solution manual for principles & automatic process control by smith & corripio download at http://sharecash.org/download.php?file=1424912

Smith & Corripio, 3rd edition Problem 1-1. Automation in daily life. Identificaton of the M-D-A components, controller type, instrumentation diagram, and type of control.

(a) House air-conditioning/heating. M: Temperature element TE in thermostat TE/C

SP

D: Mercury switch in thermostat TE/C

TE/C

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

AC /H

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

(b) Cooking oven. M: Temperature element TE, usually a gas-filled bulb

S

SP TC

TE

D: Temperature controller TC. A: Solenoid S that operates the heating element in the oven

S

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

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

(d) Automatic sprinkler system for fires.

SP

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.

TE/C

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

Water main

(e) Automatic cruise speed control. 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.

SP SC

S Air

ST

Transmission

Engine

(f) Refrigerator. M: temperature sensor TE, usually a gas-filled bulb D: Temperature controller C, mechanically linked to the sensor

SP TE/C

S

A: Solenoid S that turnsd the refrigeration compressor on and off The controller is On/Off and the control is feedback. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 1-2. Automatic shower temperature diagram. M: temperature sensor TE, a gas-filled bulb D: temperature controller TC, mechanilly integrated to the sensor, but with a signa output

SP TC TE

A: solenoid operated control valve on the hot water line. The cold water valve is operated manually.

S Hot water

Cold water

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Smith & Corripio, 3rd. edition Problem 2-1. Derivation of Laplace transforms from its definition ⌠ F( s) =  ⌡

∞

f ( t) e

− st

dt

0

(a)

⌠ F( s) =  ⌡

f ( t) = t

∞

t⋅ e

− st

By parts:

dt

u=t

dv = e

0

du = dt 1 ⌠ + ⋅ s ⌡

−t − st F( s) = e ⋅ ∞ s 0

∞

e

− st

v=

− st

dt

−1 − st e s

1 − st e ⋅ ∞ 2 s 0

dt = 0 − 0 −

0

1

=

2

s

F( s) =

(b)

f ( t) = e

− at

∞

e

− at − st

e

0

⌠ dt =  ⌡

∞

e

− ( s + a)t

dt =

0

−1 s+a

e

− ( s + a)t

⋅ ∞

1

=

s+a

0

1

F( s) =

(c)

f ( t) = cos⋅ ωt

⌠  − st cos⋅ ωt⋅ e dt =   0 ⌡

⌠ F( s) =  ⌡

2

s

where a is constant ⌠ F( s) =  ⌡

1

∞

∞

e

i⋅ ωt

−e

− i⋅ ωt

e

2

− st

s+a

dt

0



1 ⌠  = 2 ⌡



=

=

1

∞

e

− ( s − i⋅ ω )t

0



∞

e

− ( s + i⋅ ω )t

dt

0

−1

e 2  s − i⋅ ω  1

⌠ dt +  ⌡

1

2  s − i⋅ ω

− ( s− i⋅ ω )t

+

⋅ ∞ 0

1

+

−1 s + i⋅ ω

e

   

− ( s+ i⋅ ω )t

 = s − i⋅ ω + s + i⋅ ω =

s + i⋅ ω 

2 ⋅ ( s − i⋅ ω ) ( s + i⋅ ω )

  0 

⋅ ∞

2⋅ s

(2

2⋅ s + ω

)

2

F( s) =

=

s 2

s +ω s 2

s +ω

2

2

(d) f ( t) = e− atcoss⋅ ωt

⌠  − at − st e cos⋅ ωt⋅ e dt =   0 ⌡

⌠ F( s) =  ⌡

∞



=

1 ⌠  2 ⌡



=

1

∞

e

− ( s + a+ i⋅ ω )t

0

∞

e

0 ∞

⌠ dt +  ⌡

e

=

⋅

i⋅ ωt

+e

−1

2  s + a + i⋅ ω

e

− ( s + a− i⋅ ω ) t

− ( s+ a+ i⋅ ω )t

⋅ ∞

+

0

− i⋅ ωt

e

2

dt

0



=

− at e

− st

dt

   

−1 s + a − i⋅ ω

e

− ( s+ a− i⋅ ω )t

  0 

⋅ ∞

1

1 1  = s + a − i⋅ ω + s + a + i⋅ ω +  2  s + a + i⋅ ω s + a − i⋅ ω  2 ( s + a + i⋅ ω ) ( s + a − i⋅ ω ) 2 ( s + a) 2 ⋅ ( s + a) + ω 2

2



=

s+a 2

( s + a) + ω

2

F( s) =

s+a 2

( s + a) + ω

2

All the results match results in Table 2-1.1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-2. Derive Laplace transforms from the properties and Table 2-1.1 (a)

f ( t) = u( t) + 2 ⋅ t + 3⋅ t

2

(

F( s) = L u ( t) + 2 ⋅ t + 3 ⋅ t =

1

+ 2⋅

s

1

+ 3⋅

2

s

) = L(u(t)) + 2⋅L(t) + 3⋅L(t2)

2

2!

1

F( s) =

3

s

s

2

+

2

+

s

6 3

s

Used the linearity property.

(b)

f ( t) = e

(u(t) + 2⋅t + 3⋅t2)

− 2⋅ t

(

)

2

F( s) = L u ( t) + 2 ⋅ t + 3 ⋅ t ⋅

 1 + 2 + 6 ⋅ s 2 3 s s  s+2 

= s+2

=

1

2

+

s+2

( s + 2)

2

( s + 2)

(

F( s) = L u ( t) + e =

1 s

+

− 2t

1 s+2

− 2⋅ e

1

− 2⋅

s+2

1 s+1

=

1 s

+

1

−

s+2

( − t) + L(t⋅e− t) = 1 −

F( s) = L( u ( t) ) − L e

s

1 s

− 2s

(

Let g ( t) = u ( t) 1 − e G( s) = e

− 2s  1

Used the real translation theorem and linearity.

( s + 2)

3

sin⋅ t

1

s+1 +

s+1

− 2t

s − 2 ( s + 2) + 

2

1

F( s) =

Used the linearity property.

F( s) = e

( s + 2)

6

+

2

) = L(u(t)) + L(e− 2t) − 2⋅L(e− t)

F( s) =

(e) f ( t) = u( t − 2)  1 − e− 2( t−2) sin( t − 2) 

2

+

−t

Used the linearity property.

(d) f ( t) = u( t) − e− t + t⋅ e− t

3

F( s) =

Used the complex translation theorem.

(c) f ( t) = u( t) + e− 2t − 2e− t

6

+

)

1 s

1 s+2

2

−

s+1

1

+

( s + 1) −

1 s+1

2

+

1 ( s + 1)

2

Then f ( t) = g ( t − 2 )

  1

F( s) = e

− 2s 1

1  s −  2 ( s + 2) + 1  

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-3. Initial and final value check of solutions to Problem 2-2 (a) f ( t) = u( t) + 2 ⋅ t + 3t2

1

F( s) =

+

s

2

6

+

2

3

s

s

Initial value: lim t→0

(u(t) + 2t + 3t2) = 1

s⋅ 

lim s→∞

2 1 6  s + 2 + 3 = s lim →∞ s s  

1 + 2 + 6  = 1  2 s s  

Final value: lim t→∞

(u(t) + 2t + 3t2) = ∞ (

)

(b) f ( t) = e− 2t u( t) + 2t + 3t2

1 + 2 + 6  = ∞  2 s s→0 s 

Check!

lim

F( s) =

1 s+2

2

+

( s + 2)

6

+

2

( s + 2)

3

Initial value: lim

e

(u(t) + 2t + 3t2)

− 2t

t→0

= 1( 1 + 0 + 0) = 1

lim

s

s→∞

2 1 6 = ∞ + s + 2 + 2 3 ∞ ( s + 2) ( s + 2)  

Must apply L'Hopital's rule: 2 6 1 + =1  1 2( s + 2) + 2 3 ( s + 2)  

lim Final value: lim

e

s→∞

(u(t) + 2⋅t + 3t2) = 0⋅∞

− 2t

t→∞

L'Hopital's rule: 0 2 6t  lim  + + =0  2t 2t 2t t → ∞  2e 2e 2e 

lim s s→0

(c) f ( t) = u( t) + e− 2t − 2e− t

2 1 6 =0 + s + 2 + 2 2 ( s + 2) ( s + 3)   Check!

F( s) =

1 s

+

1

−

s+2

2 s+1

Initial value: lim t→0

(u(t) + e− 2t − 2e− t) = (1 + 1 − 2) + 0

lim

s

1

s→∞ s

+

1 s+2

−

2

= ∞

s + 1

∞

L'Hopital's rule: lim s→∞

1 + 1 − 2  = 0  1 1 

Final value: lim t→∞

(u(t) + e− 2t − 2e− t) = 1 + 0 + 0 = 1

lim s

s→0

1

s

+

1 s+2

−

2

=1+0+0=1

s + 1

(d) f ( t) = u( t) − e− t + t⋅ e− t

1

F( s) =

s

−

1 s+1

+

1 ( s + 1)

2

Initial value: lim t→0

(u(t) − e− t + t⋅e− t) = 1 − 1 + 0⋅1 = 0

s

1 1 1 = ∞ s − s+1 + 2 ∞ ( s + 1)  

lim s→∞

L'Hopital's rule: 1 1 − 1 + =1−1+0=0   1 2( s + 1)  

lim Final value: lim t→∞

(u(t) − e− t + t⋅e− t) = 1 − 0 + ∞⋅0

s→∞

s 1 − s + =1−0+0=1  2 s+1 s→0 ( s + 1)  lim

L'Hopital's rule:

1 − 0 + 1  = 1  t t→∞  1⋅ e  (e) f ( t) = u( t − 2)  1 − e− 2( t−2) sin( t − 2)  lim

Check! F( s) = e

− 2s 1

1  s −  2 ( s + 1) + 1  

The test of the delayed fnction is not useful. Better to test the term in brackets, g(t): Initial value: lim t→0

(1 − e− 2tsin⋅t) = 1

s

lim s→∞

1 1 =1−0=1 s −  2 ( s + 1) + 1  

Final value: lim t→∞

 1 − e− 2tsin⋅ ( t)  = 1

lim s s→0

1 1 =1+0=1 s −  2 ( s + 1) + 1   Check!

