)

+ +

+-

Xo(s)X

o

set(s)

KT

KT

Xin(s)

(b) Transfer functions by fit 3, two-point method.

∆xm

txm

20 1,tx

m0 1,

− %:= ∆xm

8− %= K1

∆xm

∆m:= K

TK

1⋅ 1.6−

%TO

%CO=

txm

0 1,% 0.283 ∆x

m⋅+ 72.736 %= t

1tx

m8 0,

txm

9 0,tx

m8 0,

−

72.736 txm

8 1,−

txm

9 1,tx

m8 1,

−⋅+:=

txm

0 1,% 0.632 ∆x

m⋅+ 69.944 %= t

2tx

m12 0,

txm

13 0,tx

m12 0,

−

69.944 txm

12 1,−

txm

13 1,tx

m12 1,

−⋅+:=

t1

7.877= t2

12.212= τ1

1.5 t2

t1

−( )min:= t01

t2

min⋅ τ1

−:= τ1

6.5 min= t01

5.71 min=

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


is unlawful.

Gc

s( ) 0.64−%CO

%TO1

1

19 s⋅+=

τI

19 min=Kc

0.64−%CO

%TO=τ

I3.33 t

01⋅:=K

c

0.9

KT

K1

⋅

t01

τ1

1−

:=From Table 7-2.1:

(e) Quarter-decay response tuning of a PI controller.Offset

KT

0.789− %=

Offset 2− %TO=Offset

0 K2

1⋅ %−

1 KT

Kc

⋅ K1

⋅+:=Offset

KT∆x

o

set⋅ K

2∆x

in⋅−

1 KT

Kc

⋅ K1

⋅+=

Gc

s( ) 0.64−%CO

%TO=K

c0.64−

%CO

%TO=K

c

0.902

KT

K1

⋅

t01

τ1

1−

:=From Table 7-2.2:

(d) Gain of a proportional controller for minimum IAE response and offset to a 1% increase in inlet moisture.

The controller must direct acting: an increase in moisture increases the controller output; this increases the speed of the pump and the rate of ferric chloride addition; filtration becomes more efficient and the moisture content of the product decreases.

t01

τ1

0.878=

The loop is difficult to control by feedback control because its ratio of dead time to time constant is high:

(c) Discuss the controllability of the loop and the controller action.

The control valve fails closed (air-to-open) to prevent overflowing the absorber on instrument powerfailure.

Control valve: Kv

5gal

min %CO⋅=K

v

fmax

100%CO:=

Transmitter (AT): KT

0.5%TO

ppm=K

T

100%TO

ymax

ymin

−:=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

C(s)

+ +

+-

Yo(s)

Yo

set(s)

KT

KT

Yin(s)

Kv

F(s)

(b) Block diagram of the loop and transfer function of each block. Use fit-3 on response data (two-point method).

Negligible lag.

fmax

500gal

min:=

Control valve, assumed linear.

Negligible lag.

ymax

200ppm:=ymin

0ppm:=

Transmitter (AT):

AT AC

Air In

Air Out

NH3 solution

Water In

SP

(a) Design a control loop to control the air outlet composition.Problem 7-16. Composition control of an absorber.

ppm 106−

:=Smith & Corripio, 3rd edition

The controller is direct acting (negative gain): increasing outlet gas composition increases the controller output; this opens the valve increasing the flow of water to he absorber and absorbing more ammonia. The ammonia composition in the outlet gas decreases.

Gc

s( ) 23−%CO

%TO=K

c23−

%CO

%TO=K

c

1

KT

Kv

⋅ K1

⋅

t01

τ1

1−

:=From Table 7-2.1:

60 50−( )ppm 10 ppm=

(c) Quarter decay ratio tuning for proportional controller and offset to a set-point change of

G1

s( )0.035− e

0.46s

0.95 s⋅ 1+

ppm min⋅

gal=

t01

0.46 min=τ1

0.95 min=t01

t2

sec⋅ τ1

−:=τ1

1.5 t2

t1

−( )sec:=Fit 3

t2

84.58=t2

ty7 0, ty

8 0, ty7 0,−( )

yo2

ppm1−

⋅ ty7 1,−

ty8 1, ty

7 1,−⋅+:=

yo2

51.12 ppm=yo2

ty0 1, ppm 0.632 ∆y

o⋅+:=

t1

46.7=t1

ty3 0, ty

4 0, ty3 0,−( )

yo1

ppm1−

⋅ ty3 1,−

ty4 1, ty

3 1,−⋅+:=

yo1

50.5 ppm=yo1

ty0 1, ppm 0.283 ∆y

o⋅+:=

K1

0.035−ppm min⋅

gal=K

1

∆yo

∆f:=

∆yo

1.77 ppm=∆yo

ty16 1, ty

0 1,−( )ppm:=

ty

0

20

30

40

50

60

70

80

90

100

110

120

130

140

160

180

250

50.00

50.00

50.12

50.30

50.60

50.77

50.90

51.05

51.20

51.26

51.35

51.48

51.55

51.63

51.76

51.77

51.77

:=

First column is time in seconds•Second column is outlet ammonia ppm•

∆f 50−gal

min:=

Response to a step change in inlet water flow of

Offset

KT

10⋅ ppm

1 KT

Kc

⋅ Kv

⋅ K1

⋅+:= Offset 1.641 %TO=

Offset

KT

3.3 ppm=

(d) Quarter decay ratio tuning of series PID controller and offset.

From Table 7-2.1: Kć

1.2

KT

Kv

⋅ K1

⋅

t01

τ1

1−

:= τÍ

2 t01

⋅:= τ´D

t01

2:=

Kć

28−%CO

%TO= τ´

I0.93 min= τ´

D0.23 min= G

cs( ) 28−

%CO

%TO1

1

0.93s+ 0.23s 1+( )=




is unlawful.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

C(s)

+ -

+-

To(s)T

o

set(s)

KT

KT

Fin(s)

%CO

TT

%TO

scfh

FF

%TO

TC FurnaceTC


The controller must be reverse acting (positive gain): increasing temperature decreases the controller output; this closes the valve reducing the fuel flow and the outlet coil temperature.

The control valve must fail closed (air-to-open) to prevent overheating the furnace on instrument power failure.

tt

0

0.5

1.0

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

7.0

8.0

9.0

10.0

11.0

12.0

14.0

20.0

425.0

425.0

425.0

425.0

426.4

428.5

430.6

432.4

434.0

435.3

436.6

437.6

439.4

440.7

441.7

442.5

443.0

443.5

444.1

445.0

:=

(a) Block diagram of the loop, fail-safe position of the valve, and controller action.

First column is •time in minutesSecond column •is temperature in ºF

∆m 5%CO:=Response to step change of:

Tmax

500degF:=

TT

TCProcess

air

FuelAir

SP

Tmin

300degF:=Transmitter (TT):

Problem 7-17. Temperature control of a furnace.

degF R:=Smith & Corripio, 3rd edition

(c) Quarter decay ratio tuning of series PID controller.


1.2

KT

K1

⋅

t01

τ1

1−

:= τÍ

2 t01

⋅:= τ´D

t01

2:=

Kć

1%CO

%TO= τ´

I4.5 min= τ´

D1.1 min= G

cs( ) 1

%CO

%TO1

1

4.5s+ 1.1s 1+( )=

(d) Synthesis tuning of series PID controller for 5% overshoot.


0.5

KT

K1

⋅

t01

τ1

1−

:= τÍ

τ1

:= τ´D

t01

2:=

Kć

0.42%CO

%TO= τ´

I3.8 min= τ´

D1.1 min= G

cs( ) 0.42

%CO

%TO⋅ 1

1

3.8s+ 1.1s 1+( )=

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purpose



is unlawful.

(b) Transfer functions using fit 3 (two-point method).

Transmitter (TT): KT

100%TO

Tmax

Tmin

−:= K

T0.5

%TO

degF=

∆T tt19 1, tt

0 1,−( )degF:= K1

∆T

∆m:= K

TK

1⋅ 2

%TO

%CO=

T1

tt0 1, degF 0.283∆T+:= T

1430.66 degF=

t1

tt6 0, tt

7 0, tt6 0,−( )

T1

degF1−

⋅ tt6 1,−

tt7 1, tt

6 1,−⋅+:= t

13.517=

T2

tt0 1, degF 0.632∆T+:= T

2437.64 degF=

t2

tt11 0, tt

12 0, tt11 0,−( )

T2

degF1−

⋅ tt11 1,−

tt12 1, tt

11 1,−⋅+:= t

26.022=

Process by fit 3: τ1

1.5 t2

t1

−( )min⋅:= t01

t2

min⋅ τ1

−:= τ1

3.76 min= t01

2.26 min=

KT

G1

s( )⋅2e

2.26− s

3.76s 1+

%TO

%CO=




is unlawful.

Gc

s( ) 23%CO

%TO1

1

0.56s+ 0.14s 1+( )=τ´

D0.14 min=τ´

I0.56 min=K´

c23

%CO

%TO=

τ´D

Tu

8:=τ´

I

Tu

2:=

K´c

Kcu

1.7:=From Table 7-1.1:

Tu

1.116min:=

Kcu

38.9%CO

%TO:=From the solution to Problem 6-24, the ultimate gain and period are:

Quarter decay ratio tuning parameters of series PID temperature controller TC.

LTLC

vp2(t)

T3

set(t)

fs(t) T1(t)

T3(t)

hset

(t)

Steam

Condensate

3 ft

h(t)

TT

TC

p1(t)

p2 = 40 psia

5 ft

N2

AOAO

Tp

3(t)

Problem 7-18. Temperature control of oil heater of Problem 6-24.


The valve must fail closed (air-to-open) to prevent by-passing too much hot oil on instrument power failure that would overheat the reactor.

Cvmax

195gal

min psi⋅:=

From Fig. C-10.1, p. 532, a 4-in valve is required.

Cvmax

119gal

min psi⋅=C

vmax200% f

d⋅

Gf

∆pv

⋅:=

(a) Size control valve for 100% overcapacity. Valve fail-safe position and controlleaction.

Tu

24min:=Kcu

16%CO

%TO:=Closed-loop test on temperature loop:

∆T 4.4degF:=∆vp 5%:=Open-loop test on temperature loop:

T 275degF:=(constant)fpump

400gal

min:=

Design conditions:

Gf

0.881=Gf

ρ ft3

⋅

62.4lb:=ρ 55

lb

ft3

:=Oil density:

Tmax

350degF:=Tmin

150degF:=Temperature transmitter (TT):

fd

200gal

min:=∆p

v10psi:=α 50:=Control valve, equal percentage:Problem data:

TT

TC

LT LC

SP

SP

Reactants

Products

Water

Steam

Problem 7-19. Temperature control of exothermic catalytic reactor.


(d) Calculate the process gain atb design conditions, including the control valve an

Kv

7.824gal

min %CO⋅=K

v

ln α( ) fd

⋅

100%CO:=

Valve fails closed. Controller is direct acting. (See part (a).)

KT

0.5%TO

degF=K

T

100%TO

Tmax

Tmin

−:=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

C(s)

+ +

+-

To(s)T

set(s)

KT

KT

Fin(s)

%CO

TT

%TO

lb/hr

FF

%TO

TC FurnaceTC

(c) Block diagram of the loop.

fv

306.6gal

min=k

v3.284=f

v

kv

1 kv

+fpump

:=

kv

Cvmax

∆pv

Gf

⋅1

fpump

fd

−⋅:=

∆pv

Gf

fpump

fvmax

−

fpump

fd

−

fvmax

Cvmax

=Flow when fully opened:Let

vp 69.6 %=Cv

59.4gal

min psi⋅=vp 1

ln

Cv

Cvmax

ln α( )+:=Cv

fd

Gf

∆pv

⋅:=At design conditions:

Cv

Cvmax

αvp 1−

⋅=∆pva

∆pv

fpump

fv

−

fpump

fd

−

2

⋅= Gf

fv

Cv

2

⋅=fv

Cv

∆pva

Gf

⋅=

Assume the pressure drop through the boler tubes varies with the square of oil flow through the tubes•the pump flow is constant as the valve position changes•the pressure drop across the valve is the same as the pressure drop across the boiler tubes.•

(b) Valve position at design conditions and maximum flow through the valve when fully opened.

The controller must be reverse acting (positive gain): an increase in reactor temperature decreases the controller output closing the by-pass valve; this reduces the by-pass flow of hot oil decreasing the oil temperature and the reactor temperature.




is unlawful.

Offset

KT

2.21− degF=Offset 1.11− %TO=Offset

KT

10− degF( )⋅

1 KT

Kc

⋅ K1

⋅+:=

Gc

s( ) 8%CO

%TO=K

c8

%CO

%TO=K

c

Kcu

2:=From Table 7-1.1:

(f) Quarter-decay ratio tuning of proportional temperature controller and offset for aset point change of -10ºF.

Gc

s( ) 9.4%CO

%TO1

0.083

s+ 3s 1+(=τ´

D3 min=

1

τ´I

0.083repeats

min=K´

c9.4

%CO

%TO=

repeats 1:=τ´D

Tu

8:=τ´

I

Tu

2:=K´

c

Kcu


(e) Quarter decay tuning parameters for series PID temperature controller.

KT

K1

⋅ 0.44%TO

%CO=K

188 degF=K

1

∆T

∆vp:=

The gain of the valve is included in K1, because the step change is in valve position.

( ) p g g gthe temperature transmitter.

Smith & Corripio, 3rd edition weight% %:=

Problem 7-20. Composition control of a double-effect evaporator

SP

SP

LC

AC

LT

AT

SP

LCLT

Vapors

SP

FC

FT

Vapors

ProductFeed

Steam

Cond.

12

13

1213

Problem data: Feed 50000lb

hr:= x

F5weight%:= x

min10weight%:= x

max35weight%:=

Open loop step response in feed composition: ∆xF

0.75weight%:=

21

21.5

22

22.5

23

23.5

24

24.5

25

0 100 200 300 400 500 600 700

Time, sec

Prod

uct c

ompo

sitio

n, w

t%

Open-loop step response to change in controller output: ∆m 2.5%CO:=

21

21.5

22

22.5

23

23.5

24

24.5

25

0 100 200 300 400 500 600 700 800 900 1000

Time, sec

Prod

uct c

ompo

sitio

n, w

t%

t01

2.91 min=

KT

G1

s( )⋅5.12e

2.91− s

4.08s 1+

%TO

%CO=

Change in feed composition: K2

24.7 21.5−( )weight%

∆xF

:= K1

1.28= KT

K2

⋅ 17.07%TO

weight%=

21.5weight% 0.283 24.7 21.5−( )⋅ weight%+ 22.41 weight%= t1

143sec:=


237sec:=

τ2

1.5 t2

t1

−( ) min

60sec:= t

02t2

τ2

−( ) min

60sec:= τ

22.35 min= t

021.6 min=

KT

G2

s( )⋅17.07e

1.6− s

2.35s 1+

%TO

weight%=

Note: Students should be encouraged to try also fits 1 and 2 and compare the answers.

The control valve must fail closed (air-to-open) to prevent overheating the evaporator on instrument power failure.

The controller must be reverse acting (positive gain): increasing product composition decreases

(a) Block diagram of the composition control loop, transfer functions, cntrol valve fasafe position, and controller action.

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

C(s)

+ +

+-

X(s)Xset

(s)

KT

KT

XF(s)

%CO

AT

%TOwt%

%TO

ACAC

wt%

wt%

Analyzer transmitter: KT

100%TO

xmax

xmin

−:= K

T4

%TO

weight%=

Determine process transfer functions by fit 3:

Change in controller output: K1

24.7 21.5−( )weight%

∆m:= K

11.28

weight%

%CO= K

TK

1⋅ 5.12

%TO

%CO=


256sec:=


419sec:=

τ1

1.5 t2

t1

−( ) min

60sec:= t

01t2

τ1

−( ) min

60sec:= τ

14.08 min=




is unlawful.

τI

7.5 min=Kc

0.34%CO

%TO=

τI

τ1

:=Kc

0.5

KT

K1

⋅

t01

τ1

1−

:=From Table 7-4.1:

(c) Controller synthesis tuning for 5% overshoot of PI composition controller.This is over twice the gain and 25% faster reset than with fit 3 parameters.

τI

7.2 min=Kc

0.61%CO

%TO=τ

I3.33 t

01⋅:=K

c

0.9

KT

K1

⋅

t01

τ1

1−

:=From Table 7-2.1:

τ1

7.5 min=τ

1580sec t

01−( ) min

60sec:=t

01130sec:=

Quarter decay ratio tuning is based on fit 1 parameters. From the figure above:

τI

9.7 min=Kc

0.25%CO

%TO=τ

I3.33 t

01⋅:=K

c

0.9

KT

K1

⋅

t01

τ1

1−

:=From Table 7-2.1:

(b) Quarter-decay ratio tuning of PI composition controller.

the controller output closing the steam control valve; this decreases the rate of evaporation reducing the product composition.

Kcu

8.0%CO

%TO:= T

u14min:=

(a) Size control valve for 100% overcapacity, valve gain at design flow, valve fail-safe position.

∆pv

p1

p2

− ∆pL

−:= Gf

1:= kL

∆pL

Gf

fcw

2⋅

:= ∆pv

5 psi=k

L8.163 10

5−× psi

min

gal

2

⋅=

Cvmax

200% fcw

⋅G

f

∆pv

⋅:= Cvmax

313.05gal

min psi⋅= C

vfcw

Gf

∆pv

⋅:= Cv

156.5gal

min psi⋅=

From Fig. C-10.1, page 532, a 6-in valve is required: Cvmax

400gal

min psi⋅:=

Valve gain, Eq. 5-2.27, page 171: Kv

ln α( )−

100%CO

fcw

1 kL

Cv

2⋅+

:= Kv

4.564−gal

min %CO⋅=

The valve must fail open (air-to-close) to prevent overheating the reactor on loss of instrument power. This is why the gain is negative.

(b) Block diagram of the control loop ad total process gain.

psia psi:=Smith & Corripio, 3rd edition

Problem 7-21. Temperature control of stirred-tank reactor.Design conditions: p

130psia:= p

215psia:=

TTTC11

Coolant

Feed

ProductP1

P2

TR

210degF:= fcw

350gal

min:=

Coil pressure drop: ∆pL

10psi:=

Temperature transmitter: Tmin

190degF:=

Tmax

230degF:=

Equal-percenage valve: α 50:=

Open-loop test: ∆fcw

10gal

min:=

∆TR

5.2− degF:=

Closed-loop test:




is unlawful.

Gc

s( )100%CO

21%TO1

0.14

s+ 1.75s 1+( )=τ´

D1.75 min=

1

τ´I

0.14repeat

min=

100%CO

K´c

21 %TO=τ´D

Tu

8:=τ´

I

Tu

2:=K´

c

Kcu


(c) Quarter-decay ratio tuning of PID temperature controller and controller action.

repeat 1:=K

TK

1⋅ K

v⋅ 5.933

%TO

%CO=Total process gain:

K1

0.52−degF min⋅

gal=K

T2.5

%TO

degF=K

1

∆TR

∆fcw

:=KT

100%TO

Tmax

Tmin

−:=

Gc(s) G1(s)

G2(s)

R(s) E(s) M(s)

C(s)

+ +

+-

TR(s)

Tset(s)

KT

KT

Fin(s)

%CO

TT

%TO

lb/hr

FF

%TO

TCTC

Gv(s)F

cw(s)

gpm

∆m 8%CO:=Open-loop step response to change in controller output:

(b) Process transfer functions from open-loop step responses by fit 2.