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-4. Laplace transform of a pulse by real translation theorem f ( t) = H⋅ u ( t) − H⋅ u ( t − T) F( s) = H⋅

1 s

− H⋅ e

− sT 1

⋅

s

= H⋅

1−e s

− sT

F( s) =

H s

(1 − e− sT)

( )

− t− t0

Problem 2-5. Delayed versus non-delayed function

τ

f ( t) = e

(a) Function is non-zero for all values of t > 0: t0

f ( t) = e

τ

e

t0

t0

−t τ

F( s) = e

τ

1

=

1

s+

τ ⋅e

t0

(from Table 2-1.1)

τ

τ ⋅e

F( s) =

τ ⋅s + 1

τ

τ ⋅s + 1

τ

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

( )

− t− t0

(

τ

)

f ( t) = u t − t0 e

⌠   F( s) =  ⌡

∞

( )

− t− t0

(

τ

)

u t − t0 e

e

− st

dt

0 ∞

Let

λ = t − t0

∞

⌠ ⌠ −λ  1   − s + λ − s λ + t − t ⋅ s  0 0  τ τ dλ = e F( s) =  u( λ ) e e ⋅ e  dλ ⌡ ⌡− t 0

(

)

0

=e

− t0⋅ s

 −1 ⋅ ⋅e 

− s+

s+

1 τ

⋅λ

⋅ ∞

1

e

=

s+

0

τ

− t0⋅ s

=

1

τ ⋅e

τ ⋅s + 1

τ

The result to part (b) agrees with the real translation theorem. t0 := 1

Sketch the functions:

τ := 1

u ( t) :=

F( s) =

0 if t < 0 1 if t ≥ 0

t0

f ( t) := e

τ

τ ⋅e

− t0⋅ s

τ ⋅s + 1

( )

− t− t0

−t

⋅e

− t0⋅ s

(

)

fd ( t) := u t − t0 ⋅ e

τ

2

τ

2 fd( t)

f ( t)

0

0

2

4 t

0

0

2

4 t

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-6. Solution of differential equations by Laplace transforms Input function:

X( t) = u ( t)

X( s) =

d⋅ y ( t)

(a)

dt

Laplace transform: Solve for Y(s):

2 y( 0) = 5 x( 0) = 3 d ⋅ Y( t)

+ 2 Y( t) = 5 X( t)

dt

1

5

Y( s) =

A1 =

=

s+2 s

5

lim

−5 s+1

+

2

9⋅

d ⋅ y( t) 2

+ 18⋅

dt

A1

+

s+2

5 d⋅ y ( t)

1

Y( 0 ) = y ( 0 ) − y ( 0 ) = 0

s

s A2 =

Invert:

s

X( t) = x ( t) − x ( 0 )

A2

= −2.5

s→−2 s

Y( s) =

5

lim

s→0 s+ 2

Y( t) = −2.5e

−t

= 2.5

+ 2.5 u ( t)

(Table 2-1.1)

+ 4 y ( t) = 8 x ( t) − 4

dt

4⋅ y( 0) = 8 x( 0) − 4

Initial steady state: Subtract:

Y( t) = y ( t) − y ( 0 )

sY( s) − Y( 0 ) + 2 Y( s) = 5 X( s) = 5 ⋅

Partial fractions:

(b)

(Table 2-1.1)

s

+ 2 y ( t) = 5 x ( t) + 3

Initial steady state: Subtract:

1

2

9⋅

d ⋅ Y( t) 2

+ 18⋅

dt

d ⋅ Y( t)

+ 4 Y( t) = 8 X( t)

dt

Y( t) = y ( t) − y ( 0 )

Y( 0 ) = 0

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

2

9s Y( s) + 18s⋅ Y( s) + 4 Y( s) = 8 X( s) = 8 ⋅ Solve for Y(s):

1

8

Y( s) =

2

9s + 18s + 4

r1 :=

s

r2 := Expand in partial fractions:

A1 =

Y( s) =

lim

=

2

18 − 4 ⋅ 9 ⋅ 4 2⋅ 9

r1 = −0.255

2

−18 −

9 ( s + 0.255 ) ( s + 1.745 )s 8

s

−18 +

8

s → − 0.255 9 ( s + 1.745 )s

1

18 − 4 ⋅ 9 ⋅ 4 2⋅ 9 =

A1 s + 0.255

+

r2 = −1.745 A2 s + 1.745

8 9 ⋅ ( −0.255 + 1.745 ) ⋅ ( −0.255 )

+

A3 s

= −2.342

A2 = A3 =

8

lim

−2.342 s + 0.255

Y( t) = −2.342 e 2

(c)

9⋅

d ⋅ y( t) 2

d ⋅ Y( t)

d ⋅ Y( t) dt

dt

9 ( 0.255 ) ( 1.745 ) +

+ 0.342e

= 0.342

= 2.0

2 s

− 1.745t

+ 2 u ( t)

+ 4 y ( t) = 8 x ( t) − 4

dt

+ 9⋅

2

s + 1.745

d⋅ y ( t)

+ 9⋅

2

9⋅

0.342

− 0.255t

dt Subtract initial steady state:

+

8

=

s → 0 9 ( s + 0.255 ) ( s + 1.745 )

Invert with Table 2-1.1:

+ 4 Y( t) = 8 X( t)

Y( 0 ) = 0

(9s2 + 9s + 4)Y(s) = 8 X(s) = 8⋅ 1s

Laplace transform:

2

−9 +

Find roots:

r1 :=

Solve for Y(s), expand:

Y( s) =

9 − 4⋅ 9⋅ 4

r2 :=

2⋅ 9

A1 s + 0.5 − 0.441i

A3 =

s → 0 9s2 + 9s + 4

−1 + 1.134i s + 0.5 − 0.441i

Y( t) = ( −1 + 1.134i)e

A2

+

s + 0.5 + 0.441i

9⋅

d ⋅ y( t) 2

+ 12⋅

dt Subtract initial steady state:

2

9⋅

+

A3

d ⋅ Y( t) 2

dt

+ 12⋅

−1 − 1.134i s + 0.5 + 0.441i

+ ( −1 − 1.134i)e

d⋅ y ( t) dt d ⋅ Y( t) dt

s = −1 + 1.134i

=2

+

( − 0.5+ 0.441i)t

2

(d)

r1 = −0.5 + 0.441i r2 = −0.5 − 0.441i

9 ( 2 ⋅ 0.441i) ( −0.5 + 0.441i)

8

lim

Y( s) =

2⋅ 9

8

s → − 0.5+ 0.441i 9 ( s + 0.5 + 0.441i) s

A2 = −1 − 1.134i

9 − 4⋅ 9⋅ 4

9 ( s + 0.5 − 0.441i) ( s + 0.5 + 0.441 )s

8

lim

2

−9 −

8

=

Invert using Table 2-1.1:

8 9 ( −1.745 + 0.255 ) ( −1.745 )

8

lim

Y( s) =

A1 =

=

s → − 1.745 9 ( s + 0.255 )s

+

2 s

( − 0.5− 0.441i)t

+ 2 u( t)

+ 4 y ( t) = 8 x ( t) − 4

+ 4 Y( t) = 8 X( t)

Y( 0 ) = 0

(9s2 + 12s + 4)Y(s) = 8 X(s) = 8⋅ 1s

Laplace transform: Find roots:

r1 :=

2

−12 +

12 − 4 ⋅ 9 ⋅ 4

r2 :=

2⋅ 9

2

−12 −

12 − 4 ⋅ 9 ⋅ 4

r1 = −0.667

2⋅ 9

r2 = −0.667 Solve for Y(s) and expand:

8

Y( s) =

2

9 ( s + 0.667 ) s A1 =

8

lim

s → − 0.667 9s

=

−4

A2 =

3

A3 =

Invert using Table 2-1.1:

s + 0.667

+

A3 s

−8  8  = lim = −2  ds  9s  s → − 0.667 9s2 8

lim

s → 0 9 ( s + 0.667 ) 2

d ⋅ y( t) 3

2

+ 7⋅

d ⋅ y ( t) 2

+ 21⋅

=2

d ⋅ y( t)

dt

3

2⋅

A2

d

s → − 0.667

dt Subtract initial steady state:

2

+

 −4 t − 2 e− 0.667t + 2 u ( t)  3 

Y( t) =

2⋅

( s + 0.667 )

lim

3

(e)

A1

=

d ⋅ Y( t) 3

2

+ 7⋅

dt

d ⋅ Y( t) 2

+ 21⋅

+ 9 y( t) = 3 x( t)

dt d ⋅ Y( t) dt

dt

+ 9 Y( t) = 3 X( t) Y( 0 ) = 0

(2s3 + 7s2 + 21s + 9)Y(s) = 3 X(s) = 3⋅ 1s

Laplace transform:

  9    −1.5 − 2.598i  21      polyroots = −1.5 + 2.598i  7   −0.5    2  

Find roots:

Solve for Y(s) and expand:

Y( s) =

= A1 =

3 2 ( s + 1.5 − 2.598i) ( s + 1.5 + 2.598i) ( s + 0.5)s A1 s + 1.5 − 2.598i 3

lim

s → − 1.5+ 2.598i 2 ( s + 1.5 + 2.598i) ( s + 0.5)s

3 2 ( 2 ⋅ 2.598i) ( −1 + 2.598i) ( −1.5 + 2.598i)

+

A2 s + 1.5 + 2.598i

+

A3 s + 0.5

= 0.027 + 0.022i

= 0.027 + 0.022i

A2 = 0.027 − 0.022i

+

A4 s

A3 =

3

lim

s → − 0.5 2 ( s + 1.5 − 2.598i) ( s + 1.5 + 2.598i)s

3 2 ( 1 − 2.598i) ( 1 + 2.598i) ( −0.5)

= −0.387

Y( s) =

A4 =

0.027 + 0.022i s + 1.5 − 2.598i

= −0.387

lim

3 3

=

2

s → 0 2s + 7s + 21s + 9

+

0.027 − 0.022i s + 1.5 + 2.598i

+

−0.387 s + 0.5

1 3

+

1 1 3 s

Invert using Table 2-1.1: Y( t) = ( 0.027 + 0.022i)e

( − 1.5+ 2.598i)t

+ ( 0.027 − 0.022i)e

( − 1.5− 2.598i)t

− 0.387e

− 0.5t

+

1 3

u( t)

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Smith & Corripio, 3rd edition Problem 2-7. Solve Problem 2-6(d) with different forcing functions 2

9⋅

d ⋅ Y( t) 2

+ 12⋅

d ⋅ Y( t)

+ 4 Y( t) = 8 X( t)

dt

dt

Y( 0 ) = 0

−t

(a) Forcing function:

3

X( t) = e

From Table 2-1.1:

1

X( s) =

s+

(9s2 + 12s + 4) s + 1  

3

8

=

2

A2 =

8

s → − 0.667 9 ( s + 0.333 )

lim

d

 

=

−8

Invert using Table 2-1.1: Y( t) =

−8

( s + 0.667 )

2

+

−8 −8 s + 0.667

s + 0.667

+

A3 s + 0.333

8

+

=8

= −8

s → − 0.667 9 ( s + 0.333 ) 2

1

A2

s → − 0.333 9 ( s + 0.667 ) 2

8 s + 0.333

 −8 t − 8 e− 0.667t + 8e− 0.333t  3  − ( t− 1)

(b) Forcing function:

2

+

lim

 = lim 

8

3

( s + 0.667 )

A3 =

3

s → − 0.667 ds  9 ( s + 0.333 ) 

Y( s) =

A1

=

9 ( s + 0.667 ) ( s + 0.333 ) lim

3

8

Y( s) =

A1 =

1

X( t) = u ( t − 1 ) e

3

X( s) =

e

−s

s+

1

(Real translation theorem)

3

The partial fraction expansion of the undelayed signal is the same: Y( s) =

1 8 8  −8  e− s − + 3 2 s + 0.667 s + 0.333   ( s + 0.667 ) 

Apply the real translation theorem in reverse to this solution: Y( t) = u ( t − 1 ) ⋅  

−8

 3

⋅ ( t − 1 ) − 8 ⋅ e



− 0.667⋅ ( t− 1)

+ 8⋅ e

− 0.333⋅ ( t− 1) 