K

T

25

%TO

weight%

=K

T

100%TO

x

max

x

min

−:=Transmitter AT:

Gc(s) G1

(s)

G2

(s)

R(s) E(s) M(s)

C(s)

+ +

+-

X(s)

Xset

(s)

KT

KT

XF(s)

%CO

AT

%TO

wt%

%TO

ACAC

wt%

wt%

(a) Block diagram of the moisture control loop.

x

max

5weight%:=

x

min

1weight%:=Transmitter AT:x 3weight%:=x

F

15weight%:=Design conditions:

AT

AC

Feed

Fuel

Air

Stack

Dry

phospahates

Problem 7-22. Solids moisture control of a phosphates pebbles drier


2.5

3

3.5

4

4.5

5

0 50 100 150 200 250 300 350 400

Time, sec

Prod

uct m

oist

ure,

wt %

∆x 4.5 3−( )weight%:= K

1

∆x

∆m

:= K

1

0.188

weight%

%CO

= K

T

K

1

⋅ 4.688

%TO

%CO

=

From the graph:t

01

70sec

min

60sec

⋅:= 3.0weight% 0.632∆x+ 3.95 weight%= t

2

190sec:=

τ1

t

2

min

60sec

⋅ t

01

−:= t

01

1.17 min= τ1

2 min= K

T

G

1

s( )⋅4.688e

1.17− s

2.0s 1+

%CO

%TO

=

Open-loop step response to change in inlet moisture: ∆x

F

3weight%:=

The controller is reverse acting (positive gain): increasing product moisture content decreases the controller output; this decreases the table feeder speed and the feed rate reducing the moisture input to the drier and the moisture content of the product.

G

c

s( )

100%CO

199%TO

1

1

1.5s

+ 0.52s+=τD

0.52 min=τI

1.5 min=100%CO

K

c

199 %TO=

τD

0.482 τ1

⋅t

01

τ1

1.137

⋅:=τI

τ1

0.878

t

01

τ1

0.749

:=K

c

1.435

K

T

K

1

⋅

t

01

τ1

0.921−

:=From Table 7-2.2:

(c) Minimum IAE tuning of parallel PID moisture controller on disturbance inputs acontroller action.

K

T

G

2

s( )⋅16.67e

1.08− s

1.75s 1+

%CO

weight%

=τ2

1.75 min=t

02

1.08 min=τ2

t

2

min

60sec

⋅ t

02

−:=

t

2

170sec:=3.0weight% 0.632∆x+ 4.26 weight%=t

02

65sec

min

60sec

⋅:=From the graph:

K

T

K

2

⋅ 16.67

%TO

weight%

=K

2

0.667

weight%

weight%

=K

2

∆x

∆x

F

:=∆x 5.0 3.0−( )weight%:=

2.5

3

3.5

4

4.5

5

5.5

0 50 100 150 200 250 300 350 400

Time, sec

Prod

uct m

oist

ure,

wt%


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.

50%CO ∆m+ 57.11 %CO=If the initial controller output is 50%CO, the final steady state is:

∆m 7.11 %CO=∆m

K

2

−

K

1

2− weight%( ):=

∆m K

1

⋅ K

2

2− weight%( )⋅+ 0=For the change in outlet moisture to be zero:

(e) Controller output required to avoid offset for the disturbance of part (d).

3weight%

Offset

K

T

− 2.51 weight%=Final steady state moisture control of the product:

Offset

K

T

0.49 weight%=Offset 12.28 %TO=Offset

K

T

0⋅ K

T

K

2

⋅ 2− weight%( )⋅−

1 K

T

K

c

⋅ K

1

⋅+:=

K

c

0.37

%CO

%TO

=K

c

1

K

T

K

1

⋅

t

01

τ1

1−

:=From Table 7-2.1:

(d) New moisture content of the product when the feed moisture content decreaseby 2 weight%. Controller is proportional only tuned for quarter decay ratio respons


Problem 7-23. Level control by manipulatiion of inlet flow.

f (t)

h(t)

f (t)

LC

LT

SP

i

o

Control valve fails closed.

Block diagram of the level control loop and required controller action.

Gc(s)

1/AsCset

(s) E(s) M(s)

C(s)

+ -

+

-H(s)

KT

Fo(s)

%CO

LT

%TO

ft

3

/min

ft

%TO

LC

Gv(s)Fi(s)

ft

3

/min

1/As

The only difference between this diagram and the one of Fig. 7-3.1 is that the controller manipulates the inlet flow instead of the outlet flow.

G

v

s( )

K

v

τv

s⋅ 1+=

The controller must be reverse acting (positive gain): increasing level decreases the controller output; this closes the control valve decreasing the inlet flow and the level drops.

Closed-loop transfer function.

C s( )

K G

c

s( )⋅

s τv

s⋅ 1+( )⋅ K G

c

s( )⋅+C

set

s( )

K

u

τv

s⋅ 1+( )⋅

s τv

s⋅ 1+( )⋅ K G

c

s( )⋅+F

o

s( )−=

where K

K

v

K

T

⋅

A

%TO

%CO min⋅= K

u

K

T

A

%TO

ft

3

=

The formulas of Section 7-3 apply to this case also. Only the action of the controller is different.



C

vmax

200% f

1d

⋅1Pa

1.45 10

4−⋅ psi

m

3

1000kg g⋅ h

d

⋅⋅

264.2gal

m

3

⋅60sec

min

⋅:=Size the valve:

f

d

0.013

m

3

s

=g 9.807

m

s

2

=G

f

ρ m

3

⋅

1000kg

=

3 eqns. 3 unks.∆p

v

t( ) ρ g⋅ h t( )⋅1.45 10

4−⋅ psi

Pa

=

Assume valve exit is at the level of the bottom of the tank and at the same pressure.

2 eqns. 3 unks. (∆pv)f

1

t( ) C

vmax

vp t( )⋅∆p

v

t( )

G

f

⋅m

3

264.2gal

min

60sec

=Valve equation:

1 eqn. 2 unks. (h, f1)π D

2

⋅

4

ρ⋅d h t( )⋅

dt

⋅ ρ f

i

t( )⋅ ρ f

o

t( )⋅− ρ f

1

t( )⋅−=Mass balance:

The model of Section 4-1.1 must be modified to account for the variable valve position.

(a) Model as in Section 4-1.1, draw the block diagram, determine the transfer functions and the maximum gain of a proportional controller for non-oscillatory response. Determine the effective time constants of the closed-loop at that gain, and the offset caused by a 0.001 m3/s.

G

c

s( ) K

c

=

Proportional controller:

τv

5s:=

f

1d

0.003

m

3

s

:=

Control valve sized for twice the design flow of

h

max

3m:=h

min

1m:=

Level transmitter:

h

d

2m:=D 3m:=

Problem data:

fi(t)

h(t)fo(t)

LCLT

SP

f1(t)

Problem 7-24. Comparison of tank level dynamic models of Section 4-1.1



G

2

s( )

K

2

τ s⋅ 1+=G

1

s( )

K

1

τ s⋅ 1+=where

H s( ) G

1

s( ) F

i

s( ) F

o

s( )−( )⋅ G

2

s( ) VP s( )−=Laplace transform and rearrange:

K

2

0.156

m

%CO

=τ min⋅

60s

157.08 min=

K

1

1.3 10

3

×s

m

2

=τ 9425 s=K

2

2 h

d

⋅

vp

d

:=K

1

2 h

d

⋅

k

v

vp

d

⋅:=τ

π D

2

⋅

4

2 h

d

⋅

k

v

vp

d

⋅⋅:=where

τd H t( )⋅

dt

⋅ H t( )+ K

1

F

i

t( )⋅ K

1

F

o

t( )⋅− K

2

VP t( )⋅−=

C

vmax

56.4

gal

min psi⋅= From Fig. C-10.1, p. 532, a 3-in valve is required.

C

vmax

110

gal

min psi⋅:=

Let f

1

t( ) k

v

vp t( )⋅ h t( )⋅=

where k

v

C

vmax

1000kg

m

3

g⋅1.45 10

4−⋅ psi

Pa

⋅⋅m

3

264.2gal

⋅min

60s

:= k

v

8.275 10

3−×

m

3

s m⋅=

Valve position at design conditions:vp

d

f

1d

k

v

h

d

⋅:= vp

d

25.6 %=

Substitute and simplify:π D

2

⋅

4

d h t( )⋅

dt

⋅ f

i

t( ) f

o

t( )− k

v

vp t( )⋅ h t( )⋅−=


π D

2

⋅

4

d H t( )⋅

dt

⋅ F

i

t( ) F

o

t( )− k

v

h

d

⋅ VP t( )−k

v

vp

d

⋅

2 h

d

⋅H t( )−= H 0( ) 0=

Rearrange into standard firts-order form:

As the (negative) controller gain is increased, this term decreases. When the term is negative the roots are complex and the response is oscillatory. The maximum gain at which the response is not oscillatory is when the term is zero:

τv

2

2 τv

⋅ τ⋅+ τ2

+ 4 τv

⋅ τ⋅− 4τv

τ⋅ K

p

⋅ K

c

⋅+ τv

τ−( )2 4 τv

⋅ τ⋅ K

p

⋅ K

c

⋅+=

The term inside the radical:

r

1

τv

τ+( )− τv

τ+( )2 4 τv

⋅ τ⋅ 1 K

p

K

c

⋅−( )⋅−+

2 τv

⋅ τ⋅=Roots of the caracteristic equation:

τv

τ⋅ s

2

⋅ τv

τ+( )s+ 1+ K

p

K

c

⋅− 0=Characteristic equation of the loop:

C s( )

K

p

− K

c

⋅

τv

s⋅ 1+( ) τ s⋅ 1+( )⋅ K

p

K

c

⋅−C

set

s( )

K

u

τv

s⋅ 1+( )⋅

τv

s⋅ 1+( ) τ s⋅ 1+( )⋅ K

p

K

c

⋅−F

i

s( ) F

o

s( )−( )⋅+=

Closed-loop transfer function:

The control valve must fail closed (air-to-open) to prevent emptying the tank on instrument power failure.

The controller is direct acting (negative gain): increasing level increases the controller output opening the valve and increasing the flow out of the tank; this decreases the level.

K

u

6.667 10

4

×%TO s⋅

m

3

=K

p

7.801

%TO

%CO

=K

u

K

T

K

1

⋅:=K

p

K

T

K

2

⋅:=Let

K

T

50

%CO

m

=K

T

100%TO

h

max

h

min

−:=Level transmitter LT:

G

v

s( )

1

τv

s⋅ 1+=

Valve positioner:G

c

s( ) K

c

=Proportional controller:

Gc(s)Cset

(s) E(s) M(s)

C(s)

+-

+

-H(s)

KT

Fi(s) - Fo(s)

%CO

LT

%TO

m

3

/sec

m

%TO

LC

Gv(s)

VP(s)

%VP

G2

(s)

G1

(s)

K

u

0.071

%TO

%CO m

3

⋅=

K

p

8.278 10

4−×

%TO

%CO sec⋅=K

u

4

πD

2

K

T

:=K

p

4

πD

2

k

v

h

d

⋅ K

T

⋅:=Let

The block diagram is the same as in part (a) with these transfer functions.

G

2

s( )

4

π D

2

⋅

k

v

h

d

⋅

s

⋅=G

1

s( )

4

πD

2

1

s

=where

G

1

s( ) F

i

s( ) F

o

s( )−( )⋅ G

2

s( ) VP s( )−=

H s( )

4

πD

2

1

s

F

i

s( ) F

o

s( )− k

v

h

d

⋅ VP s( )−( )=Laplace transform:

H 0( ) 0=π D

2

⋅

4

d H t( )⋅

dt

⋅ F

i

t( ) F

o

t( )− k

v

h

d

⋅ VP t( )−=

Express this linear equation in terms of deviation variables:

π D

2

⋅

4

d h t( )⋅

dt

⋅ f

i

t( ) f

o

t( )− k

v

h

d

⋅ vp t( )⋅−=Substitute into mass balance:

f

1

t( ) C

vmax

vp t( )⋅∆p

v

G

f

⋅m

3

264.2gal

min

60sec

= k

v

h

d

⋅ vp t( )⋅=

The model now neglects the effect of the level on the flow out of the tank:

(b) Repeat part (a) modeling the tank as in section 7-3.1 (integrating process).

Offset

K

T

2.826 10

3−× m=

Offset 0.141 %TO=Offset

K

u

1 K

p

K

cmax

⋅−0.001

m

3

s

:=Offset for change in inlet flow:

τe

9.995 s=τe

2 τv

⋅ τ⋅

τv

τ+:=

The equivalent time constants of the closed loop at this gain are two identical roots at:

K

cmax

60.3−%CO

%TO

=K

cmax

τv

τ−( )2−

4 τv

⋅ τ⋅ K

p

⋅:=



These results are identical to those of part (a),showing that for the purposes of the level controller, the effect of the level on the outlet flow is negligible. Notice that the results are the same although the models look quite different.

Offset

K

T

2.829 10

3−× m=

Offset 0.141 %TO=Offset

K

u

K

p

− K

cmax

⋅0.001

m

3

s

:=The offset for the change in inlet flow is:

τe

10 s=τe

2 τv

⋅:=

The equivalent time constants of the closed loop are identical for this gain (the radical is zero):

K

cmax

60.4−%CO

%TO

=K

cmax

1−

4 τv

⋅ K

p

⋅:=

The controller is, like before, direct acting (negative gain). As the negative controller gain increases in magnitude, the term inside the radical decreases and, when it becomes negative, the roots are complex and the response is oscillatory. At the maximum gain for non-oscillatory response, the term in the radical is zero:

1 4 τv

⋅ K

p

⋅ K

c

⋅+Term in the radical:

r

1

1− 1 4 τv

⋅ K

p

− K

c

⋅( )⋅−+

2 τv

⋅=Roots of the characteristic equation:

τv

s

2

⋅ s+ K

p

K

c


C s( )

K

p

− K

c

⋅

s τv

s⋅ 1+( )⋅ K

p

K

c

⋅−C

set

s( )

K

u

τv

s⋅ 1+( )⋅

s τv

s⋅ 1+( )⋅ K

p

K

c

⋅−F

i

s( ) F

o

s( )−( )+=


A τI

⋅ τv

⋅ s

3

⋅ A τI

⋅ s

2

⋅+ K

c

K

v

⋅ K

T

⋅ τI

⋅ s⋅− K

c

K

v

⋅ K

T


This means there is no offset for either set-point changes or disturbances.

Offset ∆c

set

∆c−= 0=∆c

K

c

− K

v

⋅ K

T

⋅

0 K

c

K

v

⋅ K

T

⋅−∆c

set 0

0 K

c

K

v

⋅ K

T

⋅−∆f

i

+= ∆c

set=

To obtain the steady-state transfer functions, set s = 0:

K

T

τI

⋅ s⋅ τv

s⋅ 1+( )⋅

A τI

⋅ s

2

⋅ τv

s⋅ 1+( )⋅ K

c

τI

s⋅ 1+( )⋅ K

v

⋅ K

T

⋅−F

i

s( )⋅+

C s( )

K

c

− τI

s⋅ 1+( )⋅ K

v

⋅ K

T

⋅

A τI

⋅ s

2

⋅ τv

s⋅ 1+( )⋅ K

c

τI

s⋅ 1+( )⋅ K

v

⋅ K

T

⋅−C

set

⋅ s( )⋅=

Substitute and rearrange:

C s( )

G

c

s( )− G

2

s( )⋅

1 G

c

s( ) G

2

s( )⋅−C

set

s( )

G

1

s( )

1 G

c

s( ) G

2

s( )⋅−F

i

s( )+=Closed-loop transfer function:

(a) Closed-loop transfer function, characteristic equation, and offset.

G

1

s( )

K

T

A s⋅=G

2

s( )

K

v

K

T

⋅

A s⋅ τv

s⋅ 1+( )⋅=G

c

s( ) K

c

1

1

τI

s⋅+⋅=PI controller:

Gc(s)

Cset

(s) E(s) M(s) C(s)

+-

+

-

Fi(s)

%CO

%TO

%TO

LC

G2

(s)

G1

(s)

Problem 7-25. Proportional-integral level control.




C s( ) C

set

s( )

1

K

c

− K

v

⋅

τI

s⋅

τI

s⋅ 1+F

i

s( )+=

So the dominant time constant is equal to the integral time of the controller. This pole however cancels the zero in the numerator for set point changes:

For very high controller gains:

A τI

⋅

K

c

− K

v

⋅ K

T

⋅s

2

τI

s⋅+ 1+ τI

s⋅ 1+=

K

c

− K

v

⋅ K

T

⋅

A

4

τI

<

As the controller gain is negative (direct acting controller), the roots are complex as long as the term in parenthesis is positive, that is, at low controller gains. The response will be oscillatory for:

K

c

K

v

⋅ K

T

⋅ τI

⋅( )2 4 K

c

⋅ K

v

⋅ K

T

⋅ A⋅ τI

⋅+ K

c

K

v

⋅ K

T

⋅ τI

⋅ K

c

K

v

⋅ K

T

⋅ τI

⋅ 4A+( )⋅=

The response i oscillatory when the term inside the radical is negative (complex conjugate roots).

r

1

K

c

K

v

⋅ K

T

⋅ τI

⋅ K

c

K

v

⋅ K

T

⋅ τI

⋅( )2 4A τI

⋅ K

v

− K

T

⋅ K

c

⋅( )⋅−+

2AτI

=Roots:

A τI

⋅ s

2

⋅ K

c

K

v

⋅ K

T

⋅ τI

⋅ s⋅− K

v

K

T

⋅ K

c

⋅− 0=Characteristic equation:

(c) For negligible valve time constant, determine the limits of the controller gain forwhich the loop is oscillatory. Dominant time constant at high controller gain.

ωK

c

− K

v

⋅ K

T

⋅

A τv

⋅=where

C t( ) A sin ω t⋅ φ+( )⋅=

Root r1 is cancelled by the zero in the numerator of the transfer function. The response of the level is oscillatory with no damping, as the controller gain is negative (direct acting controller). The loop gain does not affect the nature of the response, only the frequency of the oscillations that increases as the square root of the loop gain. The response is:

r

3

i−K

v

− K

T

⋅ K

c

⋅

A τv

⋅⋅=r

2

i

K

v

− K

T

⋅ K

c

⋅

A τv

⋅⋅=r

1

1−

τv

=

A τv

⋅ s

2

⋅ τv

s⋅ 1+( ) K

v

K

T

⋅ K

c

⋅ τv

s⋅ 1+( )⋅− τv

s⋅ 1+( ) A τv

⋅ s

2

⋅ K

v

K

T

⋅ K

c

⋅−= 0=τI

τv

=For

(b) Roots of the characteristic equation when the integral time is set equal to the valve time constant. Level response under these conditions.

The control valve fails closed (air-to-open) to prevent overflowing the evaporator on instrument ait failure.

The controller is reverse acting (positive gain): increasing level decreases controller output to close the valve and reduce the feed flow to the evaporator; this decreases the level.

Gc(s)Cset

(s) E(s) M(s)

C(s)

+ -

+

-H(s)

KT

Fp(s)

%CO

LT

%TO

ft

3

/min

ft

%TO

LC

Gv(s) G1

(s)

G1

(s)

Ws(s)

ft

3

/min

G2

(s)

-Fi(s)

lb/min


Solution:

∆p

v

5psi:=Assume

∆f 80

lb

min

=∆f 10% f

d

⋅:=

Disturbance, feed flow:

τv

2sec:=

Valve sized for 100% overcapacity, linear.