 

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Smith & Corripio, 3rd edition Problem 2-8. Response characteristics of the equations of Problem 2-6 d⋅ y ( t)

(a)

dt

+ 2 y ( t) = 5 x ( t) + 3 2 y( 0) = 5 x( 0) + 3

Initial steady state: d ⋅ Y( t)

Subtract:

dt

+ 2 Y( t) = 5 X( t)

Y( t) = y ( t) − y ( 0 ) X( t) = x ( t) − x ( 0 )

s⋅ Y( s) + 2 Y( s) = 5 X( s) Y( 0 ) = y ( 0 ) − y ( 0 ) = 0 A 5 1 + terms of X(s) Y( s) = X( s) = s+2 s+2

Laplace transform: Solve for Y(s):

Y( t) = A1 ⋅ e

Invert using Table 2-1.1:

− 2t

+ terms of X(t)

The response is stable and monotonic.The dominant and only root is

−5

Time for response to decay to within 0.67% of its initial value: Final steady-state value for unit step input:

lim s⋅

(Final value theorem)

s→0

5

s+2 s

2

(b)

9⋅

d ⋅ y( t) 2

+ 18⋅

d⋅ y ( t)

2

9⋅

d ⋅ Y( t)

+ 18⋅

2

d ⋅ Y( t) dt

dt

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

8

r 5

→

2

= 2.5 min

= 2.5

+ 4 y ( t) = 8 x ( t) − 4

dt

dt Subtract initial steady state:

1

−1

r := −2 min

+ 4 Y( t) = 8 X( t)

Y( 0 ) = 0

X( s)

2

9s + 18s + 4 Find roots:

r1 :=

−18 +

2

18 − 4 ⋅ 9 ⋅ 4

r2 :=

2 ⋅ 9 min

−18 −

2

18 − 4 ⋅ 9 ⋅ 4 2 ⋅ 9 min

−1

r1 = −0.255 min

−1

r2 = −1.745 min Invert using Table 2-1.1:

Y( t) = A1 ⋅ e

− 0.255t

+ A2 ⋅ e

− 1.745t

+ terms of X(s) −1

r1 = −0.255 min −5 = 19.6 min r1

The response is stable and monotonic. The domnant root is: Time for the response to decay to 0.67% of its initial value:

Final steady-state value for unit step input: (Final value theorem)

lim s⋅ s→0

8 2

9s + 18s + 4

1 s

→2

2

(c)

9⋅

d ⋅ y( t)

+ 9⋅

2

d⋅ y ( t)

2

Subtract initial steady state:

9⋅

d ⋅ Y( t) 2

+ 9⋅

d ⋅ Y( t)

8

Y( s) =

+ 4 Y( t) = 8 X( t)

dt

dt Laplace transform and solve for Y(s):

+ 4 y ( t) = 8 x ( t) − 4

dt

dt

Y( 0 ) = 0

X( s)

2

9s + 9s + 4 Find the roots:

r1 :=

−9 +

2

9 − 4⋅ 9⋅ 4 2 ⋅ 9 min

2

−9 −

r2 :=

9 − 4⋅ 9⋅ 4 2 ⋅ 9 min

−1

r1 = −0.5 + 0.441i min

−1

r2 = −0.5 − 0.441i min Invert using Table 2-3.1:

Y( t) = D⋅ e

− 0.5t

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:

2π

T :=

T = 14.25 min

−1

0.441min Decay ratio:

e

− 0.5min

−1

T

= 0.00081

−5

Time for oscillations to die:

−1

= 10 min

−0.5min Final steady state value for a unit step imput:

s→0

(Final value theorem) 9⋅

d ⋅ y( t) 2

+ 12⋅

9⋅

d ⋅ Y( t) 2

s

2

d⋅ y ( t)

+ 12⋅

d ⋅ Y( t) dt

dt

→2

+ 4 y ( t) = 8 x ( t) − 4

dt

dt 2

Subtract initial steady state:

1

9s + 9s + 4

2

(d)

8

lim s⋅

+ 4 Y( t) = 8 X( t) Y( 0 ) = 0

Laplace transform and solve for Y(s):

Y( s) =

8

X( s)

2

9s + 12s + 4 Find roots:

r1 :=

−12 +

2

12 − 4 ⋅ 9 ⋅ 4 2 ⋅ 9 min

r2 :=

−12 −

2

12 − 4 ⋅ 9 ⋅ 4 2 ⋅ 9 min

−1

r1 = −0.667 min

−1

r2 = −0.667 min Invert using Table 2-1.1:

(

) − 0.667t

Y( t) = A1 ⋅ t + A2 e

+ terms of X(t)

−1

r1 = −0.667 min −5 Time required for the response to decay within 0.67% of its initial value: = 7.5 min r1

The response is stable and monotonic. The dominant root is

8

lim s⋅

Final steady state value for a unit step input:

s→0

(Final value theorem) 2⋅

d ⋅ y( t)

9s + 12s + 4

3

2

d ⋅ y ( t)

+ 7⋅

dt 2⋅

d ⋅ Y( t) 3

2

+ 7⋅

d ⋅ Y( t)

+ 21⋅

2

dt Laplace transform and solve for Y(s):

3

+ 9 y( t) = 3 x( t)

dt d ⋅ Y( t)

dt 3

Y( s) =

d ⋅ y( t)

dt

3

Subtract initial steady state:

+ 21⋅

2

→2

s

2

3

(e)

1

dt

+ 9 Y( t) = 3 X( t)

X( s)

2

2s + 7s + 21s + 9 Find roots:

  9   21 −1 r := polyroots    min  7   2  

 −1.5 − 2.598i  −1 r =  −1.5 + 2.598i min  −0.5  

The response is stable and oscillatory. The dominant root is The period of the oscillations is:

−1

r = −0.5 min 2

2π

T :=

T = 2.42 min

−1

2.598min Decay ratio:

e

− 1.5 min

−1

T

= 0.027

Time for response to die out:

−5 r

= 10 min

2

Final steady state value for a unit step input: (Final value theorem)

lim s⋅ s→0

3 3

2

2s + 7s + 21s + 9

1 s

→

1 3

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Smith & Corripio, 3rd edition Problem 2-9. Second-Order Response: Bird Mobile M := 50gm y 0 := −27cm

Problem data:

Solution: Force balance: M⋅

d⋅ v ( t) dt

= −M ⋅ g − k ⋅ y ( t) + f ( t) d⋅ y ( t)

Velocity:

dt

Initial steady state:

-ky(t)

= v( t) 0 = −M ⋅ g − k ⋅ y 0

-Mg

Subtract and substitute:

f(t)

2

M⋅

d ⋅ Y( t) 2

y(t) y=0

= −k ⋅ Y( t) + f ( t)

dt

Y( 0 ) = 0 Value of k:

k :=

Laplace transform:

Solve for Y(s):

−M ⋅ g y0

k = 1.816

N m

2

M ⋅ s Y( s) + k ⋅ Y( s) = F( s) Y( s) =

1 2

F( s) =

M⋅ s + k

A1 s − i⋅

k M

+

A2 k

s + i⋅

M + terms of F(s) θ := 0

Invert using Table 2-3.1:

 k t⋅ s + θ   M 

Y( t) := D⋅ sin

The mobile will oscillate forever with a period of

D := 1 + terms of f(t)

T := 2π ⋅

M k

T = 1.043 s

1

Y ( t)

0

1 0

2

4 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: 2

M⋅

d ⋅ Y( t) 2

= −k ⋅ Y( t) − b ⋅

dt

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

F( s)

2

M⋅ s + b⋅ s + k Invert:

−b

Y( t) = D⋅ e

2M

r1 =

−b +

2

b − 4M ⋅ k 2M

=

−b 2M

+ i⋅

k M

−

b

2

4M

2

2

⋅t

2  k  b + terms of f(t) − t+θ  M 2 4M  

b < 4M ⋅ k

sin

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Smith & Corripio, 3rd edition Problem 2-10. Responses of general first-order differential equation τ⋅

d ⋅ Y( t) dt

+ Y( t) = K⋅ X( t)

Laplace transform and solve for Y(s):

(a) Unit impulse:

X( t) = δ ( t) Y( s) =

Y( s) =

Y( 0 ) = 0

K τ ⋅s + 1

X( s)

From Table 2-1.1:

X( s) = 1

K τ ⋅s + 1 −t

Invert using Table 2-1.1:

K τ Y( t) := e τ 1

Y ( t)

(b) Pulse of Fig. 2-1.1b

0.5

0

From Example 2-1.1b:

0

2

4 t

X( s) = H⋅

Substitute:

A1 =

Y( s) =

lim s→

K τ ⋅s + 1

K⋅ H − 1 τ ⋅s

⋅ H⋅

1−e

− sT

s 1−e

− sT

=

s

= −K⋅ H

A2 =

(

K⋅ H

lim

s → 0 τ ⋅s + 1

τ

Y( s) = K⋅ H

 

1 s

−

1 s+

)

A2   A1 − sT + 1−e  s s + 1 τ  

1 τ

(

= K⋅ H

)

 1 − e− sT 

Invert using Table 2-1.1, and the real translation theorem:

KH := 1 τ := 1

T := 1

H := 1

−t − ( t− T)        τ τ Y( t) := KH⋅  u ( t) − e − u ( t − T) ⋅  1 − e 

X( t) := H⋅ ( u ( t) − u ( t − T) )

1 Y( t) X( t)

0.5

0 0

2

4 t

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Smith & Corripio, 3rd edition Problem 2-11. Response of an integrating process A⋅ Laplace transform and solve for H(s):

H( s) F( s)

Response to a unit step in flow:

A := 1

H( s) =

= f ( t)

dt

H( s) =

Transfer function of the tank:

Substitute:

d ⋅ h( t)

1 A⋅ s =

F( s)

1 A⋅ s

f ( t) = u( t) 1 1 A 2 s

F( s) =

1 s

Invert using Table 2-1.1:

(Table 2-1.1)

h ( t) :=

1 A

t

10

f(t) h( t)

h(t)

h( 0) = 0

5

0

0

5

10

t

The tank is an integrating process because its ouput, the level, is the time integral of its input, the inlet flow. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-12. Second-order differeential equations of Problem 2-6. 2 2 d ⋅ Y( t )

τ ⋅

Standard form of the second-order equation:

2

+ 2⋅ ζ ⋅ τ ⋅

d ⋅ Y( t) dt

dt 2

(b)

9⋅

d ⋅ y( t) 2

+ 18⋅

d⋅ y ( t)

2

Subtract the initial steady state:

9⋅

d ⋅ Y( t) 2

+ 18⋅

+ 4 y ( t) = 8 x ( t) − 4

dt

dt

d ⋅ Y( t) dt

dt

+ Y( t) = K⋅ X( t)

+ 4 Y( t) = 8 X( t)

Y( 0 ) = 0

2

9 d ⋅ Y( t) 18 d ⋅ Y( t) ⋅ + ⋅ + Y( t) = 2 X( t) 2 4 dt 4 dt

Divide by Y(t) coefficient: Match coeffients to standard form:

τ :=

Equivalent time constants: Find roots:

r1 :=

r2 :=

9

18 − 4 ⋅ 9 ⋅ 4

−1

2

18 − 4 ⋅ 9 ⋅ 4

−1

τ e2 :=

2

(b)

9⋅

d ⋅ y( t) 2

dt

+ 9⋅

−1

τ e1 = 3.927 min

r1 −1

τ e2 = 0.573 min

r2

5 ⋅ τ e1 = 19.64 min

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

d⋅ y ( t) dt

+ 4 y ( t) = 8 x ( t) − 4

2

9 d ⋅ Y( t) 9 d ⋅ Y( t) ⋅ + ⋅ + Y( t) = 2 X( t) 2 4 dt 4 dt

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

Compare coefficients to standard form:

τ e1 :=

r2 = −1.745 min

2 ⋅ 9 min

ζ = 1.5

Overdamped.

r1 = −0.255 min

2 ⋅ 9 min −18 −

4⋅ 2⋅ τ

K := 2

2

−18 +

18min

min τ = 1.5 min ζ :=

4

τ :=

9 4

min τ = 1.5 min ζ :=

9min 4⋅ 2⋅ τ

Y( 0 ) = 0

ζ = 0.75

K := 2 Find roots:

r1 :=

2

−9 +

9 − 4⋅ 9⋅ 4 2 ⋅ 9 min

Frequency of oscillations:

ω := 0.441

Underdamped.