∆h

T

4ft:=A 10ft

2

:=

ρ 98

lb

ft

3

:=f

d

800

lb

min

:=

SPLC LT

AT

Vapors

Product

Feed

Steam

Condensate

Problem data:

Problem 7-26. Level control of an evaporator.


Characteristic equation of the loop:A s⋅ τ

v

s⋅ 1+( )⋅ K

c

K

v

⋅+ A τv

⋅ s

2

⋅ A s⋅+ K

c

K

v

⋅ K

T

⋅+= 0=

Roots of the characteristic equation:r

1

A− A

2

4 A⋅ τv

⋅ K

c

⋅ K

v

⋅ K

T

⋅−+

2 A⋅ τv

⋅=

r

2

A− A

2

4 A⋅ τv

⋅ K

c

⋅ K

v

⋅ K

T

⋅−−

2 A⋅ τv

⋅=

The roots are non-oscillatory as long as the term in the radical is positive (real roots). The maximum controller gain for which this is so is:

K

cmax

A

4 τv

⋅ K

v

⋅ K

T

⋅:= K

cmax

11.4

%CO

%TO

=

At this gain the term in the radical is zero and the effective time constants are identical to each other and equal to:

τe

2 τv

⋅:= τe

4 s=

Offset for a 10% change in the product and vapor flow:Offset 0

K

T

−

K

cmax

K

v

⋅ K

T

⋅

∆f

ρ−:=

Offset

K

T

0.011 ft=Offset 0.272 %TO=



Level transmitter LT:K

T

100%TO

∆h

T

:= K

T

25

%CO

ft

= G

c

s( ) K

c

=

Size the control valve:C

vmax

200%

f

d

ρ⋅

7.48gal

ft

3

ρ ft

3

⋅

62.4lb ∆p

v

⋅⋅:= C

vmax

68.4

gal

min psi⋅=

From Fig. C-10.1, page 532, a 3-in valve is required.C

vmax

110

gal

min psi⋅:=

Valve gain:K

v

C

vmax

100%CO

∆p

v

62.4⋅ lb⋅

ρ ft

3

⋅⋅

ft

3

7.48gal

:= K

v

0.262

ft

3

min %CO⋅=

G

v

s( )

K

v

τv

s⋅ 1+= G

1

s( )

1

A s⋅= G

2

s( )

E

ρ

1

A s⋅= E = evaporator economy, lb

vapors/lb steam.


C s( )

K

c

K

v

⋅ K

T

⋅

A s⋅ τv

s⋅ 1+( )⋅ K

c

K

v

⋅ K

T

⋅+C

set

s( )

K

T

A s⋅ τv

s⋅ 1+( )⋅ K

c

K

v

⋅ K

T

⋅+F

p

s( )

E

ρW

s

s( )+−=



τ´

D

τ2

=τ´

I

τ1

=K´

c

τ1

K τc

τ3

+( )⋅=

This is a series PID controller with the following tuning formulas:

G

c

s( )

τ1

K τc

τ3

+( )⋅1

1

τ1

s⋅+ τ

2

s⋅ 1+( )=

G

c

s( )

τ1

s⋅ 1+( ) τ2

s⋅ 1+( )⋅

K 1 τ3

s⋅−( )⋅

1 τ3

s⋅−

τc

s⋅ 1+ 1− τ3

s⋅+=

Substitue and reaarrange:

C s( )

R s( )

1 τ3

s⋅−

τc

s⋅ 1+=To avoid a postive pole in the controller, define:

This contains a positive pole which results in an unstable response when it doesn't exactly match the time constant of the transfer function.

G

c

s( )

τ1

s⋅ 1+( ) τ2

s⋅ 1+( )⋅

K 1 τ3

s⋅−( )⋅

1

τc

s⋅⋅=

The controller transfer function is:

C s( )

R s( )

1

τc

s⋅ 1+=For teh standard closed-loop response:

G

c

s( )

1

G s( )

C s( )

R s( )

1

C s( )

R s( )

−

=Synthesis formula, Eq. 7-4.2, page 258:Solution:

G s( )

K 1 τ3

s⋅−( )⋅

τ1

s⋅ 1+( ) τ2

s⋅ 1+( )⋅=

Problem 7-27. Sythesize controller for process with inverse response.


G s( )

2.05e

2.75− s

8.25s 1+

%TO

%CO

=

t

0

2.75 min=τ 8.25 min=t

0

t

2

τ−:=τ 1.5 t

2

t

1

−( )⋅:=

t

2

12 1−( )min:=64.8%TO 0.632∆c+ 67.391 %TO=

t

1

6.5 1−( )min:=64.8%TO 0.283∆c+ 65.96 %TO=Fit 3 model:

K

p

2.05

%TO

%CO

=

K

p

∆c

2%CO

:=

∆c 4.1 %TO=

∆c 68.9 64.8−( )%TO:=

Open-loop response to a 2%CO step change in the signal to the valve at 1 minute:

This reactor is simulated in Problem 13-5 and the temperature control loop is simulated in Problem 13-21. For the development of the simulation and the Simulink block diagrams, see the solutions to those problems.

Problem 7-28. Simulation of temperature control loop for the non-isothermalreactor of Section 4-2.3.


By comparison, the gain is slightly smaller and the integral is 60% slower than for quarter-decay ratio response.

G

c

s( ) 1.5

%CO

%TO

1

1

8.3s

+ 1.4s 1+( )=τ´

D

1.4 min=τ´

I

8.3 min=

K´

c

1.5

%CO

%TO

=τ´

D

t

0

2

:=τ´

I

τ:=K´

c

τ

K

p

t

0

⋅:=

From Table 7-4.1:

(b) Synthesis tuning with τc = 0 of series PID temperature controller.

Notice the inverse response of the reactor temperature to the change in reactants flow.

The controller output increases to close the coolant valve which is air-to-close (fails opened).

Response to a -0.2 ft3/min change in rectants flow at 1 min:

G

c

s( ) 1.8

%CO

%TO

1

1

5.5s

+ 1.4s 1+( )=τ´

D

1.4 min=τ´

I

5.5 min=

K´

c

1.8

%CO

%TO

=τ´

D

t

0

2

:=τ´

I

2 t

0

⋅:=K´

c

1.2

K

p

t

0

τ

1−

:=From Table 7-2.1:

(a) Quater-decay ratio tuning of series PID temperature controller.

The responses to a -0.2 ft3/min change in reactants flow at 1 min are:

Notice that the return to the set point is much slower than with quarter-decay ratio tuning.




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.

(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.

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.


Steady state model: Total mass balance: wC ww+ wP=

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

Solve for xP: 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

wPxP

w

FC

FT

FC

FT

SP

SPCaustic

Water

AC

31

2

Problem data:

wP 40klb

hr:= xP 30mass%:=

xC 50mass%:=

Assumptions:Perfect mixing•Constant mass•

Manipulated variables: Flows of water and caustic

wC and ww

Controlled variables: Product flow and mass fraction: wP and xP


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

mass% hr⋅

klb

KOL1 1,

µ1 1,1.25−=The gain is then:1

µ1 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.•

xPµ

0.6

0.4

0.4

0.6=µ

wC

wP

ww

wP

ww

wP

wC

wP

:=wP

wwwCRelative gains:

xPKOL

1

0.5

1

0.75−=KOL

1

ww xC⋅

wP2

klb

hr mass%⋅

1

wC− xC⋅

wP2

klb

hr mass%⋅:=

wP

wwwCOpen loop gains:

wC 24klb

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⋅


ww− xC⋅


wC− xC⋅


=

wC

wC ww+

ww

wC ww+

ww

wC ww+

wC

wC ww+

=

wP

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 1BTU

lb degF⋅:=ρ 8.33

lb

gal:=

T2 80degF:=T1 170degF:=

To 110degF:=fo 3gpm:=

Problem data:

TETC

Hotwater

Coldwater

S

SP

S

FE FCSP

f1 f2

fo

T2T1

To

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

Smith & Corripio, 3rd edition gpmgal

min:=

degF R:=


degF

gpm

KOL1 0,

µ1 0,30=

KOL0 1,

µ0 1,1.5=

The closed-loop gains are:1

µ1 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:

Toµ

0.333

0.667

0.667

0.333=µ

f1

fo

f2

fo

f2

fo

f1

fo

:=fo

f2f1

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

f2f1

(c) Relative gains and pairing that minimize interaction.

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

ToKOL

1

20

1

10−=KOL

1

f2 T1 T2−( )⋅

fo2

gpm

degF

1

f1− T1 T2−( )⋅

fo2

gpm

degF:=

fo

f2f1

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=

Economy: wv E wS⋅=

Solute balance: wF xF⋅ wP xP⋅=

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

wP E wS⋅+( )wP

⋅= xF 1E wS⋅

wP+⋅=

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

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

mP mSOpen-loop gains:

wF

KOL

δwF

δmP

δxP

δmP

δwF

δmS

δxP

δmS

=

KvP

KvP xF⋅E− wS⋅

wP2

⋅

KvS E⋅

KvS xF⋅E

wP⋅

=

xP


Problem 12-4. Control of an evaporator.

SP

SPSP

FC

FT

LC ACLT

ATFeed

Product

Vapors

Steam

mS

m PwF

xP

wP

wS

wv

xF

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.

Total mass balance: wF wP wv+=

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*WS), control throughput with the signal to the •valve on the product line and the composition with the signal to the steam valveif 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.


KOL1

0.417−

0.9

0.563=

xP

The first row is in klb/klb, and the second is in mass%/(klb/hr).Relative gains:

wP wS

wFµ

wP

wP E wS⋅+

E wS⋅

wP E wS⋅+

E wS⋅

wP E wS⋅+

wP

wP E wS⋅+

:= µ0.6

0.4

0.4

0.6=

xP

Pairing that minimizes interaction iscontrol throughput with the product rate•control composition with steam rate•

When the feed flow loop is closed, the product composition loop gain changes by1

µ1 1,1− 66.7 %= The closed-loop gain of the composition loop is:

mass% hr⋅

klb

KOL1 1,

µ1 1,0.938=



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

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

hr:= E 0.9

klb

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

wFKOL

1

xF− E⋅ wS⋅

wP2

klb

hr mass%⋅

E

xF E⋅

wP

klb

hr mass%⋅:=

Relative gains:

The first column is in mole%/(klb/hr) and the second in mole%/(lbmole/hr).

yDKOL

3.14−

1.71−

0.51

0.676=KOL

xB0 1,xB0 0,

−

wS0 1,wS0 0,

−

yD0 1,yD0 0,

−

wS0 1,wS0 0,

−

xB0 2,xB0 0,

−

wR0 2,wR0 0,

−

yD0 2,yD0 0,

−

wR0 2,wR0 0,

−

:=xB

wRwS

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 Test 1 Test 2Base

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.

SteamBottoms

DistillateFeed

Condenser

LC

PC

LC

AC

AC

1

2

1

2

LT

AT

LT

AT

PT

Reflux

Problem 12-6. Distillation product composition control.


wS wR

xBµ

KOL0 0,KOL1 1,⋅

KOL

KOL0 1,KOL1 0,⋅−

KOL

KOL0 1,KOL1 0,⋅−

KOL

KOL0 0,KOL1 1,⋅

KOL

:= µ1.697

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,

µ1 1,0.398=

mole% hr⋅

lbmole


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

dayf

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 volumeConstant standar density•

fp 60.0kbl

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 run3. 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

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

B0 0,

B0 0,

B2 0,

B0 0,

B1 0,

B0 0,

B0 1,

B2 1,

B2 1,

B2 1,

B1 1,

B2 1,

B0 2,

B1 2,

B2 2,

B1 2,

B1 2,

B1 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

kblyp 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 KOL1−

:= 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

KOL0 0,B

0 0,⋅

KOL1 0,B

0 1,⋅

KOL2 0,B

0 2,⋅

KOL0 1,B

1 0,⋅

KOL1 1,B

1 1,⋅

KOL2 1,B

1 2,⋅

KOL0 2,B

2 0,⋅

KOL1 2,B

2 1,⋅

KOL2 2,B

2 2,⋅

:=xp

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 KOL1−

:=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/sG 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− e320− s

166s 1+

21−

10 s⋅ 1+

0.050−

10 s⋅ 1+

1.033−

47s 1+

=

TOR, N-mTime is in sec

SW, kg/sMW, kg/sSF, kg/s

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

Manipulated variables:SF = solids flowMW = mill water flowSW = slurry tank water

Controlled variables:TOR = mill torqueFML = flow from millDCF = cyclone feed density AT

XTFT AC

XC FC

DCF

TORFML

Mill

Cyclone SW

SF

MW

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


Block diagram of decoupled system:

DCF, kg/liter

FML, liters/secKOL D⋅

32.372

0

0

0

0.258

0

0−

0

0.771−

=

TOR, N-m Decoupled system gains: kg/seckg/seckg/sec

mDCFmFMLmTOR

All variables in kg/sec

SWset

MWset

SFkg/s

MWkg/s

SWkg/s

µ

KOL0 0,B

0 0,⋅

KOL1 0,B

0 1,⋅

KOL2 0,B

0 2,⋅

KOL0 1,B

1 0,⋅

KOL1 1,B

1 1,⋅

KOL2 1,B

1 2,⋅

KOL0 2,B

2 0,⋅

KOL1 2,B

2 1,⋅

KOL2 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

B1 0,

B0 0,

B2 0,

B0 0,

B0 1,

B1 1,

1

B2 1,

B1 1,

B0 2,

B2 2,

B1 2,

B2 2,

1

:= D

1

0.407−

1.163

1.664−

1

2.144−

0.244

0.053−

1

=

R1(s)

R2(s)

R3(s)

M1(s)

M2(s)

M3(s)

SF

MW

SW

TOR

FML

DCF

GC1(s)

GC2(s)

GC3(s)

G11(s)

G12(s)

G13(s)

G21(s)

G22(s)

G23(s)

G31(s)

G32(s)

G33(s)

1.664-

0.244

0.407

0.053

1.163

2.144-

-

-

+

+

+

++

+++

+++

+++

+

+

+

-

-

-

Instrumentation diagram of decoupler:

AT

XTFT

ACXC FC

DCF

TORFML

Mill

SW

SF

MW

XY FY AY

-1.664

0..244-0.407 -0.053

1.163

-2.144

TORSP FMLSP DCFSP

33 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.


xBµ

1.217

0.217−

0.217−

1.217=

yD

mSmR

µ

KOL0 0,KOL1 1,⋅

KOL

KOL−( )0 1,

KOL1 0,⋅

KOL

KOL−( )0 1,

KOL1 0,⋅

KOL

KOL0 0,KOL1 1,⋅

KOL

:=

xBKOL

3.0−

2.5

0.75

3.5−

%TO

%CO:=

yD

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

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 0.35s+MR 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 0.35s+MS 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

mRManipulated variables:Reflux flowSteam flow to reboiler

Controlled variables:Distillate purityBottoms product purity

SteamBottoms

DistillateFeed

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.


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

0.751 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

2.51 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:

G11(s)

G12(s)

G21(s)

G22(s)

D1(s)

D2(s)

Gc1(s)

Gc2(s)

R1(s)

R2(s)

MR(s)

MS(s)

YD(s)

XB(s)

+

+

+

+

+

+

+

+

+

+

-

-


D2 s( )

GP2 2,− s( )

GP2 1,s( )

= 0.9−

1.1

1.5s 1+

2.0s 1+e

1 0.3−( )s−=GP2 2,s( ) GP2 1,

s( ) D2 s( )⋅+ M2 s( ) 0=

D1 s( )

GP1 1,− s( )

GP1 2,s( )

= 0.81−

1.2

2.4s 1+

1.4s 1+e

0.6 1.1−( )s−=GP1 2,s( ) D1 s( )⋅ GP1 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.

c2µ

1.234−

2.234

2.234

1.234−=µ

KOL0 0,KOL1 1,⋅

KOL

KOL0 1,− KOL1 0,

⋅

KOL

KOL0 1,− KOL1 0,

⋅

KOL

KOL0 0,KOL1 1,⋅

KOL

:=c1

m2m1

Relative gains by Eq. 12-2.7, page 418:

KOL0.81

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( )

0.81e0.6− s

1.4s 1+

1.1e0.3− s

1.5s 1+

1.2e1.1− s

2.4s 1+

0.9es−

2s 1+

=

C s( ) GP s( ) M s( ) GPU s( ) U s( )⋅+=

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


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:

GP12(s)

D1(s)

D2(s)

Gc1(s)

Gc2(s)

R1(s)

R2(s)

M2(s)

M1(s)

C1(s)

C2(s)

+

+

+

+

++

++

+

+

-

-

GP11(s)

GU1(s)

GU2(s)

GP22(s)

GP21(s)+

+

U(s)


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


Simplify: So, the system would be decoupled.C2 s( ) Gv2 s( ) G22 s( )

G12 s( )

G11 s( )G21 s( )⋅−⋅ M2 s( )⋅=


w1 50lb

min= 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=

xw1 x1⋅ w2 x2⋅+

w1 w2+=

∆xw1 w2+( )x1 w1 x1⋅− w2 x2⋅−

w1 w2+( )2Kv1 ∆m1⋅

w1 w2+( )x2 w1 x1⋅− w2 x2⋅−

w1 w2+( )2Kv2 ∆m2⋅+= 0=

w2 x1 x2−( )⋅ Kv1⋅ ∆m1⋅ w1 x2 x1−( )⋅ Kv2⋅ ∆m2⋅+ 0= ∆m2

Kv1 w2⋅

Kv2 w1⋅∆m1=

Kv1

wmax

100%CO:= Kv2

wmax

100%CO:= Kv1 1.5

lb

min %CO⋅= Kv2 1.5

lb

min %CO⋅=

The gains of the decouplers are: Kv1−

Kv21−=

Kv1 w2⋅

Kv2 w1⋅1=


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

ACSP

AT

SP

wx

11 wx

wx

22

m

m

1

2V

FC

FT

Design conditions:

x1 10mass%:= x2 30mass%:=

w 100lb

min:= x 20mass%:=

Valves are linear with constant pressure drop.

wmax 150lb

min:=

Analyzer transmitter AT:

xmin 5mass%:= xmax 35mass%:=

At the design conditions: w1 wx x1−

x2 x1−⋅:= w2 w w1−:=

+

SP

AT

SP

wx

11 wx

wx

22

m1

m2

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

Kv133.33 %CO=

w2

Kv233.33 %CO=


RatioxF

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

wFmax

100%CO E⋅ KvS⋅mFC

KvP

E KvS⋅mP−=wF

set wFmax

100%CO=

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

wFset

E KvS⋅

KvP

E KvS⋅mP−=wS

wFset

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

SPSP

FC

FT

LC ACLT

ATFeed

Product

Vapors

Steam

mS

m PwF

xP

wP

wS

wv

xF

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


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:

RatioxF

xPset

wFmax

100%CO KvP⋅=

mP Ratio mFC⋅= where

Instrumentation diagram:

SP

SP

SP

FC

FT

LC

AC

LT

ATFeed

Product

Vapors

Steam

mS

m P

wF

xP

wP

wS

xF

FY

AY

3

X

mFC

RatioFY: 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.


f t( ) 0dynes=y 0( ) 100cm=For this problem:

k 1816gm

s2=k

M g⋅27cm

:=g 980.7cm

s2=M 50gm:=From the solution of problem 2-9:

y t( )and a second integartion givesd y t( )⋅

dtIntegration of this second derivative results in:

d2 y t( )

dt2g−

f t( )M

+kM

y t( )−=Solve for the highest derivative:

Md2 y t( )

dt2⋅ M− g⋅ k y t( )⋅− f t( )+=

The differential equation representing the motion of the bird mobile is, from Problem 2-9:

13-1. Simulation of Bird Mobile of Problem 2-9.

Solutions to Problems 13-1 to 13-17

Chapter 13. Simulation of Process Control Systems

The period of oscillation is, as in the solution to problem 2-9:

Period 2 π⋅Mk

⋅:=

Period 1.043 s=

The number of complete cycles in 10 seconds is:

10sPeriod

9.592=

The simulation plot shows the same result.


Period 1.795 s=Period2 π⋅

gL

:=gL

3.501 Hz=The frequency of oscillation is:

(Table 2-1.1)x t( ) x 0( ) cosgL

t=r2 i−gL

⋅=r1. igL

⋅=Roots:

s2 gL

+ X s( ) 0=The solution of the differential equation:

x 0( ) 0.1m=L 0.8m:=M 0.5kg:=g 9.807m

s2=

d2 x t( )

dt2gL

− x t( )=

Substitute and simplify to obtain:

tan θ( ) sin θ( )= x t( )L

=

For small angles θ, from the geometry:

M− g⋅ tan θ( )⋅ Md2 x t( )

dt2⋅=

Application of Newton's Second Law of Motion:

Mgx(t)

L

Mg sin2

2

13-2. Simulation of a Pendulum

The number of oscillations in 10 s is:

10sPeriod

5.572=



wi 0.194kgs

=wi Ao 2M 601300⋅ Pa

Rg T⋅⋅ 601300Pa po−( )⋅:=

Assume the compressor is initially off and comes on after 200 s (five time constants) with the exact flow required to maintain the initial pressure:

τ 42.895 s=τ

V 2M

Rg T⋅⋅ 601300⋅ Pa 500000⋅ Pa⋅

Ao 2 601300⋅ Pa po−( )⋅:=

From the linearization of Problem 2-23, we know that the time constant is:

p 0( ) 500000 101300+( )Pa=T 70 273.16+( )K:=Rg 8.314Pa m3⋅

mole K⋅⋅:=

po 101300Pa:=M 29gm

mole:=Ao 0.785cm2

:=V 1.5m3:=Problem parameters:

d p t( )⋅

dt

Rg T⋅

V M⋅wi t( ) wo t( )−( )=Substitute and solve for dp(t)/dt:

ρ t( )M

Rg T⋅p t( )=Ideal gas law, assuming constant temperature:

wo t( ) Ao 2 ρ t( ) p t( ) po−( )⋅=Flow through the orifice:

Vdρ t( )

dtwi t( ) wo t( )−=

In Problem 2-23 the mass balance on the tank produced the following equation:

13-3. Simulation of Punctured Air Tank of Problem 2-23.

As predicted by the linearized model, the pressure reaches steady state in about 200 s.


Run the simulation for 25 hrs (five time constants). Simulate the oven as a step function from an inital temperature of 535ºR to 800ºR.

d T t( )⋅

dtσ ε⋅ A⋅M cv⋅

Ts t( )4 T t( )4−=

Integrate the differential equation :

τ 5.16 hr=τM cv⋅

4 σ⋅ ε⋅ A⋅ 535R( )3⋅

:=

By the linearization done in Problem 2-24, the time constant of the turkey is:

σ 0.1718 10 8−⋅

BTU

hr ft2⋅ R4:=T 0( ) 535R=cv 0.95

BTUlb R⋅

:=

ε 0.6:=Ts 800R:=A 3.5ft2:=M 12lb:=The parameters, given in this problem are:

M cv⋅dT t( )

dt⋅ σ ε⋅ A Ts t( )4 T t( )4

−⋅=

From the solution to Problem 2-24, the differential equation obtained from an energy balace on the turkey is:

13-4. Simulation of the turkey temperature response of Problem 2-24.

From the response, the time constant is much less than 5 hr. This is because the time constant gets smaller with temperature. At 800ºR it is:

τM cv⋅

4 σ⋅ ε⋅ A⋅ 800R( )3⋅

:= τ 1.54 hr=

From the response, the actual time constant seems to be about 2 hr, which is more in line with how long it takes to cook a turkey.


E 27820BTU

lbmole:=

Rg 1.987BTU

lbmole R⋅:= ρ 55

lb

ft3:= Cp 0.88

BTUlb R⋅

:= A 36ft2:=fc 0.8771ft3

min:=

ρc 62.4lb

ft3:= ∆Hr 12000−

BTUlbmole

:= U 75BTU

hr ft2⋅ R⋅:= Vc 1.56ft3:= Cpc 1

BTUlb R⋅

:=

Check that initial conditions are at steady state (derivatives = 0):

rA ko e

E−

Rg 678.9⋅ R⋅ 0.2068

lbmole

ft3

2⋅:= rA 0.039

lbmole

ft3 min⋅=

fV

cAi 0.2068lbmole

ft3−⋅ rA− 5.87− 10 4−

×lbmole

ft3 min⋅=

fV

Ti 678.9R−( )∆Hrρ Cp⋅

rA⋅−U A⋅

V ρ⋅ Cp⋅678.9 602.7−( )R− 7.915− 10 3−

×R

min=

fcVc

Tci 602.7R−( ) U A⋅Vc ρc⋅ Cpc⋅

678.9 602.7−( )R+ 0.027−R

min=

The following is the Simulink diagram for the reactor:

13-5. Non-isothermal Chemical Reactor of Section 4-2.3lbmole 453.59mole:=Rearranging the model equations from Section 4-2.3:

d cA t( )⋅

dtf t( )V

cAi t( ) cA t( )−( ) rA t( )−= cA 0( ) 0.2068lbmole

ft3=

rA t( ) ko e

E−

Rg T t( )⋅⋅ cA

2⋅ t( )=

d T t( )⋅

dtf t( )V

Ti t( ) T t( )−( )∆Hrρ Cp⋅

rA t( )⋅−U A⋅

V ρ⋅ Cp⋅T t( ) Tc t( )−( )−= T 0( ) 678.9R=

d Tc t( )⋅

dt

fc t( )

VcTci t( ) Tc t( )−( ) U A⋅

Vc ρc⋅ Cpc⋅T t( ) Tc t( )−( )+= Tc 0( ) 602.7R=


ft3:= Ti 633.5R:= f 1.3364

ft3

min:= Tci 540R:=

Parameters:V 13.46ft3:= ko 8.33 108

⋅ft3

lbmole min⋅:=

The following are the responses for a 0.25 ft3/min increase in process flow at 1 minute followed by a 0.1 ft3/min increase in coolant flow at 30 minutes.

Observe the inverse response in the reactor temperature for the change in process flow.


The folowing is the Simulink diagram for the mixer:

f2 37.5galmin

=f2 f f1−:=

f1 62.5galmin

=f1 fcA2 0.025mole cm 3−

⋅−

cA2 cA1−⋅:=

f1 cA1⋅ f2 cA2⋅+ f cA⋅− 0=At the initial steady state:

(assuming constant volume)f1 f2+ f=Total mass balance:

Ah 200gal:=f 100galmin

:=cA2 0.05mole

cm3:=cA1 0.01

mole

cm3:=Problem parameters:

cA 0( ) 0.025mole

cm3=

d cA t( )⋅

dt

f1 t( ) cA1 t( )⋅ f2 t( ) cA2 t( )⋅+ f t( ) cA t( )⋅−

A h⋅=

The model equation, from the solution to Problem 3-1:

13-6. Mixing Process of Problem 3-1

The responses to a step increase in f1 from 62.5 to 67.5 GPM at 1 minute:

The concentration response is typical first-order with a time constant of approximately 2 min.


13-7. Feedback control of composition in mixer of Problem 3-1

Introduce the following blocks from the Chapter 13c public nodels:Figure 13-4.1, F401Vlv1, control valve with time constant of 1 min, linear, maximum flow of •100 gpm, and initial condition of 37.5%C.O.Figure 13-4.3, F403PI, PI controller with initial condition of 37.5% CO.•Figure 13-4.7, F407Trmr, transmitter with 1 min time constant, range of 0 to 1 mole/cm3, •and initial condition of 0.025 mole/cm3

The controller was tuned for quarter decay ratio response with a gain of 20%CO/%TO and an intgral time of 1.5 min.

This is the Simulink diagram of the loop (the mixer block is the one from Problem 13-5):

The response to a 5 gpm increase in f1 at 1 minute is:

The outlet flow increases by 5 gpm at 1 min and then the controller increases f2 to bring the outlet concentration back up to the set point.

The high controller gain rsults in a very minor deviation of the outlet concentration from its set point.


The responses to a 5 ft3/min step increase in process flow are:

The Simulink diagram is given by:

k 1 min 1−=k

f cA1 0.5lbmole ft 3−⋅−⋅

V 0.5⋅ lbmole ft 3−⋅

:=Initial conditions at steady state:

Di 5.5in:=Lp 400ft:=V 150ft3:=Problem parameters:

cA1 2lbmole

ft3:=f 50

ft3

min:=Design conditions:

cA3 t( ) cA2 t to−( )=

cA2 0( ) 0.5lbmole

ft3=

d cA2 t( )⋅

dtfV

cA1 t( ) cA2 t( )−( ) kcA2 t( )−=

The model equations, from the solution to Problem 3-2, are:

13-8. Isothermal reactor of Problem 3-2

The concentration response shows a time constant of about 0.75 min, a dead time of a little over 1 min, and a steady state change of 0.38 lbmole/ft3.

The values from the linear model are:

τV

f k V⋅+:= τ 0.75 min=

toπ Di

2⋅ Lp⋅

4f:= to 1.32 min=

∆cA2

cA1 0.5lbmole

ft3−

f k V⋅+5⋅

ft3

min:=

∆cA2 0.038lbmole

ft3=


The responses to a 0.1 step increase in feed composition are:

The Simulink diagram for this problem is:

z0 0.513=

y0 0.625=z0V y0⋅ L 0.4⋅+

F:=y0

0.4⋅

1 1−( )0.4+:=

At initial steady state:

2.5:=M 500kmole:=Problem parameters:

V 5kmole

s=V F L−:=L 5

kmoles

:=F 10kmole

s:=Design conditions:

y t( )x t( )⋅

1 1−( )x t( )+=

F V L+=

x 0( ) 0.4=d x t( )⋅

dt1M

F z t( )⋅ V y t( )⋅− L x t( )⋅−( )=

Rearranging the model equation developed in Problem 3-11: kmole 1000mole:=13-9. Flash drum of Problem 3-11

These are typical first-order responses with a time constant of about 50 s and a gain on x of about 1 which match the results of the linear model in the solution of Problem 3-11.


The responses to a 20 ft3/min increase in inlet flow are:

The Simulink diagram for the tray is:

As the response is fast, convert time units to s by multiplying the derivative by 60 s/min.

h0 0.136 ft=

h0fo

0.415 w⋅ 2 g⋅⋅

1

1.5

:=fo fi:=fi 30ft3

min:=Initial steady state conditions:

S 11.2ft2:=w 3ft:=Problem parameters:

fo t( ) 0.415 w⋅ h t( )1.5⋅ 2 g⋅⋅=

d h t( )⋅

dt1S

fi t( ) fo t( )−( )=

The model equations from the solution to Problem 3-12 are:

13-10. Distillation tray of Problem 3-12

The first-order response has a time constant of approximately 2 s and the steady-state change is about 0.054 ft. The time costant matches the one from the linerized model from the solution of Problem 3-12. Using the gain from that solution, the steady-state change in level should be:

20ft3 min 1−⋅

331.4ft2 min 1−⋅

0.06 ft= close!

The students can check the results for the change in 10 ft3/min.


From Table 7-1.1, for a series PID controller tuned for quarter decay ratio response:

KcKcu1.7

:= τITu2

:= τDTu8

:= Kc 147−%CO%TO

= τI 1.5 min= τD 0.38 min=

Control valve (from F401Vlv1): Kv 0.0542m3

min %CO⋅:= τv 0.1min:= (Solution of Problem

6-11)

Initial position:f2Kv

44.28 %CO=

Transmitter (from F407Trmr): τT 3min:= Initial output: 50 20−

70 20−60 %TO=

The series PID Controller block is taken from the Public Model Library, F405PIDs

The Simulink block diagram for the blender conentration control loop is:

13.11. Blending tank of Problems 3-18 and 6-11 The diagram for the blender is essentially the same as for Problem 13-6 with slightly different notation and the following parameter and design values:

%CO %:=c1 80

kg

m3:= c2 30

kg

m3:= c0 50

kg

m3:= f 4

m3

min:= V 40m3

:= %TO %:=

At the intial steady state: f f1 f2+= f c0⋅ f1 c1⋅ f2 c2⋅+=

f2 fc0 c1−

c2 c1−⋅:= f1 f f2−:= f1 1.6

m3

min= f2 2.4

m3

min=

From the results of Problem 6-11, the ultimate gain and period are:

Kcu 250−%CO%TO

:= Tu 3.01min:=

The responses to a 0.1 m3/min increase in f1 are:

The decay ratio is somewhat greater than 1/4. Students may adjust the controller tuning parameters to improve the response.

Notice that the concentration can be controlled very tightly.

f2 f1:= f1 3m3

min=

Cv1f1

h10:= Cv2

f2

h20:= Cv1 1.897

m2.5

min=

Cv2 1.897m2.5

min=

The linearized gains and time constants are:

K12 h10⋅

Cv1:= τ1

2 A1⋅ h10⋅

Cv1:= K1 1.667

min

m2= τ1 15 min=

K2Cv1Cv2

h20h10

⋅:= τ22 A2⋅ h20⋅

Cv2:= K2 1= τ2 15 min=


13-12. Non-interacting tanks in series of Fig. 4-1.1The model equations developed in Section 4-1.1 are:

d h1 t( )⋅

dt1

A1fi t( ) fo t( )− f1 t( )−( )= h10 2.5m:=

d h2 t( )⋅

dt1

A2f1 t( ) f2 t( )−( )= h20 2.5m:=

f1 t( ) Cv1 h1 t( )⋅= f2 t( ) Cv2 h2 t( )⋅=

Design conditions: fi 5m3

min:= fo 2

m3

min:=

Problem parameters: A1 9m2:= A2 9m2

:=

At initial steady state: f1 fi fo−:=

The responses to a 0.2 m3/min step increase in inlet flow are:

The responses of the level and flow for the first tank are first-order with a time constant of 15 min. The gains are 1.0 for the flows and the steady-state changes in level are about 0.35 m, as predicted by the linear model:

K1 0.2⋅m3

min0.333 m=

The responses for the second tank are second order with the same steady-state change in level, meaning that the gain K2 is unity as predicted by the linear model.


13-13. Interacting tanks of Fig. 4-2.1

The model equations developed in Section 4-2.1 are the same as for problem 13-12, except for the flow betwen the tanks:

f1 t( ) Cv1 h1 t( ) h2 t( )−⋅=

Design conditions are the same except for the initial condition in tank 1: h10 5m:=

At the initial steady state: Cv1f1

h10 h20−:= Cv1 1.897

m2.5

min=

It can be shown that the gain of the inlet flow on the level in the second tank is the same as K1 in Problem 13-12.

K1 1.667min

m2=

The following is the Simulink diagram for the interacting tanks is series:


The change in the level in the second tank is the same as in Problem 13-12. Students may want to study the effect of reducing the resistance between the two tanks by changing the initial condition on h1 and recalculating Cv1. For example, for h10 = 2.6 m,

Cv1f1

2.6m h20−:=

Cv1 9.487m2.5

min=

The response of the second tank becomes first-order and the two tanks behave as a single tank.



τ2 5 min=τ1 5 min=K2 0=τ2V2

fA fB+:=τ1

V1fA

:=K2fB

fA fB+:=

K1 1=K1fA

fA fB+:=This problem is linear with a gains and time constants:

T4 500 K=T2 500 K=T4fA T2⋅ fB T3⋅+

fA fB+:=T2 T1:=At initial steady state:

T3 500K:=T1 500K:=fB 0m3

min:=V2 5m3

:=V1 5m3:=fA 1

m3

min:=

Design conditions:

d T4 t( )⋅

dt1

V2fA T2 t( )⋅ fB T3 t( )⋅+ fA fB+( ) T4 t( )⋅−=

d T2 t( )⋅

dt

fAV1

T1 t( ) T2 t( )−( )=

The model equations developed in Section 4-1.2 are:

13-14. Non-interacting thermal tanks in series of Fig. 4-1.5


The resposes to a 10 K step increase in inlet temperature are:

The response for the first tank is first-order with a unity gain and a time constant of 5 min, matching the theoretical model. The response for the second tank is second-order also with unity ain for these conditions.

Students may study the effect of changing flows fA and fB and temperature T3 on these responses.

13-15. Interacting thermal tanks of Fig. 4-2.4

The model equations developed in Section 4-2.2, ignoring fB, are:

d T1 t( )⋅

dt1

V1fA T1 t( )⋅ fR T4 t( )⋅+ fA fR+( ) T2 t( )⋅−=

d T4 t( )⋅

dt1

V2fA fR+( ) T1 t( ) T4 t( )−( )⋅=

The design conditions and problem parameters are the same as in Problem 13-14, plus the recycle flow:

fR 1m3

min:= (The results for fR = 0 are identical to those of Prob. 13-14)

Note: In the model of Section 4-2.2, the recycle flow is assumed to be 0.2*(f A + fB).


The temperature responses for a 10 K step increase in inlet temperature are:

The students should study the effect of the recycle flow on the responses. As the recycle flow is increased, the temperatures in the two tanks approach each other and the two tanks behave as one perfectly mixed tank with the combined volume of the two tanks. They should notice that increasing the recycle flow does not appreciably change the time to steady state, or the gain.


The Simulink diagram for the reactors is:

cA20 0.767lbmole

ft3=cA20

f1 cA10⋅

f1 k2 V2⋅+:=

cA10 1.726lbmole

ft3=cA10

fo cAo⋅

f1 k1 V1⋅+ fRf1

f1 k2 V2⋅+⋅−

:=

cA20f1 cA10⋅

f1 k2 V2⋅+=

f1 20ft3

min=f1 fo fR+:=At the initial steady state:

fR 10ft3

min:=For the base case let:

V2 125ft3:=V1 125ft3:=fo 10ft3

min:=

k2 0.2min 1−:=k1 0.2min 1−

:=cAo 7lbmole

ft3:=Design conditions from Problem 6-15:

cA2 0( ) cA20=d cA2 t( )⋅

dt

f1V2

cA1 t( ) cA2 t( )−( ) k2 cA2 t( )⋅−=

cA1 0( ) cA10=d cA1 t( )⋅

dt1

V1fo cAo t( )⋅ fR cA2 t( )⋅+ f1 cA1 t( )⋅−( ) k1 cA1 t( )⋅−=

The model equatios are developed in the solution to Problem 4-9:

13-16. Reactors with recycle of Problems 4-9 and 6-15

The responses to a 0.5 lbmole/ft3 step increase in inlet concentration with a recycle flow of 10 ft3/min are:

Students shall study the effect of changing the recycle flow as indicated in the statement of the problem. Notice that the initial steady state conditions vary with the recycle flow.