−1

r1 = −0.5 + 0.441i min rad min

Period of oscillations:

T :=

2π ω

T = 14.25 min

Decay ratio:

Rise time:

e

− 0.5 min

T 4

−1

T

= 0.00081

− 0.5min

Percent overshoot:

= 3.56 min

−1

2

e −5

Settling time:

T

−1

= 2.8 %

= 10 min

−0.5min 2

(c)

9⋅

d ⋅ y( t) 2

+ 12⋅

dt Subtract initial steady state and divide by the coefficient of Y(t):

d⋅ y ( t)

+ 4 y ( t) = 8 x ( t) − 4

dt

2

9 d ⋅ Y( t) d ⋅ Y( t) ⋅ + 3⋅ + Y( t) = 2 X( t) 2 4 dt dt Y( 0 ) = 0

Compare coefficients to standard form:

τ :=

9 4

min

τ = 1.5 min K := 2

Equivalent time constants:

Find roots:

r1 :=

r2 :=

−12 +

2

12 − 4 ⋅ 9 ⋅ 4 2 ⋅ 9 min

−12 −

−1

r1 = −0.667 min

ζ :=

3min 2⋅ τ

ζ =1

Critically damped.

τ e1 :=

−1 r1

τ e1 = 1.5 min

2

12 − 4 ⋅ 9 ⋅ 4 2 ⋅ 9 min

−1

r2 = −0.667 min

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

τ e2 :=

−1 r2

τ e2 = 1.5 min

5 ⋅ τ e1 = 7.5 min

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Smith & Corripio, 3rd edition Problem 2-13. Partial fraction expansion coefficients for Eqs. 2-5.10 to 2-5.13 Step response, over-damped second-order differential equation: X( s) = Y( s) =

A1 =

K

(τ e1⋅ s + 1)(τ e2⋅ s + 1) lim s→

A3 =

lim

∆x

A1

=

s

s+

K⋅ ∆x

K⋅ ∆x

)(

s → 0 τ e1⋅ s + 1 τ e2⋅ s + 1



Y( s) = K⋅ ∆x 

−τ e1

1

)

+

s+

τ e1

s

A3

+

1

s

τ e2

−K⋅ ∆x ⋅ τ e1

=

− 1 τ ⋅τ ⋅s + 1  s e1 e2  τ e1  τ e2 

(

1

A2

∆x

A2 =

τ e1 − τ e2

−K⋅ ∆x ⋅ τ e2 τ e2 − τ e1

= K⋅ ∆x

−

τ e2

1

+

1

1 s τ e2 − τ e1  τ e1 − τ e2 s + 1 s+  τ e1 τ e2   − t −t  Invert using Table 2-1.1: (2-5.10) τ e1 τ e2 τ e1 τ e2   Y( t) = K⋅ ∆x u ( t) − e − e  τ e1 − τ e2 τ e2 − τ e1  

Step response for the critically damped case: Y( s) =

A1 =

s→

A2 =

K⋅ ∆x

lim −1

2

=

τ s

K

∆x

(τ ⋅ s + 1)2

s

A1

=

s +  

−K⋅ ∆x

1

2

A2 s+

τ

A3 =

τ

+

lim

1

+

d

K⋅ ∆x

s → 0 (τ ⋅ s + 1)2

−K⋅ ∆x  K⋅ ∆x  = lim = −K⋅ ∆x − 1 τ 2 s2  s→

 − 1 ds τ 2s  s→ τ

τ

Y( s) = K⋅ ∆x 

−1

1

τ   s +  

1 τ

2

−

s

τ

τ

lim

A3

 1  + 1  1 s s + τ   

= K⋅ ∆x

Invert using Table 2-1.1:

  Y( t) = K⋅ ∆x  u ( t) − 

Ramp response for the over-damped case:

 t + 1 e  τ 

− t

  

τ

(2-5.11)

r

X( s) =

2

s Y( s) =

A1 =

A3 =

A4 =

K

(τ e1⋅ s + 1)⋅ (τ e2⋅ s + 1) s2

− 1 τ ⋅ τ ⋅  s + 1  ⋅ s2 s→ e1 e2  τ e1  τ e2 

lim

K⋅ r

(

)(

s → 0 τ e1⋅ s + 1 ⋅ τ e2⋅ s + 1

lim s→0

A1

=

1

s+

)

=

A2

+

s+

τ e1

2

K⋅ r

lim

r

1 τ e2

+

A3 2

s

+

A4 s

2

K⋅ r⋅ τ e1

A2 =

τ e1 − τ e2

K⋅ r⋅ τ e2

τ e2 − τ e1

= K⋅ r

 = lim K⋅ r⋅ −τ e1⋅ ( τ e2⋅ s + 1) − τ e2⋅ ( τ e1⋅ s + 1) ds ( τ e1⋅ s + 1 ) ⋅ ( τ e2⋅ s + 1 )    s→0 (τ e1⋅ s + 1)2 (τ e2⋅ s + 1)2 d

K⋅ r

⋅ 

(

)

= K⋅ r −τ e1 − τ e2

2  τ 2 τ e2 τ e1 + τ e2  e1 1 1 1  Y( s) = K⋅ r + + −  τ e1 − τ e2  1 1 2 s τ e2 − τ e1 s s+ s+  τ e1 τ e2  

Invert using Table 2-1.1: −t −t   2 2 τ e2 τ e1 τ e1 τ e2 Y( t) = K⋅ r e + e + t − ( τ e1 + τ e2)  τ e1 − τ e2  τ e2 − τ e1  

Ramp response for critically damped case: Y( s) =

K

r

(τ ⋅ s + 1)2

s

2

=

A1 2 s + 1    τ

+

A2 s+

1 τ

+

A3 2

s

+

A4 s

(2-5.12)

A1 =

A2 =

2 2

= K⋅ r

A3 =

−1 τ s s→ τ

lim

K⋅ r

s → 0 (τ ⋅ s + 1)2

= K⋅ r

K⋅ r  K⋅ r  = lim −2 ⋅ = 2 ⋅ K⋅ r⋅ τ 2 3 ds  2 2 − 1 τ s τ s  s → d

lim s→

A4 =

K⋅ r

lim

−1

lim s→0

τ

τ

 K⋅ r  = lim −2⋅ K⋅ r⋅ τ = −2 ⋅ K⋅ r⋅ τ 2 ds  ( τ ⋅ s + 1) 3  ( τ ⋅ s + 1)  s → 0 d

1 2⋅ τ  2⋅ τ  1 + + −  s    1  2 s + 1 s2 τ s + τ    

Y( s) = K⋅ r

Invert using Table 2-1.1:

−t     τ Y( t) = K⋅ r⋅ ( t + 2 ⋅ τ ) e + t − 2 ⋅ τ

(2-5.13)

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-14. Derive step reponse of n lags in series ∆x

K

Y( s) =

n

∏ (τ k⋅ s + 1)

s

=

A0 s

n

+

K⋅ ∆x

lim

s

∑

k=1 s +

1 τk

= K⋅ ∆x

n

s→0

∆x

Ak

k= 1

A0 =

X( s) =

∏ (τ k⋅ s + 1)

k= 1

−t

Invert using Table 2-1.1:

n

∑

Y( t) = K⋅ ∆x ⋅ u ( t) +

Ak ⋅ e

τk

k= 1

Ak =

K⋅ ∆x

lim s→

−1

n

τk

s⋅

∏

j = 1⋅ ( j≠ k)

1 τk Substitute:

⋅

1 τk

n− 1

n

⋅ τ k⋅

−1

s + 1 ⋅ τj  τ j j= 1 

∏

−K⋅ ∆x

=

∏

j = 1⋅ ( j≠ k)

K⋅ ∆x

=

n

τk

=

n

⋅

n

∏

j = 1( j≠ k)

−K⋅ ∆x ⋅ τ k n

( τ k − τ j) ∏

 −1 + 1  ⋅ τj τ  k τj  j = 1

∏

n− 1

( τ k − τ j)

j = 1( j≠ k)

− t  n− 1   n τk τk  Y( t) = K⋅ ∆x u ( t) − e    n k=1  (τ k − τ j)   j = 1( j≠ k)  

∑

(2-5.23)

∏

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Smith & Corripio, 3rd edition Problem 2-15. Transfer function of second-order interacting systems. Y( s) =

k1

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1) − k2

k1

X( s) =

(

2

X( s)

)

τ 1⋅ τ 2⋅ s + τ 1 + τ 2 s + 1 − k2

Standard form of the second-order differential equaton, Eq. 2-5.4: Y( s) =

K

X( s)

2 2

τ s + 2ζ ⋅ τ ⋅ s + 1 k1

Rerrange interacting equation: Y( s) =

Comparing coefficients:

Gain:

K=

1− k2

 τ 1⋅ τ 2  2 τ 1 + τ 2 s + s+1  1 − k2  1 − k2  k1

Time constant:

1 − k2 ζ =

Damping ratio:

Find the roots of the denominator:

(

2

τ1 + τ2

(

2⋅ τ ⋅ 1 − k2

)

=

X( s)

τ =

τ 1⋅ τ 2 1 − k2

τ1 + τ2

(

2⋅ τ 1⋅ τ 2⋅ 1 − k2

)

)

τ 1⋅ τ 2⋅ s + τ 1 + τ 2 s + 1 − k2 = 0

r1 =

(

)

− τ1 + τ2 +

(τ 1 + τ 2)2 − 4τ 1⋅ τ 2(1 − k2) 2⋅ τ 1⋅ τ 2

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

(τ 1 + τ 2)2 − 4τ 1⋅ τ 2⋅ (1 − k2) = τ 12 + 2τ 1⋅ τ 2 + τ 22 − 4τ 1⋅ τ 2 + 4τ 1⋅ τ 2⋅ k2 2

2

= τ 1 − 2τ 1 ⋅ τ 2 + τ 2 + 4τ 1 ⋅ τ 2 ⋅ k 2

(

= τ1 − τ2

)2 + 4τ 1⋅ τ 2⋅ k2

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

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

r1 =

(

)

− τ1 + τ2 +

(τ 1 + τ 2)2 − 4τ 1⋅ τ 2(1 − k2) 2⋅ τ 1⋅ τ 2

If τ1 , τ2 , and k 2 are positive, and if k 2 < 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: 2⋅ τ 1⋅ τ 2

τ e1 = τ1 + τ2 −

(τ 1 − τ 2)2 + 4τ 1⋅ τ 2⋅ k2

2⋅ τ 1⋅ τ 2

τ e1 = τ1 + τ2 +

(τ 1 − τ 2)2 + 4τ 1⋅ τ 2⋅ k2

The first of these is the dominant time constant. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-16. Transfer function of a second-order feedback control loop C( s) =

Kc

Kc

R( s) =

( 3s + 1 ) ⋅ ( s + 1 ) + Kc

2

3s + 4s + 1 + Kc

This is a second-order process with a proportional controller. Standard second-order transfer function, Eq. 2-5.4:

K

C( s) =

R( s)

2

τ + 2ζ ⋅ τ ⋅ s + 1 Rearrange feedback loop transfer function and compare coefficients: Kc 1+ K c

C( s) =

3 1 + Kc

Gain:

K=

Kc 1 + Kc

s +

1 + Kc

R( s) s+1

3

Time constant: τ =

1 + Kc

ζ =

Damping ratio:

4

2

(

4

2 ⋅ τ ⋅ 1 + Kc

)

=

2

(

3 ⋅ 1 + Kc

)

Ranges of the controller gain for which the response is:

(i) Overdamped:

ζ >1

(ii) Underdamped:

0 1 + Kc 3

1 Kc < 3

ζ cannot be negative for positive K c Cannot be undamped for finite K c.