The Simulnk block diagram for the extractor is:

f2 14088.6m3

min=

f2Ka V⋅ c10 me c20⋅−( )⋅

2 c20⋅:=

c20 1.2776 10 4−×

kmole

m3=

c202− f1⋅ ci c10−( )⋅ Ka V⋅ c10⋅+

Ka me⋅ V⋅:=

c10 0.04kmole

m3=c10 1 Rec−( ) ci⋅:=At the initial steady state:

V 25m3:=Ka 3.646min 1−

:=me 3.95:=Problem parameters:

Rec 90%:=ci 0.4kmole

m3:=f1 5

m3


c2 0( ) c20=d c2 t( )⋅

dtKa c1 t( ) me c2 t( )⋅−( )⋅

2 f2 t( )⋅

Vc2 t( )−=

c1 0( ) c10=d c1 t( )⋅

dt

2 f1 t( )⋅

Vci t( ) c1 t( )−( ) Ka c1 t( ) me c2 t( )⋅−( )⋅−=

The model equations developed in the solution to Problem 4-6 are:

13-17. Extraction process of Problem 4-6

The responses to a 1000 m3/min step increase in solvent flow are:

Obviously the problem parameters are unreasonable. The large solvent flow makes for an almost instantaneous response of the extract composition. The effect on the raffinate composition is negligible, as the extract composition is essentially zero under the design conditions.

Ask students to try more reasonable parameter values:

me 0.95:= Ka 209min 1−:=


The Simulink diagram for the cooler is:

fc0 0.172m

3

min=fc0

U A⋅

ρc cpc⋅

T0 Tc0−( )Tc0 Tci−( )

:=

Tc0 35.5 degC=Tc0 T0

f ρ⋅ cp⋅

U A⋅Ti T0−( )−:=At the initial steady state:

cpc 4.2kJ

kg degC⋅:=ρc 1000

kg

m3

:=cp 3.8kJ

kg degC⋅:=

ρ 800kg

m3

:=Vc 1.1m3

:=A 4m2

:=U 200kJ

min m2degC⋅

:=V 5m3

:=Problem parameters:

Tci 25degC:=T0 45degC:=f 0.1m

3

min:=

Ti 70degC:=Design conditions:

Tc 0( ) Tc0=d Tc t( )⋅

dt

fc t( )

VcTci Tc t( )−( ) U A⋅

Vc ρc⋅ cpc⋅T t( ) Tc t( )−( )+=

T 0( ) T0=d T t( )⋅

dt

f t( )

VTi t( ) T t( )−( ) U A⋅

V ρ⋅ cp⋅T t( ) Tc t( )−( )−=

kJ 1000joule:=degC K:=The model equations, from the solution to Problem 4-7, are:

13-18. Temperature control of stirred tank cooler of Problems 4-7 and 6-19


Chapter 13. Simulation of Process Control Systems (continued)

fc0

0.008m

3

min %CO⋅

21.5 %CO=Controller output:

45 20−

70 20−50 %TO=Transmitter output:Initial conditions:

τD 0.95 min=

τI 3.8 min=Kc 44−%CO

%TO=τD

Tu

8:=τI

Tu

2:=Kc

Kcu


Tu 7.59min:=Kcu 75.4−%CO

%TO:=

From the solution to Problem 6-19:%TO %:=%CO %:=

To complete the temperature control loop, install, from the Public Model Library:A flow control loop, f401Vlv1, with a gain of 0.008 m3/min-%CO, negligible time constant.•A temperature transmitter, f407Trmr, with a range of 20 to 70 C and a time constant of 0.6 min.•A series PID controller, f405PIDs, tuned for quarter dcay ratio•

The responses to a 0.2 m3/min step increase in process flow at 5 min, and a 2 C step increase in set point at 30 min, are:

The controller output saturates for the change in process flow and is barely able to return the temperature to the set point. The controller output also temporarily saturates on the change in set point, but is able to recover. Saturation is a form of nonlinear behavior. In this case it causes the decay ratio to be much less than 1/4.

Encourage your students to figure out what needs to be changed so that the controller does not saturate.


V 5gal:=

kv1.954 10⋅

100%CO

lb

min:= kv 0.0618

lb

min %CO⋅=

Steam enthalpy, saturated at 1 atm., referenced to 32 F, is: 1150.4 BTU/lb

hs 1150.4BTU

lbcp 32⋅ degF+:= hs 1176

BTU

lb=

At the initial steady state: w2

f1 ρ⋅ cp⋅ T30 T1−( )⋅

hs cp T30⋅−:= w2 2.518

lb

min=

vp0

w2

kv:=

vp0 40.8 %CO=

Note: The value of w2 does not match the one in the problem statement.

Simulate the valve by inserting the block f401Vlv1 from the Public Model Library, linear, with a gain kv and a time constant of

τv 4s:= τv 0.067 min=

The Simulink block diagram is:

gpmgal

min:= degF R:=13-19. Direct contact heater of Problem 4-5


d T3 t( )⋅

dt

1

Vf1 t( ) T1 t( )⋅ f3 t( ) T3 t( )−( )⋅

hs

V ρ⋅ cp⋅w2 t( )⋅+= T3 0( ) T30=

w2 t( ) 1.954 10 vp t( )⋅= kv vp t( )⋅=

f3 t( ) f1 t( )w2 t( )

ρ+=

Substitute to eliminate f.3(t):d T3 t( )⋅

dt

f1 t( )

VT1 t( ) T3 t( )−( )

w2 t( )

V ρ⋅ cp⋅hs cp T3 t( )⋅−( )+=

Design conditions: f1 25gpm:= T1 60degF:= T30 80degF:=

Problem parameters: ρ 7lb

gal:= cp 0.8

BTU

lb degF⋅:=


The response to a 5 gpm step increase in process flow at 0.1 min and a 5 %CO step increase in sgnal to the steam control valve at 1 min, are

The time constant of the mixer is matches the value of 0.2 min obtained by linearization; the steady state changes are a bit less than predicted by the linear model.

The respons to the process flow is first-order and the one to the signal to the valve is second-order. This is because of the lag in the valve.

This is the plot also obtained with the following Simulink diagram:


0 5 10 15 20 250

2 105

4 105

6 105

fs p1 ∆p psi⋅( )scfh

∆p

fs p1 ∆p( ) 836scfh

gpm

R

psi⋅ Cv⋅ Cf⋅

p1

G T⋅⋅ y p1 ∆p( ) 0.148 y p1 ∆p( )3−⋅:=

y p1 ∆p( ) 1.63

Cf

∆p

p1⋅:=Functions for valve flow:

T 259 460+( )R:=Cf 0.8:=G 0.621:=p1 34.7psia:=Cv 440gpm

psi:=

scfhft

3

hr:=psia psi:=Problem parameters from Example 5-2.2:

GMw

29=y

1.63

Cf

∆p

p1⋅=

fs 836 Cv⋅ Cf⋅p1

G T⋅⋅ y 0.148 y

3⋅−( )=

The flow through a gas control valve can be calculated from Eq. 5-2.3:

13-20. Gas flow control valve of Example 5-2.2

This Simulink block can simulate any gas valve with variable inlet pressure and pressure drop. To obtain the flow in scf/min, divide the Cv,max/100 by 60 min/hr.

The Simulink block diagram for the temperature control loop is:

τD 0.84 min=τI 3.4 min=Kc 5.9%CO

%TO=τD

Tu

8:=τI

Tu

2:=Kc

Kcu

1.7:=

From Table 7-1.1, the quarter decay rato tuning parameters are:

Tu 6.7min:=Kcu 10%CO

%TO:=

With the lags on the valve and the transmitter, the loop can be made unstable--it wasn't for the conditions of Problem 6-12. The ultimate gain and period ar determined from the simulation:

678.9 640−

700 640−64.8 %TO=Transmitter initial condition:

ln 0.5( )

ln 50( )− 17.72 %CO=Initial position:2 0.8771⋅

ft3

min1.754

ft3

min=Valve maximum flow:

To the block developed in Problem 13-5 to simulate the reactor, we add the following blocks from the Public Model Library:

f401Vlv1: equal precentage control valve with a time constant of 0.1 min and sized for 100% •overcapacity, air-to-closef405PIDs: a series PID controller tuned for quarter decay ratio response•f407Trmr: transmitter with a range of 640 to 700 R and a time constant of 1.0 min•

13-21. Temperature control of reactor of Section -2.3 and Problem 6-12

The responses to a -0.2 ft3/min step change in reactants flow at 1 min followed by a 1 R step increase in setpoint at 30 min are:

The decay ratio is slightly greater than 1/4. The students should play with the tuning of the controller and test it for different magnitudes and directions of the input changes.


13-22. Temperature control of stirred tank heater of Example 13-4.1 with variable pressure drop across the control valve

The steam chest pressure is calculated as a function of the steam temperature using the Antoine equation for water (Reid, Prausnitz, and Sherwood, 1977, The Properties of Gases and Liquids, 3rd ed., Appendix.

p2 t( )14.7psia

760mmHge

18.30363816.44K

Ts t( ) 460R+( )K1.8R

46.13K−

−

=

The Simulink block diagram of Fig. 13-4.9 is modified as follows:

The "Gas Valve 2" is a modified version of the gas valve of Problem 13-20 in that it accepts the upstream and downstream pressures p1 and p2--instead of ∆p--and it outputs the flow in lb/min. This is done by multiplying the output by the molecular weight and dividing by 380 scf/lbmole and by 60 min/hr.

Responses to this model are compared with those of Fig. 13-4.10 by running both versions of the control loop in parallel and plotting the reponses together:

The responses to the decrease in process flow are almost identical, but those for the increase in flow are different. This is because as the steam pressure increases the capacity of the valve decreases and the flow of steam does not vary as much as when the flow is assumed to be independent of pressure. The response is then slower and less oscillatory.

This is an excellent example of the effect of a saturation nonlinearity that is often overlooked in modeling.


The Simulink diagram for the control loop is:

m0 40.65 %CO=m0

f

Kv:=Initial conditions:

Kv Ko⋅100 0−( )%TO

1 0−( )lb gal1−

⋅⋅ 1.845=Process gain:

Ko 0.0075lb

gal gpm⋅:=Kv 2.46

gpm

%CO:=From the solution to Problem 6-17:

The Simulink diagram for each reactor is the same as in Problem 13-8, with a different notation and without the transportation lag in the outlet pipe (here it is asssumed e reactors are close to each other and the tranporation lag is negligible).

Making a subsystem block of each reactor, we build the concentratioon control loop by adding, from the Public Model Library:

f401Vlv1: a linear control valve sized for 100% overcapacity and a time constant of 0.1 min•f405PIDs: a series PID controller tuned for quarter decay ratio response•f407Trmr: a concentration transmitter with a range of 0 to 1.0 lb/gal and negligible lag.•

c

4

2

1

0.5

lb

gal=c

3

f c2

⋅

f k V⋅+:=c

2

f c1

⋅

f k V⋅+:=c

1

f c0

⋅

f k V⋅+:=At the initial steady state:

V 1000gal:=k 0.1min1−

:=c0

4lb

gal:=f 100gpm:=Design conditions:

c0

t( ) ci t( )=and

For j = 1,2,3cj

0( ) cj0=d c

j⋅ t( )

dt

f t( )

Vc

j 1− t( ) cj

t( )− k cj

⋅ t( )−=

The model equation for each reactor is developed in the soution to Problem 6-17:

13-23. Concentration control of isothermal reactors in series of Problem 6-17

From a step test in controller output, the time constant is about 9 min and the dead time is negligible, so the process is very controllable. The quarter decay ratio tuning formulas impose no limit on the gain or on how small the integral time can be, so we use the synthesis formulas:

Kc adjusutable τI τ= 9min= τD 0min= (PI controller)

The responses to a 1 lb/gal increase in inlet eactants concentartion (a change in inlet flow is not possible since the flow is the manipulated variable), with Kc = 10%CO/%TO, are:

The problem will be more interesting with a 1 min time lag in the transmitter. The students can be asked try this case and to play with the tuning parameters, e.g., smaller integral time for faster response.

Notice the deviation in concentration is small. Students could also examine effect of the controller gain on the magnitude of the deviation.


T30 200degF:=

Problem parameters: U 136BTU

hr ft2degF⋅

:= D 3ft:= A 974ft π⋅ 0.5⋅ in:= A 127.5 ft2

=

ρ 53lb

ft3

:= CM 974ft 0.178⋅lb

ft0.12⋅

BTU

lb degF⋅:= CM 20.8

BTU

degF=

cp 0.45BTU

lb degF⋅:= ps 115psig:= 872.9

BTU

lb:=

At the initial steady state: f3 f1:= Ts0 T30

ρ cp⋅ f1⋅

U A⋅T30 T1−( )+:= Ts0 343.4 degF=

ws

U A⋅ Ts0 T30−( )⋅:= ws 47.48

lb

min=

The Simulink diagram for the heater is:

13-24. Temperature and level control of oil heater of Problem 6-24The model equations from the solution to Problem 6-24 are:

d h t( )⋅

dt

ft3

7.48gal

4

πD2

f1 t( ) f3 t( )−( )= h 0( ) h0=

d T3 t( )⋅

dt

f1 t( )

V t( )T1 t( ) T3 t( )−( )⋅

U A⋅

V t( ) ρ⋅ cp⋅Ts t( ) T3 t( )−( )+= T3 0( ) T30=

V t( ) 7.48gal

ft3

⋅π D

2⋅

4⋅ h t( )=

d Ts t( )⋅

dt

1

CMws t( )⋅ U A⋅ Ts t( ) T3 t( )−( )⋅−= Ts 0( ) Ts0=

Assume the pressures are constant and determine the flows as follows:

Inlet oil flow f1(t): linear valve with variable pressure drop-Model Library f402Vlv2-sized for 50% overcapacity and pressure drop:

∆pv1 p1 14.7psi+ p2−( ) ρ g⋅ h t( ) 5ft−( )⋅−=

Outlet oil flow f3(t): Step function input (it is the disturbance)

Steam flow ws(t): linear control valve with constant pressure drop-Model Library f401Vlv1-szed for 50% overcapacity.

psig psi:=

Design conditions: p1 45psig:= p2 40psia:= f1 100gpm:= T1 70degF:=

τD 0.14 min=τI 0.56 min=Kc 45%CO

%TO=

τD

Tu

8:=τI

Tu

2:=Kc

Kcu

1.7:=

Quarter decay tuning parameters from Table 7-1.1:

Tu 1.12min:=Kcu 77.1%CO

%TO:=Ultimate gain and period of temperature controller:

m10 46.67 %CO=m1032.2

1.5 46⋅:=

ms0 60.58 %CO=wsmax 78.4lb

min=ms0

ws

wsmax:=wsmax

46

41.81.5⋅ ws⋅:=

50% for both transmittersIntial conditions, from the results of Problem 6-24::

To complete the control loops, from the Public Model Library, introduce:f405PIDs: series PID temperature controller tuned for quarter decay ratio response•f407Trmr: temperature (TT-50) and level (LT-50 transmitters with 100 to 200 F and 7 to 10 ft •ranges, and 0.5 min and 0.01 min time constants, respectively.

The responses to a 5 gpm increase in inlet oil flow are:

Students should be encouraged to study the effect of problem parameters. For example, a more realistic valve time constant of 0.1 min will result in more reasonable controller parameters.

Notice the small offset in level. This is for a level controller gain of 20%CO/%TO.


13-25. Moisture control of drier of Problem 9-1. Simple feedback vs. cascade

In the solution to Problem 9-1 he drier is represented by linear transfer functions as it will be simulated here. In the absence of information, the transfer function will be the same to a change in ambient temperature as for a change in heater outlet temperature.

(a) The Simulink diagram for the single moisture feedback control loop is

(b) The Simulink diagram for the cascade control scheme using the heater outlet temperature as the intermediate variable is:


The respones to a 10ºF step increase in ambient temperature at 1 min, and a 0.3 %moisture increase in set point at 30 min, are:

The cascade response is in magenta and the feedback in yellow. To obtain these results the gain of the TC-47 controller had to be reduced to 2 %CO/%TO from the 4.14 determined in Prpblem 9-1.

The cascade reponse is less oscillatory and faster at the expense of larger changes and oscillations in the fuel flow.

The controller parameters are:

Feedback: Cascade:

Kc 0.42−%CO%TO

:= 0.56−%CO%TO

τI 3.5min:= 1.0min

τD 0.86min:= 0.26min


The feedback control responses are in yellow and the cascade responses in magenta. They show almost perfect control to the disturbance input and a slightly faster response to the set point change for the cascade scheme at the expense of very oscillatory response of the manipulated variable.

Students should be encouraged to study how adjustment of the secondary gain may reduce the oscillations of the manipulated variable and how it affects the response of the controlled variable and/or the master controller tuning.

The block diagrams are given in the solution to Problem 9-3, and the responses to unit step changes in disturbace at 1 min and set point at 75 min are:

τI 10 min=Kc 6.1%CO%TO

=τITu1.2

:=KcKcu2.2

:=

Tu 12.5min:=Kcu 13.5%CO%TO

:=

Kc2 2.2%CO%TO

:=Cascade Control. From the solution to Problem 9-3, with


=τITu1.2

:=KcKcu2.2

:=

From Table 7-1.1, the quaerte decay tuning parameters for a PI controller are:


:=

Simple Feedback Loop. From the solution to Problem 9-3, the ultimate gain and period are:

13-26. Feedback vs. cascade control of Problem 9-3

The responses to a 5ºC step increase in set point are:


τD 0.83 min=τI 3.9 min=Kc 19%CO%TO

=

τD 0.482 τ⋅t0τ

1.137

⋅:=τIτ

0.878

t0τ

0.749

⋅:=Kc1.435

Kp

t0τ

0.921−

⋅:=

τ 12.8min:=Kp 0.385%TO%CO

:=t0 2.2min:=

Cascade Control (with the secondary proportional controller gain set at 2%CO/%TO):


=

τD 0.482 τ⋅t0τ

1.137

⋅:=τIτ

0.878

t0τ

0.749

:=Kc1.435

Kp

t0τ

0.921−

:=

The tuning parameters for minimum IAE response, from Table 7-2.2:

t0 3.5min:=τ 13.5min:=Kp 0.285%TO%CO

:=

Simple Feedback Loop:

The block diagrams from the solution to Problem 9-4 are simulated as in Problem 13-25 with a parallel PID master controller and a proportional slave controller with a gain of 2%CO/%TO. From the open-loop responses to step changes in contrller output, the following parameters are obtained:

13-27. Jacketed reactor of Problem 9-4. Simple feedback vs. cascade control

The responses show a faster response for the cascade scheme (magenta) than for simple feedback (yellow), at the expense of doubling the variation in the controller output. In fact, because this linear analysis does not impose limits on the controller output, we see increases of 200 and 400%CO, which are not possible in practice. Pointing out these restrictions is an advantage of simulation which is not obtained from the anlytical analysis.