The response canot be unstable for positive K c. The time constant and damping ratio are always real and positive for positive gain. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-17. Linearization of common process model functions. (a) Enthalpy as a function of temperature, Eq. 2-6.1: (use subscript b for base value) 2

3

4

H( T( t) ) = H0 + a1 ⋅ T( t) + a2 ⋅ T ⋅ ( t) + a3 ⋅ T ( t) + a4 ⋅ T ( t) H( T( t) ) = H Tb +  a1 + 2a2 ⋅ Tb + 3a3 ⋅ Tb + 4a4 ⋅ Tb 

( )

Let

2

3

 ( T( t ) − Tb )

Hd ( Γ ( t) ) = H( T( t) ) − H Tb

( )

Γ ( t ) = T( t ) − Tb

Hd ( Γ ( t) ) =  a1 + 2a2 ⋅ Tb + 3a3 ⋅ Tb + 4a4 ⋅ Tb  2

3

 Γ ( t)

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

o

p ( T( t ) ) = e

B T( t) + C A−

( )

o

Let

B

o

p ( T( t ) ) = p Tb +

(

Tb + C

P Γ ( t) ) =

e

( T( t ) − Tb )

P ( Γ ( t ) ) = p ( T( t ) ) − p Tb o

Γ ( t ) = T( t ) − Tb

o(

)

2

B Tb+ C

o

( )

o

( ) Γ ( t) ( Tb + C ) 2 o

B ⋅ p Tb

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

y( x( t) ) =

1 + ( α − 1 ) x ( t)

( )

y x ( t) = y xb +

Let

 1 + ( α − 1 ) ⋅ x b ⋅ α − α ⋅ xb ⋅ ( α − 1) ( x ( t) − xb ) 2 ( )  1 + α − 1 x b

( )

X( t) = x ( t) − x b

Y( X( t ) ) =

Y( X( t) ) = y ( x ( t) ) − y x b

α

 1 + ( α − 1) xb 



2

X( t)

(d) Flow as a function of pressure drop, Eq. 2-6.4: f ( ∆p ( t) ) = k ⋅ ∆p ( t)

( )

f ( ∆p ( t) ) = f ∆p b + Let

k 2 ⋅ ∆p b

(∆p(t) − ∆pb) ( )

F( ∆P( t) ) = f ( ∆p ( t) ) − f ∆p b

∆P( t) = ∆p ( t) − ∆p b

F( ∆P( t) ) =

k 2 ⋅ ∆p b

∆P( t)

(e) Radiation heat transfer rate as a function of temperature, Eq. 2-6.5: 4

q ( T( t) ) = ε ⋅ σ ⋅ A⋅ T ( t)

( )

(

3

q ( T( t) ) = q Tb + 4 ⋅ ε ⋅ σ ⋅ A⋅ Tb T( t) − Tb Let

Γ ( t ) = T( t ) − Tb

)

Q( Γ ( t) ) = q ( T( t) ) − q Tb

( )

Q( Γ ( t) ) = 4 ⋅ ε ⋅ σ ⋅ A⋅ Tb ⋅ Γ ( t) 3

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-18. Linearization of radiation heat transfer--range of accuracy. q ( T) = 4ε ⋅ σ ⋅ A⋅ T

4

Use subscript "b" for base value for linearization. d ⋅ q ( T)

From the solution to Problem 2-17(e), the slope is:

dT

= 4 ⋅ ε ⋅ σ ⋅ A⋅ T

3

Temperature range for which the slope is within 5% of the slope at the base value 3

error = 4 ⋅ ε ⋅ σ ⋅ A⋅ T − 4 ⋅ ε ⋅ σ ⋅ A⋅ Tb

= 0.05⋅  4 ⋅ ε ⋅ σ ⋅ A⋅ Tb

3



3

 T  − 1 = 0.05 T  b

Simplify and rearrange:

For

Tb := 400K

Tmax :=

3

Tb := 600K

Tmax :=

3

1.05 Tb

Tmin :=

3

1.05 Tb

Tmin :=

3

3

Tmax = Tmin =

3

K := 1.8R

3



1.05 Tb = 1.0164Tb 0.95 Tb = 0.983Tb

0.95 Tb

Tmax = 407 K

Tmin = 393 K

0.95 Tb

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⋅ T −  ε ⋅ σ ⋅ A⋅ Tb + 4ε ⋅ σ ⋅ A⋅ Tb T − Tb  4

(

4

3

4

3

)

4

(

= 0.05 ε ⋅ σ ⋅ A⋅ T

)

4

4

T − 4 ⋅ Tb ⋅ T + 3Tb = 0.05T As the error is always positive, the absolute value brackets can be dropped. Rearrange into a polynomial and find its roots:

Simplify and rearrange:

0.95 

4

 −4 T +3=0 Tb Tb   T

 3    −1.014 − 1.438i  −4   −1.014 + 1.438i  polyroots  0   =   0.921   0    1.108      0.95   Ignore the complex roots. The other two roots are the lower and upper limits of the range: 0.921 ≤ For

Tb := 400K

Tmin := 0.921 ⋅ Tb

T Tb

≤ 1.108

Tmax := 1.108Tb

Tmin = 368 K

Tmax = 443 K

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-19. Equilibrium vapor composition--range of accuracy y( x) =

α⋅x

Use subscript "b" for base value for linearization.

1 + (α − 1)x d⋅ y( x)

From the solution to Problem 2-17(c):

dx

For the slope: error =

α

Upper limit:

−

2  1 + ( α − 1 ) x

α 2  1 + ( α − 1 ) x

α 2  1 + ( α − 1 ) x b

= 0.05

α 2  1 + ( α − 1 ) x b

 1 + (α − 1) xb    − 1 = 0.05  1 + (α − 1)x  2

Simplify and rearrange:

Lower limit:

=

1 + (α − 1)xb

1 + ( α − 1 ) x min

=

1 + (α − 1) xb

1 + ( α − 1 ) x max

=

(

)

(

)

x min α , x b :=

1.05

0.95

1 + (α − 1) xb −

1.05

1 + (α − 1) xb −

0.95

x max α , x b :=

1.05( α − 1 )

0.95( α − 1 )

(a)

x min( 1.1 , 0.1) = −0.143 (zero)

x max( 1.1 , 0.1) = 0.362

0 ≤ x ≤ 0.362

(b)

x min( 1.1 , 0.9) = 0.637

x max( 1.1 , 0.9) = 1.183 (one)

0.637 ≤ x ≤ 1

(c)

x min( 5 , 0.1) = 0.092

x max( 5 , 0.1) = 0.109

0.092 ≤ x ≤ 0.109

(d)

x min( 5 , 0.9) = 0.872

x max( 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 α ⋅x

error =

1+ ( α − 1)x

α ⋅ xb

1 + (α − 1) xb

+

− 1 = 0.05

α

 1 + ( α − 1) xb

2

( x − xb)

2  1 + ( α − 1) xb − 1 = 0.05 1 + ( α − 1 ) x α ⋅ x b  1 + ( α − 1 ) x b + α ⋅ x − α ⋅ x b  

α⋅x

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

2 2  1 + ( α − 1) xb α ⋅ x = 0.95 1 + ( α − 1) x  α ( α − 1) xb + α ⋅ x

0.95⋅ ( α − 1 ) ⋅ x +  0.95⋅ ( α − 1 ) ⋅ x b + 0.95 − 1 − 2 ⋅ ( α − 1 ) ⋅ x b − ( α − 1 ) ⋅ x b  2

2

2

2

2

( )  ⋅ x + 0.95⋅ α − 1 ⋅ xb

2  +  −0.05⋅ ( α − 1) 2 x − 0.05 − 2 ( α − 1) ⋅ x + 0.95( α − 1) = 0 b   x xb  xb    b

0.95( α − 1 ) 

x

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

(a)

α := 1.1

x min := 0.138x b

(b)

α := 1.1

0.95⋅ ( α − 1 )     2 0.05 0.138    ( ) ( ) − 2 α − 1   =  polyroots −0.05 α − 1 x b − xb     7.231    0.95( α − 1 )  

x b := 0.1

x max := 7.231x b

x b := 0.9

x min := 0.444x b

(c)

α := 5

x b := 0.1

x min := 0.605x b

(d)

α := 5

x b := 0.9

x min := 0.577x b

x min = 0.014

x max = 0.723

0.014 ≤ x ≤ 0.723

0.95⋅ ( α − 1 )     2 0.05 0.444    ( ) ( ) − 2 α − 1   =  polyroots −0.05 α − 1 x b − xb     2.25    0.95( α − 1 )   x max := 2.25x b

x min = 0.4

x max = 2.025

0.40 ≤ x ≤ 1

0.95⋅ ( α − 1 )     2 0.05 0.605    ( ) ( ) − 2 α − 1   =  polyroots −0.05 α − 1 x b − xb     1.653    0.95( α − 1 )   x max := 1.653x b

x min = 0.061

x max = 0.165

0.061 ≤ x ≤ 0.165

0.95⋅ ( α − 1 )     2 0.05     =  0.577  ( ) ( ) − 0.05 α − 1 x − − 2 α − 1 polyroots b x     1.732  b   0.95( α − 1 )   x max := 1.732x b

x min = 0.519

x max = 1.559

0.519 ≤ x ≤ 1

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-20. Linearization of chemical reaction rate.

(

)

2

r cA( t) , cB( t) = k ⋅ cA( t) cB( t)

Use subscript "b" for base value for linearization.