The responses to a 5 ft3/min step decrease in process flow at 3 min followed by an increase of 10ºF in inlet temperature at 30 min are:

The Simulink diagram for the ratio control scheme is obtained by a simple modification of the diagram of Fig. 13-4.11:

Ratio 3.75min %TO⋅

ft3=Ratio

ρ cp⋅ Tset Ti−( )⋅ 100%TOfmax

⋅:=

Tset 150degF:=The ratio for set point of

fmax 75lb

min:=Ti 100degF:=966

BTUlb

:=cp 0.8BTU

lb degF⋅:=ρ 68

lb

ft3:=

where the ratio is adjusted by the operator or a feedback controller to obtain the desired outlet temperature. For the conditions of the problem, the ratio is:

wset t( )⋅ρ cp⋅ Tset t( )⋅ Ti−⋅

f t( )= Ratio f t( )⋅=

If the variation in the inlet temperature diturbance is neglected, the feedforward controller calculation becomes:

13-28. Feedforward temperature control of heater of Example 13-4.2

The responses are identical for the change in process flow, but the ratio controller (magenta) does not take action on the change in inlet temperature resulting in an error of about 10ºF. This reponse is slower than the one for the flow change, so a feedback controller can adjust the ratio to maintain the outlet temperature at the set point.


The Simulink diagram for the ratio control scheme is:

Kc 166−%CO%TO

=Kc 147−%CO%TO

100%CORmax

⋅f2max

f1⋅:=

The controller gain must be adjusted by the factor the output is multiplied by:

Rmax f1max⋅

100%CO f2max⋅0.0221

1%CO

=

The signal from the flow transmitter must then be scaled by the factor:

(at 100%CO)Rmax 3:=f2max 5.42m3

min:=f1max 4

m3

min:=Transmitter ranges:

Ratio 1.5=Ratiof2f1

:=f2 2.4m3

min:=f1 1.6

m3

min:=

Design conditions:

The diagram of Problem 13-11 is modified to introduce a sesor/transmitter for the concentrated stream flow and a multiplier to represent the ratio controller. The transmitter has a ange of 0 to 4 m3/min and a time constant of 0.75 min. The multiplier allows the feedback controller to adjust the ratio.

13-29. Ratio control of blending tank of Problems 3-18 and 13-11

The responses to a 0.1 m3/min step increase in process flow at 1 min followed by a 2 kg/m3 step increase in concentrated stream composition at 15 min are:

The ratio controller (magenta) results in a smaller initial deviation than the simple feedback (yellow) for the change in process flow. The responses are essentially the same for the change in concentration.


The Simulink diagram for the cascade control loop is:

τD 0.47 min=τI 1.9 min=τDTu8

:=τITu2

:=

Kc 6%CO%TO

=KcKcu

1.7:=Quarter decay ratio tuning from Table 7-1.1:

Tu 3.75min:=Kcu 10.5%TO%CO

:=Master loop:

τI τ:=Set Kc to 2%CO/%TO to obtain fast response without much oscillation.

Synthesis tuning from Table 7-4.1:

t0 0min:=τ 3min:=K2 1.95%TO%CO

:=Slave loop:

An open-loop step test in the slave controller results in the following parameters:

The Simulink diagram of Problem 13-21 is modified to introduce, from the Public Model Library,A jacket temperature transmitter with a range of 560 to 660ºR and a time constant of 1.0 min•f403PI: a PI controller for the jacket temperature tuned by the synthesis method.•

13-30. Cascade temperature control of reactor of Section 4-2.3

The responses to a 0.2 ft3/min step decrease in reactants flow at 2 min followed by a 1ºR step increase in set point are:

The cascade responses (magenta) are faster than the ones for simple feedback control (yellow) at the expense of more oscillations in the manipulated variable.

Students should be encouraged to study the effect of the tuning parameters. Encourage them to try the formulas for tuning cascade controllers in Table 9-3.1.


The Simulink diagram for the linear feedforward controller is:

τD 1.6 min=τI 6.5 min=

Kc 2.5−%CO%TO

=τD3.25min

2:=τI 2 3.25⋅ min:=Kc

1.20.98−

3.256.75

1−:=

The feedback controller must be retuned because of the replacement of the equal-percentage valve with the coolant flow control loop. From Table 7-2.1 for quarter-decay ratio response:

Bias 83.86− %TO=Bias 1.255− 66.82⋅ %TO:=

m 0( ) m 0( ) 1.255 66.82⋅ %TO+ Bias+= m 0( )=Bias required for the same set point:

1.3364ft3

minKDT⋅ 66.82 %TO=Initial reactant transmitter output:

To start with a correct initial set point of the coolant flow controller, a bias is needed:

delay 4.5 min=

13-31. Linear feedforward control of reactor of Section 4-2.3 and Problem 13-21

For the linear feedforward controller, use Eq. 11-2.1, page 379:

FFCGD−

HD GM⋅=

The reactant flow transmitter is taken from the Public Model Library, f407Trmr, with a range 0 to 2 ft3/min and a tim constant of 0.1 min:

HDKDT

0.1 s⋅ 1+= KDT

100%TO

2ft3 min 1−⋅

:= KDT 50%TO

ft3 min 1−⋅

=

From step tests on the disturbance, the folowing parameters are obtained:

GD 61.5%TO

ft3 min 1−⋅

e 5.5− s

9s 1+⋅=

FFC61.5− %TO

KDT ft3⋅ min 1− 0.98−( )⋅

6.75s 1+

9s 1+⋅ e 5.5 3.25−( )s−=

GM 0.98−%TO%CO

e 3.25− s

6.75s 1+⋅=

61.5− %TO

KDT ft3⋅ min 1− 0.98−( )⋅1.255

%CO%TO

=Feedforward controller gain:

FFC 1.225%CO%TO

6.75s 1+

9.s 1+⋅ e 2.25− s 1+

⋅=

This calls or a lead of 6.75 min, a lag of 9 min and a delay of 2.25 min. To simplify the design, let's drop the lead-lag unit since the lead and lag are about the same, and increase the delay by the difference:

delay 2.25min 9 6.75−( )min+:=

The responses to a 0.2 ft3/min step decrease in reactants flow at 1 min are

The feedforward responses are inmagenta and the simple feedback responses are in yellow. The feedforward controller required some adjustments:

Delay 3.5min:=

Kc 5−%CO%TO

:=

τI 3min:=

The delay is necessary so that the feedforward action does not aggravate the initial inverse response of the temperature to the reactants flow. The delayed feedforward correction reduces the downward temperature deviation and the oscillation.

The reason the manipulated responses are offset is that one is adjusting the air-to-close equal percentage valve and the other the coolant flow set point.


The feedback controller will adjust the ratio in the range 0 to Rmax as its output varies from 0 to 100%CO, so the signal from the process flow transmitter must be scaled. Use flow transmitter ranges of 0 to 2 ft3/min for the process flow and 0 to 1.75 ft3/min for the coolant flow.

fc 0.876ft3

min=fc Ratio f⋅:=f 1.3364

ft3

min:=Check:

Ratio 0.655=Ratioρ cp⋅ Ti T−( )⋅ ∆Hr−( ) cAi cA−( )⋅+

ρc cpc⋅ Tc Tci−( )⋅:=

Ratio at design conditions:

cA 0.2068lbmole

ft3:=cAi 0.5975

lbmole

ft3:=

Tc 602.7R:=Tci 540R:=T 678.9R:=Ti 635R:=

13-32. Ratio control of reactor of Section 4-2.3 and Problem 13-21

Design the feedforward controller by the procedure of Section 11-6, page 395,with the reactants flow as the only major disturbance:

Control objective: T(t) = Tset(t) (reactor temperature)1.Manipulated variable: fc(t) (coolant flow to the jacket)2.Disturbances: f(t) (reactants flow)3.Steady state balances:4.

V rA t( )⋅ f t( ) cAi cA−( )⋅=

f t( ) ρ⋅ cp⋅ Ti T t( )−( )⋅ V rA t( )⋅ ∆Hr−( )⋅+ U A⋅ T t( ) Tc−( )⋅= fc t( ) ρc⋅ cpc⋅ Tc Tci−( )⋅=

Combine to eliminate rA(t):

f t( ) ρ cp⋅ Ti Tset t( )−⋅ ∆Hr−( ) cAi cA−( )⋅+⋅ fc t( ) ρc⋅ cpc⋅ Tc Tci−( )⋅=

Solve for the manipulated variable:fc t( ) Ratio t( ) f t( )⋅=

6. Feedback trim: The feedback controller will adjust the ratio when the set point or some of the other disturbances changes7. Dynamic compensation: From inspection, we would say that a net lead will probably be required because the coolant flow must overcome the lag of the jacket, while the reantants flow does not. However, from the results of Problem 13-31 we know we need a delay of 3.5 min.

lbmole 453.59mole:=

Design conditions: ρ 55lb

ft3:= cp 0.88

BTUlb degF⋅

:= ∆Hr 12000−BTU

lbmole:= ρc 62.4

lb

ft3:=

cpc 1BTU

lb degF⋅:=


Kcnew 3.3−%CO%TO

=Kcnewfcmax

Rmax f⋅Kc⋅:=Kc 5−

%CO%TO

:=

Kc KcnewRmax fmax⋅

fcmax 100⋅ %CO⋅

ffmax⋅ 100⋅ %TO=

Rmax f⋅

fcmaxKcnew⋅=

To maintain the same loop gain, the feedback controller gain must be adjusted:

fcfcmax

50 %TO=

ffmax

66.8 %TO=m0 32.8 %CO=m0RatioRmax

100⋅ %CO:=Initial conditions:

yRmax fmax⋅

fcmax 100⋅ %COr⋅ x⋅=Scaled feedforward equation:

fcmax100%TO

yRmax

100%COr⋅

fmax100%TO⋅ x⋅=

Substitute into the feedforward equation:

rRatioRmax

100⋅ %CO=yfc

fcmax100⋅ %TO=x

ffmax

100⋅ %TO=The scaled variables are:

fcmax 1.75ft3

min:=fmax 2

ft3

min:=Rmax 2:=

The responses to a 0.2 ft3/min step decrease in reactants flow at 2 min are:

The response with the ratio controller (magenta) is superior to the one for simple feedback (yellow). To improve the response the delay was reduced to 3.0 min.

The reason the manipulated variable responses are offset is that the feedback controller is adjusting the position of the air-to-close equal- percentage valve and the ratio is adjusting the coolant flow set point. This is also teh reason they move in opposite directions.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposeonly 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.

The Simulink diagram for the cascade control scheme is:

Kc1 5%CO%TO

:=Set τD1 0min:=τI1 5min:=Use synthesis tuning for the master also:

t0 0min:=τ .1 5min:=K1 3.3%TO%CO

:=

With these slave settings, an open-loop step test on the master controller output:

Kc2 5%CO%TO

:=Set atKc2 tunable=τI2 τ2:=Synthesis tuning from Table 7-4.1:

t0 0min:=τ2 6min:=K2 0.58%TO%CO

:=

%TO %:=%CO %:=From the results of an open-loop step test on the slave controler output:

To the diagram of Problem 13-23 add, from the Public Model Library:A concentration transmitter with a range of 0 to 4 lb/gal and a time constant of 1 min•f403PI: a PI controller tuned by the synthesis formula•

13-33. Cascade control of reactors in series of Problem 13-23



The responses to a 0.05 lb/gal step increase in set point at 1 min and a 1 lb/gal step increase in reactant concentration at 10 min are:

The set point response of the cacade scheme (magenta) is faster than the one for simple feedback (yellow) at the expense of slamming the control valve opened for a period of time. For the change in reactant concentration, the concentation in the third reactor moves in the opposite directon for the cascade scheme. This is because the slave controller detects the change faster and moves the valve to correct. As the change in reactants flow affects the concentration in the third tank faster than the change in inlet concentration, the concentration decreases on the drecrease in flow.


The Simulink diagram to simulate tank is:

wP t( ) wc t( ) ww t( )+=Total balance:

x 0( ) x0=d x t( )⋅

dt1M

wc t( ) xc t( )⋅ wP t( ) x t( )⋅−( )=Caustic balance:

Assume perfectly mixed tank with constant mass:

ww 16klbhr

=ww wP wc−:=

wc 24klbhr

=wcwP x0⋅

xc:=


M 10klb:=Problem parameters:

xc 50mass%:=

x0 30mass%:=wP 40klbhr

:=

mass% %:=Design conditions: klb 1000lb:=

AC

SP

AT

FCSP

FT

wPx

ww

FC

FT

FC

FT

SP

SPCaustic

Water

wc

1

2

3

3xc

13-34. Multivariable control of caustic blending tank of Problem 12-2.

With these tuning parameters, the responses to a

The controllers cannot be tuned for quarter-decayb ratio response because the dead time of the loops is zero. They were tuned onserving the simulation responses to obtain:

FC-3: Kc = 0.90%CO/%TO and integral time of 0.1 min•AC-3: Kc = -100%CO/%TO and integarl time of 10 min (the process integral time is 15 min)•

x050mass%

60 %TO=30mass%AC-3:

wP

60klb hr 1−⋅

66.667 %TO=40klbhr

FC-3:

mc0 40 %CO=mc0wc

60klb hr 1−⋅

:=24klbhr

FC-2:

mw0 26.667 %CO=mw0ww

60klb hr 1−⋅

:=16klbhr

FC-1:Initial conditions:

To complete the control loops insert, from the Public Model Library:f403PI: two PI controllers to control the product flow and composition•f407Trmr: two transmitters, one for the composition, AC-3, with a range of 0 to 50 mass% and •a time constant of 1 min, and one for the product flow, FC-3, with a range of 0 to 60 klb/hr and negligible time constantTwo flow controllers, FC-1 and FC-2, with ranges of 0 to 60 klb/hr and time constants of 0.1 •minute. These are modifications of the simple valve model and can be copied from f411ffst.

The Simulink diagram for the linear decoupler is:

Bw 26.667− %CO=Bw D21− mc0⋅:=mw0 mw0 D21 mc0⋅+ Bw+=

Bc 26.667 %CO=Bc D12− mw0⋅:=mc0 mc0 D12 mw0⋅+ Bc+=

Initial outputs: To start at the proper decoupler outputs, biases must be added to the decouplers:

D21 0.667%CO%CO

=D21wwwc

:=D21Kvc− Kxc⋅

Kvw Kxw⋅=

wwwc

=

Kxwwc−

wP=Kxc

wwwP

=∆x Kvc Kxc⋅ D21 Kvw⋅ Kxw⋅+( )∆mFC=

D12 1−%CO%CO

:=D12Kvw−

Kvc= 1−

%CO%CO

=

Kvc Kvw= 60 klb⋅ hr 1−⋅

100%CO=∆wP Kvw D12 Kvc⋅+( )∆mAC= 0=

To keep the flow constant, the caustic flow must be adjusted as follows:

(c) Linear decoupler design as in Example 12-3.1

(a) These are the responses for the correct pairing: caustic flow controls the product flow and water flow controls the composition. The composition control has less than 1/4 deacay ratio.

Part (b) with the opposite pairing is left as an exercise to the students.

Note: as these loops are so controllable, both pairings should produce about the same performance. For part (b) the composition controller must be reverse acting.

The responses to a 10 klb/hr step increase in product flow set point at 2 min and a 1 mass% step increase in product composition set point at 20 min are:

For these ideal conditions the decoupler schemes (magenta) results in perfect control for both changes.

Students should be encouraged to design and test a basic principles decoupler similar to the one of Example 12-3.4


2 eqns. 2 unk.(M,x)

Expand and substitute total balance:

M t( )d x t( )⋅

dt⋅ x t( ) wF t( ) wv t( )− wP t( )−( )⋅+ wF t( ) xF t( )⋅ wP t( ) x t( )⋅−=

Simplify: d x t( )⋅

dt1

M t( )wF t( ) xF t( ) x t( )−( )⋅ wv t( ) x t( )⋅+= x 0( ) x0=

Economy relates steam and vapor flows: wv t( ) E ws t( )⋅=

At the intial steady state:

wPwF xF⋅

x0:= wv wF wP−:= ws

wvE

:= wP 8571lbhr

= ws 43609lbhr

=

Maximum hold up: Mmax MminM0 Mmin−

hL0+:= Mmax 1100 lb=

13-35. Multivariable control of evaporator of Figure 12-3.4

wP x

wS

wF xF

Ts

T

wV

Condensate

Cond.

Design conditions:

wF 50000lbhr

:= xF 12mass%:=

x0 70mass%:= E 0.95:=

M0 747lb:= hL0 50%TO:=

Process parameters:

Mmin 394lb:=

Developent of the model equations (assume perfect mixing):

Total mass balance: d M t( )⋅

dtwF t( ) wv t( )− wP t( )−= M 0( ) M0=

Sugar balance: d M t( ) x t( )⋅( )⋅

dtwF t( ) xF t( )⋅ wP t( ) x t( )⋅−=

The evaporator is thus simulated as an integrating process and does not operate well without the level controller. The complete control system includes, from the Public Model Library:

A proportional level controller with the gain set at Kc = 20 %CO/%TO (tight control)•f407Trmr: two transmitters, one for the product composition with a range of 40 to 90 mass% •sugars and a time constant of 0.6 min, and one for the feed flow with a range of 0 to 70000 lb/hr and a negligible time constantf401Vlv1: three control valves sizd for 100% overcapacity, one on the feed, one on the product, •and one on the steamf403PI: two PI constrollers, one for the product composition manipulating the control valve on •the product and the other one on the feed flow manipulating the steam control valve

Control valve gains: Feed = 50000100%CO

lbhr

2⋅ 1000lb

hr %CO⋅=

Product =wP 2⋅

100%CO171.4

lbhr %CO⋅

=

Steam =ws 2⋅

100%CO872.2

lbhr %CO⋅

=

Initial valve positions are 50%. Initial transmitter outputs:x

x0 40mass%−

90 40−( )mass%100⋅ %TO 60 %TO=

wfwF

70000lb hr 1−⋅

100⋅ %TO 71.4 %TO=

The Simulink diagram for the evaporator is:

The Simulink diagram of the control system is

The responses to a 1000 lb/hr step increase in feed flow set point at 5 min and a 1 mass% step increase in product composition set point at 20 min are:

This is for the recommended pairing. The responses shows that an evaporator can be controlled very tightly. Normally the feed flow will not be stepped because the sudden change in steam flow will upset the steam header.

Students may be asked to try alternative pairings of the three controlled and manipulated variables.


The Simulink disgram for the control system is:

τI 60min:=Kc 3−%CO%TO

:=Bottoms composition:

τI 60min:=Kc 5%CO%TO

:=Distillate composition:

Open loop step tests on the two composition controller outputs showed no dead time on the response of the composition, so the controllers cannot be tuned by the minimum IAE formulas. Trial and error tuning resulted in the following tuning parameters:

KWsFC 1957lb

hr %TO⋅=mB0 50%CO:=KWsFC

97873lbhr

2⋅

100%TO:=Steam flow:

KLrFC 96.04lbmolehr %TO⋅

=mD0 50%CO:=KLrFC

4802lbmole

hr2⋅

100%TO:=

Reflux flow:

lbmole 453.59mole:=Flow control loop gains and initial conditions sized for 200% of design flows:

The model equations, design conditions, and process parameters are given in Example 13-5.1. To complete the composition loops add, from the Public Model Library:

Two flow control loops for the reflux and steam flows (these are copied from the other two •"valves" used on the distillate and bottoms flows).f407TRmr: two transmitters with ranges of 0 to 1.0 mole fraction and 1 min time cnstants for the •distillate and bottoms compositions.f403PI: two PI controllers to be tuned for minimum IAE.•

13-36. Control of distillation column of Example 13-5.1

The responses to a 200 lbmole/hr step increase in feed flow at 20 min are

Surprisingly, although the two controllers were tuned independently, each with he other loop opened, they both performed satisfactorily when both loops were closed.

Students may be encouraged to test the tuning and to observe the responses to other disturbances.