6

Problem parameters:

k := 0.5

m

cAb := 2

2

kmole hr Linearize: Let

kmole := 1000mole

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

kmole m

(

kmole

cBb := 1

3

)

3

m

(

2

)

r cA( t) , cB( t) = r cAb , cBb + 2k ⋅ cAb⋅ cBb cA( t) − cAb + k ⋅ cAb cB( t) − cBb CAb( t) = cA( t) − cAb CB( t) = cB( t) − cBb

(

)

2

R CA( t) , CB( t) = 2k ⋅ cAb⋅ cBb⋅ CA( t) + k ⋅ cAb ⋅ CB( t) At the given base conditions: 2 ⋅ k ⋅ cAb⋅ cBb = 2 hr

(

)

R CA( t) , CB( t) = 2hr For

cA := 3

kmole 3

m

kmole 3

m

2

k ⋅ cAb = 2 hr

CA( t) + 2hr

−1

−1

CB( t )

−1

(off by 50%) 2

cB := 2

−1

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

2

k ⋅ cA − k ⋅ cAb = 2.5 hr For

−1

−1

2 ⋅ k ⋅ cAb⋅ cB − 2 ⋅ k ⋅ cAb⋅ cBb = 2 hr

(off by 125%) −1

(off by 100%) 2

2

k ⋅ cAb − k ⋅ cAb = 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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-21. Linearization of Raoult's Law for equilibrium vapor composition. o

p ( T( t ) )

y ( T( t ) , p ( t ) , x ( t ) ) =

Raoult's Law:

p ( t)

x( t)

Linearize:

A−

o

p ( T( t ) ) = e

B T( t) + C

Use subscript "b" for base value for linearization.

( )

o

xb δ p Tb o y ( T( t ) , p ( t ) , x ( t ) ) = y Tb , p b , x b + ⋅ ⋅ p ⋅ ( T( t) ) ⋅ T( t) − Tb + x( t) − x b pb p b δT

(

)

(

)

−p

+

o

( Tb ) x b

pb

 A− B  δ  T( t) + C ⋅e =

δT

Let

B

A−

B

( Tb + C ) 2

Tb+ C

⋅e

Y( Γ ( t ) , P( t) , X( t) ) = y ( T( t) , p ( t) , x ( t) ) − y Tb , p b , x b

(

2

(

)

( p( t) − p b)

( ) ( Tb + C ) 2 o

=

B ⋅ p ⋅ Tb

)

Γ ( t ) = T( t ) − Tb

P( t) = p ( t) − p b

X( t) = x ( t) − x b

( ) Γ ( t) + p (Tb) X(t) − p (Tb)⋅ xb P( t) Y( Γ ( t ) , P( t) , X( t) ) = 2 pb 2 p b ⋅ ( Tb + C ) pb o

o

x b ⋅ B ⋅ p ⋅ Tb

degC := K Numerical values for benzene at:

mmHg :=

p b := 760mmHg

( )

o

pob = p Tb

A−

pob := e x b ⋅ B⋅ pob

(

p b ⋅ Tb + C

)

2

atm 760

mole% := %

Tb := 95degC

A := 15.9008 Let

o

B := 2788.51degC

x b := 50mole% C := 220.80degC

B Tb+ C

= 0.022

pob = 1177 mmHg

mmHg

1

pob

degC

pb

= 1.549

pob ⋅ x b pb

2

= 0.00102

1 mmHg

Y( Γ ( t ) , P( t) , X( t) ) = pob ⋅ x b pb

= 77.441 %

0.022 degC

(

Γ ( t) + 1.549 X( t) −

0.00102 mmHg

P( t)

)

y Tb , p b , x b = 77.44mole%

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-22. Linearization of reactor of Examples 2-6.4 and 2-6.1. From the results of Example 2-6.4:

τ =

V

Let

( )

k b = k Tb

d ⋅ CA( t)

Use subscript "b" for base value for linearization. fb −V⋅ k Tb E⋅ cAb K2 = K3 = 2 fb + V⋅ k Tb R⋅ Tb fb + V⋅ k Tb

( )

3

V := 2.6m

Tb := 573K

( )

( )

3

m fb := 0.002 s

cAib := 12

−1

k b := 100s

E := 22000

(

)

(

( ))

kmole 3

m

kcal kmole

( )

0 = fb ⋅ cA.b − cAb − k Tb ⋅ V⋅ cAb

From the initial steady state:

cAb :=

τ :=

Calculate parameters:

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

dt

cAib − cAb K1 = fb + V⋅ k Tb

( )

fb + V⋅ k Tb

Problem parameters:

τ⋅

fb ⋅ cAib

cAb = 9.231 × 10

fb + k b ⋅ V cAib − cAb K1 := fb + V⋅ k b

V fb + k b ⋅ V

fb K2 := fb + V⋅ k b

K3 := 1.987

kcal

2

kmole⋅ K

(

⋅ Tb ⋅ fb + V⋅ k b

)

3

m

τ = 0.01 s

K1 = 0.046

−V⋅ k b ⋅ E⋅ cAb

− 5 kmole

s⋅ kmole 6

m

K2 = 7.692 × 10

3

m fb + V⋅ k b = 260.002 s

−6 − 6 kmol

K3 = −3.113 × 10

3

m K

Linearized equation:

0.01⋅ sec ⋅

d ⋅ CA( t) dt

+ CA( t) = 0.046

kmole s 3

m

3

m

−6

F( t) + 7.692 ⋅ 10

CAi( t) − 3.113

kmole 3

Γ ( t)

m K

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-23. Pressure in a compressed air tank when punctured. Assumptions: • Air obeys ideal gas law • Constant temperature

p b := ( 500 + 101.3 )kPa Ao := 0.785cm

2

wi(t)

kPa := 1000Pa

Design conditions:

M := 29

T := 70degC

wo(t)

p(t)

po

V

kg kmole 3

V := 1.5m

3

Rg := 8.314 ⋅

kPa ⋅ m

kmole⋅ K

p o := 101.3kPa

Use subscript "b" for base value for linearization.

Solution: V⋅

Mass balance on the tank:

d ⋅ ρ ( t) dt

= wi( t) − wo ( t) 2

p( t) = ρ ( t) ⋅

Bernoulli's equation:

v ( t) 2

+ po

v( t) =

2⋅

( p( t) − p o) ρ ( t)

(

wo ( t) = ρ ( t) ⋅ Ao ⋅ v ( t) = Ao ⋅ 2 ⋅ ρ ( t) ⋅ p ( t) − p o

Flow through the orifice caused by the bullet:

ρ ( t) =

Ideal gas law:

)

M⋅ p ( t) Rg ⋅ ( T + 273K)

Substitute into mass balance: V⋅ M Rg ⋅ ( T + 273 ⋅ K)

⋅

d ⋅ p( t) dt

2⋅ M = wi( t) − Ao ⋅ p( t) p( t) − p o Rg ⋅ ( T + 273K)

(

)

Solve for the derivative: d⋅ p ( t) dt Linearize:

Let

(

)

= g wi( t) , p ( t) =

d⋅ p ( t) dt

=

P( t) = p ( t) − p b a1 =

δ⋅ g δ ⋅ wi

⋅

δ⋅g δ ⋅ wi

Rg ⋅ ( T + 273K)  V⋅ M

(wi( t) − wb) +

⋅ b

δ⋅ g δ⋅ p

( p( t) − p b)

⋅ b

Wi( t) = wi( t) − wb a1 :=

b

2⋅ M   wi( t) − Ao ⋅ R ⋅ ( T + 273K) p( t) ⋅ ( p( t) − p o) g  

Rg ⋅ ( T + 273K) V⋅ M

a1 = 65.56

kPa kg

a2 =

a2 :=

Substitute:

δ⋅g δ⋅p

⋅

= b

−Ao 2⋅ V

⋅

V

⋅

M kPa

M⋅ pb⋅ pb − po

1000Pa

(

)

⋅

Transfer function:

1 −a2

P( s) Wi( s)

=

K :=

)

2

(2pb − po)

(2⋅ pb − po)1000Pa 

m    100cm 

kPa 1

= a1 ⋅ Wi( t) + a2 ⋅ P( t)

τ :=

(

⋅ p p − p 0   2  b b

−a2

Compare to standard form of first-order equation: Then

−1

2 ⋅ Rg ⋅ ( T + 273K) 1

2 ⋅ Rg ⋅ ( T + 273 ⋅ K)

d ⋅ P( t) dt

−Ao

τ⋅

⋅

d ⋅ P( t) dt

d ⋅ P( t) dt

+ P( t) =

2

a1 −a2

a2 = −0.023 sec

Wi( t)

P( 0 ) = 0

+ P( t) = K⋅ Wi( t)

a1 −a2

−1

τ = 42.9 sec

3 kPa ⋅ sec

K = 2.8 × 10

kg

K τ ⋅s + 1

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.) K := 1.8R Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-24. Temperature of a turkey in an oven. Assumptions • Uniform turkey temperature • Negligible heat of cooking • Radiation heat transfer only

Ts(t)

T(t) M

Energy balance on the turkey: M ⋅ cv ⋅

d ⋅ T( t ) dt

= ε ⋅ σ ⋅ A⋅  Ts ( t) − T ( t)  4



4



Use subscript "b" for linearization base values. Solve for the derivative: Linearize:

d ⋅ T( t ) dt d ⋅ T( t ) dt

where

Let

a1 =

Γ s( t) = Ts( t) − Tsb

Substitute:

ε ⋅σ ⋅A  4 4 = g Ts( t) , T( t) = Ts ( t ) − T ( t )    M ⋅ cv

(

)

(

)

(

= a1 ⋅ Ts( t) − Tsb + a2 ⋅ T( t) − Tb

δ⋅g δTs

⋅

=

4⋅ ε ⋅ σ ⋅ A M ⋅ cv

b

Tsb

3

dt

= a1 ⋅ Γ s( t) + a2 ⋅ Γ ( t)

Γ ( 0) = 0 τ⋅

Standard form of the first-order differential equation: 1 −a2

⋅

d⋅ Γ ( t) dt

4 ⋅ ε ⋅ σ ⋅ A⋅ Tb

3

⋅

d ⋅ Γ ( t)

4 ⋅ ε ⋅ σ ⋅ A⋅ Tb Laplace transform:

dt

M ⋅ cv

τ = Γ ( s) Γ s( s)

=

d ⋅ Γ ( t)

+ Γ ( t) =

M ⋅ cv

Compare coefficients:

δ⋅g

a2 =

δT

⋅

=

−4 ⋅ ε ⋅ σ ⋅ A

b

M ⋅ cv

Tb

3

Γ ( t ) = T( t ) − Tb

d⋅ Γ ( t)

Divide by -a2 and rearrange:

)

3

dt a1 −a2

(base is initial steady state)

+ Γ ( t) = K⋅ Γ s( t) Γ s( t) 3

 Tsb  + Γ ( t) =  Γ s( t) Tb    Tsb  K=  Tb 

3

K τ ⋅s + 1

The input variable is the temperature of the oven wall. See problem 13-4 for the simulation. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 2-25. Slab heated by an electric heater by radiation. Assumptions: • Uniform temperature of the slab • Heat transfer by radiation only

Ts

Energy balance on the slab: M ⋅ cv ⋅

d ⋅ T( t ) dt

T(t)

= q ( t) − ε ⋅ σ ⋅ A⋅ T ( t) − Ts    4

4

C = M ⋅ cv α = ε ⋅σ ⋅A d ⋅ T( t ) 4 4 Substitute C⋅ = q ( t) − α T ( t) − Ts    dt Let

Problem parameters: C := 180

BTU R

Use subscript "b" to denote linearization base value. α := 5 ⋅ 10

− 8 BTU

hr⋅ R

Solve for the derivative:

d ⋅ T( t ) dt

dt

(

1 C

Substitute:

1 C

−4 ⋅ α ⋅ Tb

d⋅ Γ ( t) dt

α 4 4 T ( t ) − Ts   C

(

Q( t) = q ( t) − q b

a2 :=

−a2

⋅

d⋅ Γ ( t) dt

4 ⋅ α ⋅ Tb

4α ⋅ Tb

3

)

⋅

a2 = b

−3 R

C

C

δq

a1 = 5.556 × 10

= a1 ⋅ Q( t) + a2 ⋅ Γ ( t)

1

δ⋅g

a1 =

3

C

Divide by -a2 and rearrange:

τ :=

q( t) −

)

τ⋅

Standard form of first-order differential equation:

Compare coefficients:

Tb := 700R

= a 1 ⋅ q ( t ) − q b + a 2 ⋅ T( t ) − Tb

Γ ( t ) = T( t ) − Tb

a1 :=

Ts := 540R

4

= g ( q ( t ) , T( t ) ) =

d ⋅ T( t )

Linearize: Let

q(t)

3

⋅

K :=

dt

+ Γ ( t) =

d ⋅ Γ ( t) dt

a1 −a2

4α ⋅ Tb

b −1

(base is initial value)

Q( t) 1 4α ⋅ Tb

3

⋅

+ Γ ( t) = K⋅ Q( t)

+ Γ ( t) =

1

δT

a2 = −0.381 hr

BTU

Γ ( 0) = 0

d ⋅ Γ ( t)

δ⋅g

3

Q( t)

τ = 2.62 hr

K = 0.01458

R⋅ hr BTU

Transfer function:

Γ ( s) Q( s)

=

K τ ⋅s + 1

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

%TO := %

Smith & Corripio, 3rd edition

%CO := %

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

D(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

G1 ( s) = Problem parameters:

K := 0.10

K

Gc( s) = Kc

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) %TO

τ 1 := 1min

%CO

C(s)

τ 2 := 0.8min

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

C( s) R( s)

Kc⋅ =

K

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1)

1 + Kc⋅

K

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1) (

2

=

Kc⋅ K

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) + Kc⋅ K

)

τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 s + 1 + Kc⋅ K = 0

Characteristic equation:

C( s)

Closed-loop transfer function:

R( s)

0.1Kc

=

2

0.8s + 1.8s + 1 + 0.1Kc

2

0.8s + 1.8s + 1 + 0.1Kc = 0

Characteristic equation:

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

r1 =

−1.8 +

(

2

1.8 − 4 ⋅ 0.8⋅ 1 + 0.1Kc 2 ⋅ 0.8

)

=

−1.8 1.6

+

2  1.8  − 1 + 0.1Kc  0.8  1.6 

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

2  1.8  − 1 + 0.1Kc = 0  0.8  1.6 

 1   1.8  0.8  Kccd := − 1 0.1   1.6   2

Critically damped: Over-damped (real roots):

%CO Kc < 0.125 %TO

%CO Kccd = 0.125 %TO

%CO Under-damped: Kc > 0.125 %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 )/2τ1 τ2 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: %CO Kc := 0.1 %TO τ e1 =

τ e2 =

(over-damped, two equivalent time constans) −1 r1 −1 r2

%CO Kc := 0.125 %TO τ e1 =

τ e2 =

−1 r1 −1 r2

%CO Kc := 0.2 %TO

τ e1 :=

τ e2 :=

2⋅ τ 1⋅ τ 2

(τ 1 + τ 2) − ( τ 1 + τ 2)

(

)

(τ 1 + τ 2) + ( τ 1 + τ 2)

2

τ e1 = 0.935 min

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

2⋅ τ 1⋅ τ 2

(

)

τ e2 = 0.847 min

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

(critically damped, two equal real time constants) τ e1 :=

τ e2 :=

2⋅ τ 1⋅ τ 2

(τ 1 + τ 2) − ( τ 1 + τ 2)

2

(

)

(τ 1 + τ 2) + ( τ 1 + τ 2)

2

τ e1 = 0.889 min

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

2⋅ τ 1⋅ τ 2

(

)

τ e2 = 0.889 min

− 4 ⋅ τ 1 ⋅ τ 2 ⋅ 1 + Kc⋅ K

(under-damped, time constant and damping ratio) 2 2

τ s + 2ζ ⋅ τ ⋅ s + 1 =

Match coefficients:

2

τ :=

τ 1⋅ τ 2 1 + Kc⋅ K

τ 1⋅ τ 2

2

1 + Kc⋅ K ζ :=

s +

τ1 + τ2 1 + Kc⋅ K

τ1 + τ2

(

)

2 ⋅ τ ⋅ 1 + Kc⋅ K

s+1

τ = 0.886 min ζ = 0.996

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

lim t→∞

Y( t) =

lim s⋅ Y( s) s→0

R( s) =

1 s

(Table 2-1.1)

%CO Kc := 0.1 %TO

lim s⋅ s→0

Kc⋅ K 2

(

1

)

τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 ⋅ s + 1 + Kc⋅ K

s

offset := ( 1 − 0.0099)%TO

→ 9.9009900990099009901 ⋅ 10

-3

offset = 0.99 %TO

Kc⋅ K %CO 1 -2 Kc := 0.125 lim s⋅ → 1.2345679012345679012 ⋅ 10 2 %TO s → 0 s τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 ⋅ s + 1 + Kc⋅ K

(

%CO Kc := 0.2 %TO

lim s⋅ s→0

)

offset := ( 1 − 0.01235 )%TO Kc⋅ K 2

(

)

τ 1 ⋅ τ 2 ⋅ s + τ 1 + τ 2 ⋅ s + 1 + Kc⋅ K offset := ( 1 − 0.01961 )%TO

offset = 0.988 %TO 1 s

→ 1.9607843137254901961 ⋅ 10

-2

offset = 0.98 %TO

These are very large offsets because the loop gains are so small. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-2. Inverse-response second-order system with proportional controller.

D(s) R(s) + E(s)

G2(s)

M(s)

Gc(s)

+

G1(s)

+

-

%TO 6 ( 1 − s) G1 ( s) = ( s + 1 ) ⋅ ( 0.5⋅ s + 1 ) %CO

C(s)

%CO Gc( s) = Kc⋅ %TO

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

Closed-loop transfer function:

R( s)

=

2

Kc⋅ 6 ( 1 − s) ( s + 1 ) ⋅ ( 0.5s + 1 ) + Kc⋅ 6 ( 1 − s)

(

)

0.5⋅ s + 1.5 − 6Kc s + 1 + 6Kc = 0

Characteristic equation:

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

r1 =

(

)

− 1.5 − 6Kc +

(1.5 − 6Kc)

2

(

− 4 ⋅ 0.5⋅ 1 + 6Kc

2 ⋅ 0.5

) = −1.5 + 6K

c+

0.25 − 30Kc + 36Kc

The response is critically damped when the term in the radical is zero: 2

0.25 − 30Kc + 36Kc = 0

Kc := Kc :=

Over-damped (two real roots):

30 +

2

30 − 4 ⋅ 0.25⋅ 36 2 ⋅ 36

30 −

2

30 − 4 ⋅ 0.25⋅ 36 2 ⋅ 36

%CO Kc < 0.00842 %TO

Under-damped (complex conjugate roots):

0.00842

and %CO %TO

%CO Kc = 0.82491 %TO %CO Kc = 0.00842 %TO %CO Kc > 0.825 %TO < Kc < 0.825

%CO %TO

2

%CO Kc > 0.25 %TO

The response is unstable when

(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 o the gain: 0.5

1.5 − 6Kc

2

1 + 6Kc

s +

2 2

s + 1 = τ s + 2ζ ⋅ τ ⋅ s + 1

1 + 6Kc 2

%CO Kc := 0.1 %TO

τ :=

%CO Kc := 0.125 %TO

τ :=

%CO Kc := 0.2 %TO

τ :=

%CO Kc := 0.3 %TO

τ :=

0.5min

1 + 6Kc

ζ :=

2

0.5min

1 + 6Kc

ζ :=

2

0.5min

1 + 6Kc

ζ :=

2

0.5min

1 + 6Kc

ζ :=

(1.5 − 6Kc)min 2 ⋅ τ ⋅ ( 1 + 6Kc) (1.5 − 6Kc)min 2 ⋅ τ ⋅ ( 1 + 6Kc) (1.5 − 6Kc)min 2 ⋅ τ ⋅ ( 1 + 6Kc) (1.5 − 6Kc)min 2 ⋅ τ ⋅ ( 1 + 6Kc)

τ = 0.559 min

ζ = 0.503

τ = 0.535 min

ζ = 0.401

τ = 0.477 min

ζ = 0.143

τ = 0.423 min

ζ = −0.127 (unstable)

Try values that result in equivalent time constants: %CO Kc := 0.005 %TO

1min

τ e1 := 1.5 − 6 ⋅ Kc −

1min

τ e2 := 1.5 − 6 ⋅ Kc + %CO Kc := 1 %TO

0.25 − 30Kc + 36Kc

2

0.25 − 30Kc + 36Kc

2

1min

τ e1 := 1.5 − 6 ⋅ Kc −

0.25 − 30Kc + 36Kc

2

1.5 − 6 ⋅ Kc +

0.25 − 30Kc + 36Kc

τ e2 = 0.559 min

τ e1 = −0.143 min (unstable)

1min

τ e2 :=

τ e1 = 0.868 min

2

τ e2 = −0.5 min

(d) Offset for various values of the gain and a unit step change in set point. %CO Kc := 0.10 %TO

lim s⋅ s→0

Kc⋅ 6 ⋅ ( 1 − s) 2

(

)

1

0.5s + 1.5 − 6 ⋅ Kc s + 1 + 6Kc

s

→ .37500000000000000000

offset := 1 − 0.375

offset = 0.625

%CO %TO

%CO Kc := 0.125 %TO

lim s⋅ s→0

Kc⋅ 6 ⋅ ( 1 − s) 2

(

1

)

0.5s + 1.5 − 6 ⋅ Kc s + 1 + 6Kc

s

→ .42857142857142857143

offset := 1 − 0.429 %CO Kc := 0.20 %TO

lim s⋅ s→0

Kc⋅ 6 ⋅ ( 1 − s) 2

(

)

1

0.5s + 1.5 − 6 ⋅ Kc s + 1 + 6Kc

s

offset = 0.571

%CO %TO

→ .54545454545454545455

offset := 1 − 0.545

offset = 0.455

%CO %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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

K := 1.8R

Smith & Corripio, 3rd edition Problem 6-3. First-order process and proportional-integral controller.