The Simulink diagram for the blender is

V 0.450kbl:=Process parameter:

Reformate

Straight runlsolve

1

x0

y0

1

x1

y1

1

x2

y2

f

f x0

⋅

f y0

⋅

7.5

28.125

24.375

kbl

day=

Alkylate


y

5

11

3

:=x

97

80

92

:=

y0

7:=x0

87:=f 60kbl

day:=Design conditions:

3 eqns. 3 unk(f,x,y)

RVP balance: y 0( ) y0

=

d y t( )⋅

dt

1

Vf1

t( ) y1

t( )⋅ f2

t( ) y2

t( )⋅+ f3

t( ) y3

t( )⋅+ f t( ) y t( )⋅−( )⋅=

Octane balance: x 0( ) x0

=d x t( )⋅

dt

1

Vf t( ) x

1t( ) f

2t( ) x

2t( )⋅+ f

3t( ) x

3t( )⋅+ f t( ) x t( )⋅−( )=

f t( ) f1

t( ) f2

t( )+ f3

t( )+=Total balance: kbl 42000gal:=

Assuming constant volume. perfectly mixed, and constant densities, the model equations are:

13-37. Control of gasoline blending tank of Example 12-2.5

To complete the control loops, add from the Public Model Library:Three flow control loops (can be copied from f411ffst) with ranges of 0 to 150% of design flow •and time constants of 0.1 minf407Trmr: three transmitters, one for the product flow with a range of 0 to 100 kbl/day and 0.1 •min time constant, one for the octane with range of 60 to 100 octane and a 1 min time constant, and one for the RVP with range of 0 to 20 RVP and 1 min time constantf403PI: three PI controllers for the product flow, octane and RVP.•

Flow control loops gains and initial conditions:

(a) With the correct pairing, the responses to a 10 kbl/day step increase in product flow set point are at 1 min:

τI

3min:=Kc

30%CO

%TO:=Octane controller:

τI

5min:=Kc

20−%CO

%TO:=RVP controller:

τI

0.1min:=Kc

0.9%CO

%TO:=Product flow controller:

The controllers are tuned as follows:

24.375kbl

day

kT3

66.7 %TO=kT3

0.366kbl

day %TO⋅=

kT3

24.375kbl

day1.5⋅

100%TO:=Reformate:

28.125kbl

day

kT2

66.7 %TO=kT2

0.422kbl

day %TO⋅=

kT2

28.125kbl

day1.5⋅

100%TO:=Straight Run:

7.5kbl

day

kT1

66.7 %TO=kT1

0.113kbl

day %TO⋅=k

T1

7.5kbl

day1.5⋅

100%TO:=Alkylate:

0.87

kT2

kT3

⋅ 1.004%CO

%TO=0.75−

kT1

kT3

0.231−%CO

%TO=u

30.75−

kT1

kT3

v1

v2

+ 0.87

kT2

kT3

v3

+=

0.27

kT2

kT1

⋅ 1.013%CO

%TO=0.71−

kT3

kT1

⋅ 2.308−%TO

%CO=u

1v

10.71

kT3

kT1

⋅ v2

⋅− 0.27

kT2

kT1

v3

⋅+=

Substitute and solve for the scaled variables u and v:

m3

kT2

v3

⋅=m2

kT3

v2

⋅=m1

kT1

v1

⋅=f2

setk

T2u

2⋅=f

3

setk

T3u

3⋅=f

1

setk

T1u

1⋅=

where all the variables are in kbl/day. In practice the gains must be scaled to apply to signals in %CO and %TO. So we have:

f2

set0.25− m

10.29m

2− m

3+=

f3

set0.75− m

1m

2+ 0.87m

3+=

f1

setm

10.71 m

2⋅− 0.27 m

3⋅+=

The decoupler developed in Example 12-3.3 gives the gains of the decoupler on the unscaled variables:

(c) Decoupler of Example 12-3.3

These are relatively tight responses without oscillations.

Students should be encouraged to do part (b) with alternate pairings of the controlled and manipulated variables. In doing so it is important to watch the direct and reverse action of the controllers.

u2

0.25−k

T1

kT2

v1

0.29

kT3

kT2

⋅ v2

− v3

+= 0.25−k

T1

kT2

⋅ 0.067−%CO

%TO= 0.29−

kT3

kT2

⋅ 0.251−%CO

%TO=

All of the initial values must be 66.67%TO, so the follwing biases must be added:

Bias1 2.308− 1.013+( )− 66.67⋅ %TO:= Bias1 86.338 %TO=

Bias3 0.231− 1.004+( )− 66.67⋅ %TO:= Bias3 51.536− %TO=

Bias2 .067− 0.251−( )− 66.67⋅ %TO:= Bias2 21.201 %TO=

These are the gains and biases used in the following Simulink diagram of the decoupler:

With this decoupler the control system of the gas blender produces essentially perfect control. The simulation with the decoupler is in a separate file.



ρ 8.33lb

gal:=L 3170ft:=d

p1.19ft:=D 3.41ft:=Problem parameters:

fo

0( ) 0gpm=

d fo

t( )⋅

dt

π dp

2⋅

4

7.48gal

ft3

g

Lh t( ) 2.22 10

3−⋅ f

ot( )− 0.5184 10

6−⋅ f

ot( )

2−⋅=

h 0( ) 0ft=d h t( )⋅

dt

4

πD2

ft3

7.48galf1

t( ) f2

t( )+ f3

t( )+ fo

t( )−( )=Combine and simplify:

4 eqns. 4 unks.

∆pf

t( ) ρ g⋅ 2.22 103−

⋅ fo

t( ) 0.5184 106−

⋅ fo

2t( )+⋅=Pressure drop correlation:

3 eqns. 4 unks.fo

t( )π

4d

p

2v t( )

7.48gal

ft3

⋅=Outlet flow in gpm:

2 eqns. 4 unks.(v,∆pf)

d

dtρπ

4⋅ d

p

2⋅ L⋅ v t( )⋅

π

4d

p

2p

aρ g⋅ h t( )⋅+ p

a− ∆p

ft( )−( )=

Momentum balance on outlet pipe:

1 eqn. 2 unks.(h,f0)

d

dtρπ D

2⋅

4⋅ h t( )⋅⋅ ρ

ft3

7.48 gal⋅⋅ f

1t( ) f

2t( )+ f

3t( )+ f

ot( )−( )⋅=Mass balance on tank:

Development of the model equations:

Lh(t) .fo(t)

dp

P1 P2 P3

13-38. Three pump ad tank start-up problem

g 115.827 103

×ft

min2

= hmax

9ft:=

Each pump flow changes from 0 to f when the flow is turned on (step function): f 750gal

min:=

Numerical values of the coefficients: 4

π D2

⋅

ft3

7.48gal0.015

ft

gal=

a1

π dp

2⋅

4

7.48gal

ft3

g

L⋅:=

a1

304gal

ft min2

⋅=

a1

2.22⋅ 103−

⋅ft min⋅

gal0.675

1

min=

a1

0.5184⋅ 106−

⋅ ftmin

gal

2

⋅ 157.6 106−

×1

gal=

The Simulink block diagram for the tank and pipe is:

Each pump is simulated as a step test from 0 to 750 gpm and the time of the step is the time when the pump is turned on.

The Simulink diagram includes a memory device that records the maximum level in the tank for each run.

The following are the responses with pump 2 turned on at 7 min and pump 3 at 14 min:

The level in the tank reaches 10.5 ft with this sequence, which means that the tank overflows.

Students must vary the times at which pumps 2 and 3 are turned on to see if they can keep from overflowing the tank. It makes for an interesting computer game with some fundamental concepts attached.



13-39. Ecological interaction of host-parasite populations

Developemnt of the model equations:

Rat population: d Rats t( )⋅

dtR

RGt( ) R

RDt( )−= 5 Rats t( )⋅ 0.05 Rats t( )⋅ Fleas t( )⋅−=

Rats 0( ) 100=

d Fleas t( )⋅

dtR

FGt( ) R

FDt( )−= 0.2 Rats t( )⋅ Fleas t( )⋅ 20 Fleas t( )⋅−=

Fleas 0( ) 20=

The Simulink diagram to solve these equations is:

The populations of rats and fleas are as fllows:

The two populations cycle with a period of about 2/3 year for the parameters of this problem. At high flea populations the death rate of rats is higher than their growth rate, while at low rat population the death rate of the fleas is higher than their growth rate and their population decreases, allowing the rat population to grow back.

Encourage the students to investigate the effect of the four problem parameters on the population cycles.



Mc

0( ) M0

=

1 eqn. 2 unks. (Mc, wc)

Entahlpy balance: Mc

t( ) cpL

⋅d T t( )⋅

dtU A⋅ T

aT t( )−( )⋅ w

ct( )⋅−= T 0( ) T

a=

2 eqns. 3 unks. (T)

Flow through nozzle: Subcritical flow: wc

t( )πD

2

4

3600s

hr⋅

2 Mw

⋅ P t( ) P t( ) Po

−( )⋅⋅

Rg

T t( ) 273.16K+( )⋅⋅=

(the smaller of the two)

Critical flow:w

ct( )

πD2

4

3600s

hr⋅

Mw

Rg

T t( ) 273.16K+( )⋅⋅ P t( )⋅=

3 eqns. 4 unk. (P)Vapor pressure by Antoine equation:Reid, Prausnitz & Sherwood, 3rd. ed., McGraw-Hill, 1977, Appendix P T t( )( ) e

15.9611978.32

T t( ) 273.16K+ 27.01−−

101300Pa

760mmHg⋅=

4 eqns. 4 nks.

All the numbers given are in consistent units except for the diameter of the nozzle:

D 4in0.3048m

12in⋅:= D 0.102 m=

The Simulink diagram for the barge is:

degC K:= kJ 103

joule:=13-40. Environmental impact of chlorine barge accidentkmole 1000mol:=

P

Po

T

Ta

wc

Mc

Problem parameters:

D 4in:= A 212m2

:=

M0

136000kg:=

Ta

25degC:=

U 1000kJ

hr m2degC⋅

:=

Rg

8314joule

kmole K⋅:=

Po

101300Pa:=

Properties of chlorine: Mw

71.5kg

kmole:= 288

kJ

kg:= c

pL0.946

kJ

kg degC⋅:=


Chlorine mass balance: d Mc

t( )⋅

dtw

ct( )−=

The temperature is calculated in K. Time is in hr.

The responses for the barge are:

It takes a little over 3.5 hr for the barge to empty for these parameters.

Students should be encouraged to investigate which of the parameters most affects the time required to empty the barge. The heat transfer coefficient is probably the one that has the greatest effect.

Caution: When running this simulation the "Stop Time," under "Simulation Parameters," must be adjusted so that the mass Mc is never zero or negative. After that point the results are not valid.



xr4

0( ) 0=d

dtM

rt( ) x

r4t( )⋅( ) F

Catt( ) x

fCat4⋅ r

Cat1t( ) M

w4⋅−=Catalyst 1:

3 eqs. 4 unks. (xr2)

xr2

0( ) 0=d

dtM

rt( ) x

r2t( )⋅( ) F

PAt( ) x

PA⋅ r

Bt( ) M

w2⋅−=Propionic anhydride:

2 eqns. 3 unks. (xr1,rB)

xr1

0( ) xr10

=d

dtM

rt( ) x

r1t( )⋅( ) F

Catt( ) x

fCat1⋅ 2−⋅ r

Bt( )⋅ M

w1⋅=2-butanol balance:

1 eqn. 1 unk. (Mr)

Mr

0( ) Mr0

=dM

rt( )

dtF

PAt( ) F

Catt( )+=Total mass balance:

Development of the model equations:Assume

Perfect mixing of ractor and jacket contents•The reactor is initially charged with the 2-btanol, essentially pure•

Cold water

TT TT

Anhydride

Butanol

Catalyst

Hot water

13-41. Control of semi-batch reactor

Vmr

t( )

Mr

t( )

ρr

t( )

=9 eqns. 13 unks. (Vmr,ρr)

Average density: ρr

t( )

xr1

t( )

ρ1

xr2

t( )

ρ2

+x

r3t( )

ρ3

+x

rBPt( )

ρ5

+x

r4t( ) x

r5t( )+

ρ4

+=

10 eqns. 14 unks. (xrBP)

Butyl propionate: xrBP

t( ) 1 xr1

t( )− xr2

t( )− xr3

t( )− xr4

t( )− xr5

t( )−=

11 eqns. 14 unks.Reaction rates:

rB

t( ) Vmr

t( ) k1

CA

t( )⋅ k2

CCat1

t( )⋅+ k3

CCat2

t( )⋅+( )⋅ CB

t( )=

12 eqns. 18 unks. (CA,CB, Ccat1, Ccat2)

rCat1

t( ) Vmr

t( ) k4

⋅ 10

Hr

t( )−⋅ C

Cat2t( ) C

Bt( )⋅=

13 eqns. 19 unks. (Hr)

Hr

t( ) p1

CCat1

t( )⋅ p2

CCat2

t( )⋅+( )− p3

p4

T t( ) 273.16K++=

14 eqns. 19 unks.

4 eqns. 6 unks. (xr4,rCat1)

Catalyst 2: d

dtM

rt( ) x

r5t( )⋅( ) r

Cat1t( ) M

w4⋅= x

r5t( ) 0=

5 eqns. 7 unks. (xr5)

Water: d

dtM

rt( ) x

r3t( )⋅( ) F

Catt( ) x

fCat3⋅ F

PAt( ) 1 x

PA−( )⋅+ r

Bt( ) M

w3⋅+= x

r30( ) 1 x

r10−=

6 eqns. 8 unks. (xr3)Enthalpy on reactor:

d

dtM

rt( ) c

pavgt( )⋅ T

rt( )⋅( )⋅ F

PAt( ) c

p2⋅ F

Catt( ) cpcat⋅+( ) T

fT t( )−( )⋅ U

oA⋅ T

rt( ) T

jt( )−( )⋅+=

∆Hr

t( ) rB

t( )⋅− Tr

0( ) Tr0

=

7 eqns. 11 unks. (Tr,Tj,cpavg)Enthalpy on jacket:

Mj

cp3

⋅d T

jt( )⋅

dt⋅ U

oA⋅ T

rt( ) T

jt( )−( )⋅ F

cwt( ) c

p3⋅ T

jt( ) T

cw−( )⋅− F

hwt( ) c

p3⋅ T

hwT

jt( )−( )⋅+=

Tj

0( ) Tj0

= 8 eqns. 11 unks.

Volume of reactants:

Mj

415kg:= ∆Hr

80000−kJ

kmole:=

xPA

100mass%:= xr10

100mass%:=

Fcw

10000kg

hr:=

Catalyst feed composition: xfCat1

70mass%:= xfCat3

10mass%:= xfCat4

20mass%:=

Physical properties from Perry's, 7th ed., Tables 2-1 and 2-2:

2-butanolPropionic anhydride

Water Mw

74

130

18

98

130

kg

kmole:= ρ

808

1012

1000

1829

866

kg

m3

:= cp

2.876

2.345

4.187

1.424

2.345

kJ

kg degC⋅:=

Sulfuric acid

Butyl propionate

Hot water flow required for an inlet jacket temperature of 30 ºC:

Fcw

cp3

⋅ Tcw

⋅ Fhw

cp3

⋅ Thw

⋅+ Fcw

Fhw

+( )cp330⋅ degC=

Fhw

Fcw

30degC Tcw

−

Thw

30degC−⋅:=

Concentrations: CA

t( )

ρr

t( ) xr1

t( )⋅

Mw1

= CB

t( )

ρr

t( ) xr2

t( )⋅

Mw2

= CCat1

t( )

ρr

t( ) xr4

t( )⋅

Mw4

=

CCat2

t( )

ρr

t( ) xr5

t( )⋅

Mw4

= 18 eqns. 19 unks.

Reactor specific heat:

cpavg

t( ) xr1

t( ) cp1

xr2

t( ) cp2

⋅+ xr3

t( ) cp3

⋅+ xr4

t( ) xr5

t( )+( ) cp4

⋅+ xrBP

t( ) .⋅ cp5

+=

19 eqns. 19 unks.The reaction rate coefficients are Arrhenius functions of temperature:

ki

Ai

e

Ei

−

Rg

T t( ) 273.16K+( )⋅⋅= for i = 1..4

Problem parameters:

p

0.2022

0.3205

21.38−

1271

:= A

1.93 1011

⋅

1.01 1014

⋅

1.42 1014

⋅

5.05 1011

⋅

m3

kmole min⋅:= E

80480

79160

69970

76620

kJ

kmole=

Rg

8.314kJ

kmole K⋅= M

r0600kg:= T

r020degC:= T

j030degC:= T

f20degC:=

Uo

A⋅ 300kJ

min degC⋅= T

cw15degC:= T

hw90degC:=

Fhw

2500kg

hr=

The eight differential equations may be simplified as follows:d x

r1t( )⋅

dt

1

Mr

t( )F

Catt( ) x

fCat1x

r1t( )−( )⋅ F

PAt( ) x

r1t( )⋅− 2 r

Bt( )⋅ M

w1⋅−=

d xr2

t( )⋅

dt

1

Mr

t( )F

Catt( ) x

fCat2x

r2t( )−( )⋅ F

PAt( ) x

PAx

r2t( )−( )⋅+ r

Bt( ) M

w2⋅−⋅=

d xr3

t( )⋅

dt

1

Mr

t( )F

Catt( ) x

fCat3x

r3t( )−( )⋅ F

PAt( ) 1 x

PA− x

r3t( )−( )⋅+ r

Bt( ) M

w3⋅+⋅=

d xr4

t( )⋅

dt

1

Mr

t( )F

Catt( ) x

fCat4x

r4t( )−( )⋅ F

PAt( ) x

r4t( )⋅− r

Cat1t( ) M

w4⋅−=

d xr5

t( )⋅

dt

1

Mr

t( )F

Cat1t( ) F

PAt( )+( )− x

r5t( )⋅ r

Cat1t( ) M

w4⋅+=

d Tr

t( )⋅

dtF

Cat1t( ) c

pcat⋅ F

PAt( ) c

p2⋅+( ) T

fT

rt( )−( )⋅ U

oA⋅ T

rt( ) T

jt( )−( )⋅− ∆H

rrB

t( )⋅−=

1

M t( ) cpavg

t( )⋅

d Tj

t( )⋅

dt

1

Mj

Uo

A⋅

cp3

Tr

t( ) Tj

t( )−( )⋅ Fcw

t( ) Tj t( ) Tcw

−( )⋅− Fhw

t( ) Thw

Tj

t( )−( )⋅+=

d Mj

t( )⋅

dtF

Cat1t( ) F

PAt( )+=

The Simulink block diagram for the reactor is the following:

The Reactor S-fnction makes use of a file "Semibatch.m" to solve the equations. This file must be in the MATLAB "Current Directory" for the simulation to run.

The following are the temperature profiles for a batch when the catalyst is fed at 1200 kg/hr for 5 min followed by the anhydride fed at 1200 kg/hr for 30 minutes:

These trends match the trends in the CEP article by Feliu, et al. (Dec. 2003) for the same conditions.

The students must now device a control strategy to minimize the batch cycle time while satisfying the constraint that the reactor temperature must not exceed 60ºC. This constraint is violated in the base case shown here. Feliu proposes various strategies in the article.

This problem probably makes for a good term project. The instructor must decide whether to provide the model and simulation to the students or have them develop their own. If the latter please tell the students that the Arrhenius coefficients are in m3/kmole-min.



The Simulink diagram for the cooler is:

fc0 0.172m

3

min=fc0

U A⋅

ρc cpc⋅

T0 Tc0−( )Tc0 Tci−( )

:=

Tc0 35.5 degC=Tc0 T0

f ρ⋅ cp⋅

U A⋅Ti T0−( )−:=At the initial steady state:

cpc 4.2kJ

kg degC⋅:=ρc 1000

kg

m3

:=cp 3.8kJ

kg degC⋅:=

ρ 800kg

m3

:=Vc 1.1m3

:=A 4m2

:=U 200kJ

min m2degC⋅

:=V 5m3

:=Problem parameters:

Tci 25degC:=T0 45degC:=f 0.1m

3

min:=

Ti 70degC:=Design conditions:

Tc 0( ) Tc0=d Tc t( )⋅

dt

fc t( )

VcTci Tc t( )−( ) U A⋅

Vc ρc⋅ cpc⋅T t( ) Tc t( )−( )+=

T 0( ) T0=d T t( )⋅

dt

f t( )

VTi t( ) T t( )−( ) U A⋅

V ρ⋅ cp⋅T t( ) Tc t( )−( )−=

kJ 1000joule:=degC K:=The model equations, from the solution to Problem 4-7, are:

13-18. Temperature control of stirred tank cooler of Problems 4-7 and 6-19


Chapter 13. Simulation of Process Control Systems (continued)

fc0

0.008m

3

min %CO⋅

21.5 %CO=Controller output:

45 20−

70 20−50 %TO=Transmitter output:Initial conditions:

τD 0.95 min=

τI 3.8 min=Kc 44−%CO

%TO=τD

Tu

8:=τI

Tu

2:=Kc

Kcu


Tu 7.59min:=Kcu 75.4−%CO

%TO:=

From the solution to Problem 6-19:%TO %:=%CO %:=

To complete the temperature control loop, install, from the Public Model Library:A flow control loop, f401Vlv1, with a gain of 0.008 m3/min-%CO, negligible time constant.•A temperature transmitter, f407Trmr, with a range of 20 to 70 C and a time constant of 0.6 min.•A series PID controller, f405PIDs, tuned for quarter dcay ratio•

The responses to a 0.2 m3/min step increase in process flow at 5 min, and a 2 C step increase in set point at 30 min, are:

The controller output saturates for the change in process flow and is barely able to return the temperature to the set point. The controller output also temporarily saturates on the change in set point, but is able to recover. Saturation is a form of nonlinear behavior. In this case it causes the decay ratio to be much less than 1/4.

Encourage your students to figure out what needs to be changed so that the controller does not saturate.


V 5gal:=

kv1.954 10⋅

100%CO

lb

min:= kv 0.0618

lb

min %CO⋅=

Steam enthalpy, saturated at 1 atm., referenced to 32 F, is: 1150.4 BTU/lb

hs 1150.4BTU

lbcp 32⋅ degF+:= hs 1176

BTU

lb=

At the initial steady state: w2

f1 ρ⋅ cp⋅ T30 T1−( )⋅

hs cp T30⋅−:= w2 2.518

lb

min=

vp0

w2

kv:=

vp0 40.8 %CO=

Note: The value of w2 does not match the one in the problem statement.

Simulate the valve by inserting the block f401Vlv1 from the Public Model Library, linear, with a gain kv and a time constant of

τv 4s:= τv 0.067 min=

The Simulink block diagram is:

gpmgal

min:= degF R:=13-19. Direct contact heater of Problem 4-5


d T3 t( )⋅

dt

1

Vf1 t( ) T1 t( )⋅ f3 t( ) T3 t( )−( )⋅

hs

V ρ⋅ cp⋅w2 t( )⋅+= T3 0( ) T30=

w2 t( ) 1.954 10 vp t( )⋅= kv vp t( )⋅=

f3 t( ) f1 t( )w2 t( )

ρ+=

Substitute to eliminate f.3(t):d T3 t( )⋅

dt

f1 t( )

VT1 t( ) T3 t( )−( )

w2 t( )

V ρ⋅ cp⋅hs cp T3 t( )⋅−( )+=

Design conditions: f1 25gpm:= T1 60degF:= T30 80degF:=

Problem parameters: ρ 7lb

gal:= cp 0.8

BTU

lb degF⋅:=


The response to a 5 gpm step increase in process flow at 0.1 min and a 5 %CO step increase in sgnal to the steam control valve at 1 min, are

The time constant of the mixer is matches the value of 0.2 min obtained by linearization; the steady state changes are a bit less than predicted by the linear model.

The respons to the process flow is first-order and the one to the signal to the valve is second-order. This is because of the lag in the valve.

This is the plot also obtained with the following Simulink diagram:


0 5 10 15 20 250

2 .105

4 .105

6 .105

fs p1 ∆p psi⋅,( )scfh

∆p

fs p1 ∆p,( ) 836scfh

gpm

R

psi⋅ Cv⋅ Cf⋅

p1

G T⋅⋅ y p1 ∆p,( ) 0.148 y p1 ∆p,( )3

−⋅:=

y p1 ∆p,( ) 1.63

Cf

∆p

p1⋅:=Functions for valve flow:

T 259 460+( )R:=Cf 0.8:=G 0.621:=p1 34.7psia:=Cv 440gpm

psi:=

scfhft

3

hr:=psia psi:=Problem parameters from Example 5-2.2:

GMw

29=y

1.63

Cf

∆p

p1⋅=

fs 836 Cv⋅ Cf⋅p1

G T⋅⋅ y 0.148 y

3⋅−( )=

The flow through a gas control valve can be calculated from Eq. 5-2.3:

13-20. Gas flow control valve of Example 5-2.2

This Simulink block can simulate any gas valve with variable inlet pressure and pressure drop. To obtain the flow in scf/min, divide the Cv,max/100 by 60 min/hr.

The Simulink block diagram for the temperature control loop is:

τD 0.84 min=τI 3.4 min=Kc 5.9%CO

%TO=τD

Tu

8:=τI

Tu

2:=Kc

Kcu

1.7:=

From Table 7-1.1, the quarter decay rato tuning parameters are:

Tu 6.7min:=Kcu 10%CO

%TO:=

With the lags on the valve and the transmitter, the loop can be made unstable--it wasn't for the conditions of Problem 6-12. The ultimate gain and period ar determined from the simulation:

678.9 640−

700 640−64.8 %TO=Transmitter initial condition:

ln 0.5( )

ln 50( )− 17.72 %CO=Initial position:2 0.8771⋅

ft3

min1.754

ft3

min=Valve maximum flow:

To the block developed in Problem 13-5 to simulate the reactor, we add the following blocks from the Public Model Library:

f401Vlv1: equal precentage control valve with a time constant of 0.1 min and sized for 100% •overcapacity, air-to-closef405PIDs: a series PID controller tuned for quarter decay ratio response•f407Trmr: transmitter with a range of 640 to 700 R and a time constant of 1.0 min•

13-21. Temperature control of reactor of Section -2.3 and Problem 6-12

The responses to a -0.2 ft3/min step change in reactants flow at 1 min followed by a 1 R step increase in setpoint at 30 min are:

The decay ratio is slightly greater than 1/4. The students should play with the tuning of the controller and test it for different magnitudes and directions of the input changes.


13-22. Temperature control of stirred tank heater of Example 13-4.1 with variable pressure drop across the control valve

The steam chest pressure is calculated as a function of the steam temperature using the Antoine equation for water (Reid, Prausnitz, and Sherwood, 1977, The Properties of Gases and Liquids, 3rd ed., Appendix.

p2 t( )14.7psia

760mmHge

18.30363816.44K

Ts t( ) 460R+( )K

1.8R46.13K−

−

=

The Simulink block diagram of Fig. 13-4.9 is modified as follows:

The "Gas Valve 2" is a modified version of the gas valve of Problem 13-20 in that it accepts the upstream and downstream pressures p1 and p2--instead of ∆p--and it outputs the flow in lb/min. This is done by multiplying the output by the molecular weight and dividing by 380 scf/lbmole and by 60 min/hr.

Responses to this model are compared with those of Fig. 13-4.10 by running both versions of the control loop in parallel and plotting the reponses together:

The responses to the decrease in process flow are almost identical, but those for the increase in flow are different. This is because as the steam pressure increases the capacity of the valve decreases and the flow of steam does not vary as much as when the flow is assumed to be independent of pressure. The response is then slower and less oscillatory.

This is an excellent example of the effect of a saturation nonlinearity that is often overlooked in modeling.


The Simulink diagram for the control loop is:

m0 40.65 %CO=m0

f

Kv:=Initial conditions:

Kv Ko⋅100 0−( )%TO

1 0−( )lb gal1−

⋅⋅ 1.845=Process gain:

Ko 0.0075lb

gal gpm⋅:=Kv 2.46

gpm

%CO:=From the solution to Problem 6-17:

The Simulink diagram for each reactor is the same as in Problem 13-8, with a different notation and without the transportation lag in the outlet pipe (here it is asssumed e reactors are close to each other and the tranporation lag is negligible).

Making a subsystem block of each reactor, we build the concentratioon control loop by adding, from the Public Model Library:

f401Vlv1: a linear control valve sized for 100% overcapacity and a time constant of 0.1 min•f405PIDs: a series PID controller tuned for quarter decay ratio response•f407Trmr: a concentration transmitter with a range of 0 to 1.0 lb/gal and negligible lag.•

c

4

2

1

0.5

lb

gal=c

3

f c2

⋅

f k V⋅+:=c

2

f c1

⋅

f k V⋅+:=c

1

f c0

⋅

f k V⋅+:=At the initial steady state:

V 1000gal:=k 0.1min1−

:=c0

4lb

gal:=f 100gpm:=Design conditions:

c0

t( ) ci t( )=and

For j = 1,2,3cj

0( ) cj0=d c

j⋅ t( )

dt

f t( )

Vc

j 1− t( ) cj

t( )− k cj

⋅ t( )−=

The model equation for each reactor is developed in the soution to Problem 6-17:

13-23. Concentration control of isothermal reactors in series of Problem 6-17

From a step test in controller output, the time constant is about 9 min and the dead time is negligible, so the process is very controllable. The quarter decay ratio tuning formulas impose no limit on the gain or on how small the integral time can be, so we use the synthesis formulas:

Kc adjusutable τI τ= 9min= τD 0min= (PI controller)

The responses to a 1 lb/gal increase in inlet eactants concentartion (a change in inlet flow is not possible since the flow is the manipulated variable), with Kc = 10%CO/%TO, are:

The problem will be more interesting with a 1 min time lag in the transmitter. The students can be asked try this case and to play with the tuning parameters, e.g., smaller integral time for faster response.

Notice the deviation in concentration is small. Students could also examine effect of the controller gain on the magnitude of the deviation.


T30 200degF:=

Problem parameters: U 136BTU

hr ft2degF⋅

:= D 3ft:= A 974ft π⋅ 0.5⋅ in:= A 127.5 ft2

=

ρ 53lb

ft3

:= CM 974ft 0.178⋅lb

ft0.12⋅

BTU

lb degF⋅:= CM 20.8

BTU

degF=

cp 0.45BTU

lb degF⋅:= ps 115psig:= λ 872.9

BTU

lb:=

At the initial steady state: f3 f1:= Ts0 T30

ρ cp⋅ f1⋅

U A⋅T30 T1−( )+:= Ts0 343.4 degF=

ws

U A⋅ Ts0 T30−( )⋅

λ:= ws 47.48

lb

min=

The Simulink diagram for the heater is:

13-24. Temperature and level control of oil heater of Problem 6-24The model equations from the solution to Problem 6-24 are:

d h t( )⋅

dt

ft3

7.48gal

4

πD2

f1 t( ) f3 t( )−( )= h 0( ) h0=

d T3 t( )⋅

dt

f1 t( )

V t( )T1 t( ) T3 t( )−( )⋅

U A⋅

V t( ) ρ⋅ cp⋅Ts t( ) T3 t( )−( )+= T3 0( ) T30=

V t( ) 7.48gal

ft3

⋅π D

2⋅

4⋅ h t( )=

d Ts t( )⋅

dt

1

CMλ ws t( )⋅ U A⋅ Ts t( ) T3 t( )−( )⋅−= Ts 0( ) Ts0=

Assume the pressures are constant and determine the flows as follows:

Inlet oil flow f1(t): linear valve with variable pressure drop-Model Library f402Vlv2-sized for 50% overcapacity and pressure drop:

∆pv1 p1 14.7psi+ p2−( ) ρ g⋅ h t( ) 5ft−( )⋅−=

Outlet oil flow f3(t): Step function input (it is the disturbance)

Steam flow ws(t): linear control valve with constant pressure drop-Model Library f401Vlv1-szed for 50% overcapacity.

psig psi:=

Design conditions: p1 45psig:= p2 40psia:= f1 100gpm:= T1 70degF:=

τD 0.14 min=τI 0.56 min=Kc 45%CO

%TO=

τD

Tu

8:=τI

Tu

2:=Kc

Kcu

1.7:=

Quarter decay tuning parameters from Table 7-1.1:

Tu 1.12min:=Kcu 77.1%CO

%TO:=Ultimate gain and period of temperature controller:

m10 46.67 %CO=m1032.2

1.5 46⋅:=

ms0 60.58 %CO=wsmax 78.4lb

min=ms0

ws

wsmax:=wsmax

46

41.81.5⋅ ws⋅:=

50% for both transmittersIntial conditions, from the results of Problem 6-24::

To complete the control loops, from the Public Model Library, introduce:f405PIDs: series PID temperature controller tuned for quarter decay ratio response•f407Trmr: temperature (TT-50) and level (LT-50 transmitters with 100 to 200 F and 7 to 10 ft •ranges, and 0.5 min and 0.01 min time constants, respectively.

The responses to a 5 gpm increase in inlet oil flow are:

Students should be encouraged to study the effect of problem parameters. For example, a more realistic valve time constant of 0.1 min will result in more reasonable controller parameters.

Notice the small offset in level. This is for a level controller gain of 20%CO/%TO.


13-25. Moisture control of drier of Problem 9-1. Simple feedback vs. cascade

In the solution to Problem 9-1 he drier is represented by linear transfer functions as it will be simulated here. In the absence of information, the transfer function will be the same to a change in ambient temperature as for a change in heater outlet temperature.

(a) The Simulink diagram for the single moisture feedback control loop is

(b) The Simulink diagram for the cascade control scheme using the heater outlet temperature as the intermediate variable is:


The respones to a 10ºF step increase in ambient temperature at 1 min, and a 0.3 %moisture increase in set point at 30 min, are:

The cascade response is in magenta and the feedback in yellow. To obtain these results the gain of the TC-47 controller had to be reduced to 2 %CO/%TO from the 4.14 determined in Prpblem 9-1.

The cascade reponse is less oscillatory and faster at the expense of larger changes and oscillations in the fuel flow.

The controller parameters are:

Feedback: Cascade:

Kc 0.42−%CO%TO

:= 0.56−%CO%TO

τI 3.5min:= 1.0min

τD 0.86min:= 0.26min


The feedback control responses are in yellow and the cascade responses in magenta. They show almost perfect control to the disturbance input and a slightly faster response to the set point change for the cascade scheme at the expense of very oscillatory response of the manipulated variable.

Students should be encouraged to study how adjustment of the secondary gain may reduce the oscillations of the manipulated variable and how it affects the response of the controlled variable and/or the master controller tuning.

The block diagrams are given in the solution to Problem 9-3, and the responses to unit step changes in disturbace at 1 min and set point at 75 min are:


=τITu1.2

:=KcKcu2.2

:=


:=

Kc2 2.2%CO%TO

:=Cascade Control. From the solution to Problem 9-3, with


=τITu1.2

:=KcKcu2.2

:=

From Table 7-1.1, the quaerte decay tuning parameters for a PI controller are:


:=

Simple Feedback Loop. From the solution to Problem 9-3, the ultimate gain and period are:

13-26. Feedback vs. cascade control of Problem 9-3

The responses to a 5ºC step increase in set point are:



=

τD 0.482 τ⋅t0τ

1.137

⋅:=τIτ

0.878

t0τ

0.749

⋅:=Kc1.435

Kp

t0τ

0.921−

⋅:=

τ 12.8min:=Kp 0.385%TO%CO

:=t0 2.2min:=

Cascade Control (with the secondary proportional controller gain set at 2%CO/%TO):


=

τD 0.482 τ⋅t0τ

1.137

⋅:=τIτ

0.878

t0τ

0.749

:=Kc1.435

Kp

t0τ

0.921−

:=

The tuning parameters for minimum IAE response, from Table 7-2.2:

t0 3.5min:=τ 13.5min:=Kp 0.285%TO%CO

:=

Simple Feedback Loop:

The block diagrams from the solution to Problem 9-4 are simulated as in Problem 13-25 with a parallel PID master controller and a proportional slave controller with a gain of 2%CO/%TO. From the open-loop responses to step changes in contrller output, the following parameters are obtained:

13-27. Jacketed reactor of Problem 9-4. Simple feedback vs. cascade control

The responses show a faster response for the cascade scheme (magenta) than for simple feedback (yellow), at the expense of doubling the variation in the controller output. In fact, because this linear analysis does not impose limits on the controller output, we see increases of 200 and 400%CO, which are not possible in practice. Pointing out these restrictions is an advantage of simulation which is not obtained from the anlytical analysis.

The responses to a 5 ft3/min step decrease in process flow at 3 min followed by an increase of 10ºF in inlet temperature at 30 min are:

The Simulink diagram for the ratio control scheme is obtained by a simple modification of the diagram of Fig. 13-4.11:

Ratio 3.75min %TO⋅

ft3=Ratio

ρ cp⋅ Tset Ti−( )⋅

λ

100%TOfmax

⋅:=

Tset 150degF:=The ratio for set point of

fmax 75lb

min:=Ti 100degF:=λ 966

BTUlb

:=cp 0.8BTU

lb degF⋅:=ρ 68

lb

ft3:=

where the ratio is adjusted by the operator or a feedback controller to obtain the desired outlet temperature. For the conditions of the problem, the ratio is:

wset t( )⋅ρ cp⋅ Tset t( )⋅ Ti−⋅

λf t( )= Ratio f t( )⋅=

If the variation in the inlet temperature diturbance is neglected, the feedforward controller calculation becomes:

13-28. Feedforward temperature control of heater of Example 13-4.2

The responses are identical for the change in process flow, but the ratio controller (magenta) does not take action on the change in inlet temperature resulting in an error of about 10ºF. This reponse is slower than the one for the flow change, so a feedback controller can adjust the ratio to maintain the outlet temperature at the set point.



Kc 166−%CO%TO

=Kc 147−%CO%TO

100%CORmax

⋅f2max

f1⋅:=

The controller gain must be adjusted by the factor the output is multiplied by:

Rmax f1max⋅

100%CO f2max⋅0.0221

1%CO

=

The signal from the flow transmitter must then be scaled by the factor:

(at 100%CO)Rmax 3:=f2max 5.42m3

min:=f1max 4

m3

min:=Transmitter ranges:

Ratio 1.5=Ratiof2f1

:=f2 2.4m3

min:=f1 1.6

m3

min:=

Design conditions:

The diagram of Problem 13-11 is modified to introduce a sesor/transmitter for the concentrated stream flow and a multiplier to represent the ratio controller. The transmitter has a ange of 0 to 4 m3/min and a time constant of 0.75 min. The multiplier allows the feedback controller to adjust the ratio.

13-29. Ratio control of blending tank of Problems 3-18 and 13-11