D(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

G1 ( s) =

Gc( s) = Kc⋅  1 +

K τ ⋅s + 1



1



τ I⋅ s



C(s)

τ := 1

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

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

C( s) R( s)

=

2

(

) τ I⋅ s⋅ ( τ ⋅ s + 1 ) + KKc⋅ ( τ I⋅ s + 1 ) KKc⋅ τ I⋅ s + 1

(

)

τ I⋅ τ ⋅ s + 1 + KKc τ I⋅ s + KKc = 0

Characteristic equation:

Offset: the steady state gain is:

(

KKc⋅ τ I⋅ s + 1

lim

(

)

)

s → 0 τ ⋅ τ ⋅ s2 + 1 + KK τ ⋅ s + KK I c I c

=

KKc KKc

=1 (no offset)

(b) Is there an ultimate gain for this loop? 2

(

)

−τ I⋅ τ ⋅ ω u + KKcu + i 1 + KKcu ω u = 0

Substitute s = iω:

ω u := 0

KKcu := 0

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:

r1 =

(

)

− 1 + KKc ⋅ τ I +

(1 + KKc)

2

2

⋅ τ I − 4 ⋅ τ I⋅ τ ⋅ KKc

2τ I⋅ τ

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

negative: Real =

(

− 1 + KKc 2⋅ τ

)

τI > = 0.444min 1.8min τ1 + τ2

(b) Damping ratio and decay ratio with Kc equal to one half the ultimate and τI = 1 min For these values there is no ultimate gain. THe characteristic equation becomes: KKc⋅  1 +

1



1+

s

( s + 1 ) ⋅ ( 0.8s + 1 )

=1+

KKc⋅ ( s + 1 )

τ s + 2ζ ⋅ τ ⋅ s + 1 =

2

Roots: r1 =

0.8min

τ =

−1 +

ζ =

KKc

1 − 4 ⋅ 0.8⋅ KKc

These are complex for

KKc >

1min 2 ⋅ τ ⋅ KKc

r2 =

2 ⋅ 0.8 1 4 ⋅ 0.8

= 0.8s + s + KKc = 0

2 2

Standard second-order differential equation:

Damping ratio:

2

s( s + 1 ) ⋅ ( 0.8s + 1 )

−1 −

=

0.8 2 1 s + s+1 KKc KKc

1

(

)

ζ KKc :=

2 ⋅ 0.8⋅ KKc

1 2 ⋅ 0.8KKc

1 − 4 ⋅ 0.8⋅ KKc 2 ⋅ 0.8

= 0.3125

Under these conditions the decay ratio is, from Eq. 2-5.18: − 2π − 2π ⋅ ζ

DR = e

1− ζ

2⋅

2

=e

1   0.8KKc⋅ 1− 4⋅ 0.8⋅ KKc   =e

− 2π

− 2⋅ π

)

)

DR KKc := e

3.2KKc− 1

0.4

1.5

(

(

3.2⋅ KKc− 1

1

(

ζ KKc

)

DR KKc

0.2

0.5 0

5 KKc

10

0

0

5

10

KKc

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

degF := R

Problem 6-6. Design of gas flow control loop.

kscf := 1000ft

3

lbmole := 453.59mole

Design conditions: kscf

fs := 150 hr

p 1 := 150psig

T1 := 60degF

p 2 := 80psig

M := 29

psia := psi

lb

fsset(t) c(t)

α := 50

lbmole

τ v := 0.06min

psig := psi

FC m(t)

FT fs(t)

τ I := τ v

p2

p1

%CO Kc := 0.9 %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 f s(t). The valve is equal-percentage and the controller is PI.

Block diagram of the control loop:

Fsset(s)

E(s)

Ksp + -

Gc(s) C(s)

100%TO

KT :=

PI Controller:

Gc( s) = Kc⋅  1 +

fsmax := 1.5⋅ fs

Ksp := KT

fsmax



Gv(s)

1



τ I⋅ s



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

M ⋅ lbmole 29lb

y :=

1.63 Cf

⋅

p1 − p2 p 1 + 14.7psia

Fs(s)

KT

Size the flow transmitter for 150% of design flow:

Transmitter gain:

M(s)

y = 1.181

kscf fsmax = 225 hr KT = 0.444

Let

%TO⋅ hr kscf

Cf := 0.9

fy := y − 0.148y

3

fy = 0.937

Cvmax :=

(

200 ⋅ %⋅ fs⋅ G⋅ T1 + 460 ⋅ R

(

)

)

Cf ⋅ p 1 + 14.7psia ⋅ fy

gal⋅ hr 0.836kscf ⋅ min

psi

⋅

R Cvmax = 58.91

gal min⋅ psi

From Fig. C-10.1, p. 532, a 3-in Masoneilan valve is the smallest for this service: Cvmax := 110

gal min⋅ psi

Valve gain, equal-percentage, constant pressures, Eq. 5-2.24: ln( α ) Kv := f 100%CO s

kscf Kv = 5.87 hr⋅ %CO

Transfer function of the valve:

Gv ( s) =

Kv τ v⋅ s + 1

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

Fs( s) set

=

Fs ⋅ ( s)

Ksp ⋅ Kc⋅  1 +

Ksp ⋅ Gc( s) ⋅ Gv ( s) 1 + KT⋅ Gc( s) ⋅ Gv ( s)



=

τI = τv

Ksp = KT

Fs( s) set

Fs ( s) Closed-loop time constant:

τ I⋅ s

1 + KT⋅ Kc⋅  1 +



With

1

τ c :=

=

Kv

⋅

 τ v⋅ s + 1

1



Kv

τ I⋅ s τ v ⋅ s + 1 

Ksp ⋅ Kc⋅ Kv τ v ⋅ s + KT⋅ Kc⋅ Kv

τv KT⋅ Kc⋅ Kv

=

1 τ c⋅ s + 1

τ c = 0.026 min

So the closed-loop responds faster than the valve, and has no offset. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-7. Steam flow control loop. lbmole := 453.59mole

Design conditions: w := 3500

lb

p 1 := 45psig

hr

Tsh := 50degF M := 18

wset(t) c(t)

p 2 := 20psig

lb

lb wmax := 5000 hr

lbmole

FC m(t)

FT w(t)

Linear valve.

τ I := τ v

Cf := 0.8

%CO Kc := 0.5 %TO

p2

p1

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)

Ksp + -

Gc(s) C(s)

M(s) Gv(s)

W(s)

KT

Size the flow transmitter for : 100%TO

Transmitter gain:

KT :=

PI Controller:

Gc( s) = Kc⋅  1 +

Ksp := KT

wmax



1



τ I⋅ s



KT = 0.02

%TO⋅ hr lb

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

M ⋅ lbmole 29lb

y :=

1.63 Cf

⋅

p1 − p2 p 1 + 14.7psia

y = 1.319

From the steam table the saturated steam pressure at:

fy := y − 0.148y

3

p 1 + 14.7psia = 59.7 psia

fy = 0.979 Tsat := 292degF

w kscf kscf fs := ⋅ 0.380 fs = 73.889 M lbmole hr Cvmax :=

(

200 ⋅ %⋅ fs⋅ G⋅ T1 + 460 ⋅ R

(

)

T1 := Tsat + Tsh

)

Cf ⋅ p 1 + 14.7psia ⋅ fy

gal⋅ hr 0.836kscf ⋅ min

T1 = 342 degF

psi

⋅

R Cvmax = 84.3

gal min⋅ psi

From Fig. C-10.1, p. 532, a 3-in Masoneilan valve is the smallest for this service: Cvmax := 110 wvmax :=

0.836kscf ⋅ min gal⋅ hr

R

⋅

psi

p 1 + 14.7psia

⋅ Cvmax⋅ Cf ⋅

(

G⋅ T1 + 460R

)

gal min⋅ psi

M ⋅ lbmole fy ⋅ 0.380kscf lb wvmax = 9131 hr

Valve gain, linear, constant pressures, Eq. 5-2.23: wvmax Kv := 100%CO

lb Kv = 91.31 hr⋅ %CO

Transfer function of the valve:

Gv ( s) =

Kv τ v⋅ s + 1

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

W( s) set

=

W ⋅ ( s)

Ksp ⋅ Kc⋅  1 +

Ksp ⋅ Gc( s) ⋅ Gv ( s) 1 + KT⋅ Gc( s) ⋅ Gv ( s)



=

τI = τv

Ksp = KT

Fs( s) set

Fs ( s) Closed-loop time constant:

τ I⋅ s

1 + KT⋅ Kc⋅  1 +



With

1

τ c :=

=

Kv

⋅

 τ v⋅ s + 1

1



τ I⋅ s τ v ⋅ s + 1 

Ksp ⋅ Kc⋅ Kv τ v ⋅ s + KT⋅ Kc⋅ Kv

τv KT⋅ Kc⋅ Kv

Kv

=

1 τ c⋅ s + 1

KT⋅ Kc⋅ Kv = 0.913

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? Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-8. Ultimate gain and period of various process transfer functions.

D(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

Proportional controller:

(a)

G1 ( s) =

1

4

( s + 1)

(b)

( s + 1)

Real part:

2π ωu

)4 + 6min2⋅ (ω u)2 − 1

(

2

Tu = 6.28 min %CO Kcu = 4 %TO

2

s + 2s + 1 + Kc = 0

2

−ω u + 2ω u ⋅ i + 1 + Kcu = 0 + 0i

2ω u = 0 2

Tu :=

Kcu := − ω u ⋅ min 

( s + 1 ) + Kc = 0

2

Substitute s = iωu at Kc = K cu : Imaginary part:

2

−1

2

1

3

ω u := 1min

ω u − 6ω u + 1 + Kcu = 0

G1 ( s) =

2

s + 4s + 6s + 4s + 1 + Kc = 0

4

3

4

3

1 + Kc⋅ G1 ( s) = 0

ω u − 4ω u i − 6ω u + 4ω u ⋅ i + 1 + Kcu = 0 + 0i

−4 ω u + 4ω u = 0

Imaginary part:

4

( s + 1 ) + Kc = 0

4

Substitute s = iωu at Kc = K cu :

Real part:

Characteristic equation:

Gc( s) = Kc

C(s)

−ω u + 1 + Kcu = 0

ω u := 0

(There is no ultimate gain) %CO Kcu := −1 %TO

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

(c)

1 G1 ( s) = ( 4s + 1 ) ⋅ ( 2 ⋅ s + 1 ) ⋅ ( s + 1 )

8s + 14s + 7s + 1 + Kc = 0

Substitute s = iωu at Kc = K cu :

−8 ω u i − 14ω u + 7ω u ⋅ i + 1 + Kcu = 0 + 0i

Imaginary part:

Real part:

(d)

3

3

ω u :=

2

2π

2

(

)

2

(

)

Tu = 6.72 min

ωu

%CO Kcu = 11.25 %TO

3

−8 ω u i − 14ω u + 7 + 0.5Kcu ω u ⋅ i + 1 + Kcu = 0 + 0i

2

2

(

Kcu = 14ω u − 1

)

−8 ω u + 7 + 0.5Kcu ω u = 0 2

−1

ω u :=

6.5 min %CO

2

2

Kcu := 14min ⋅ ω u − 1

Kcu = 90 %TO

1 G1 ( s) = ( 4s + 1 ) ⋅ ( 0.2s + 1 ) ⋅ ( 0.1s + 1 )

3

Tu :=

2π ωu

Tu = 2.46 min

2

0.08s + 1.22s + 4.3s + 1 + Kc = 0 3

2

−0.08ω u i − 1.22ω u + 4.3ω u ⋅ i + 1 + Kcu = 0 + 0i

Substitute s = iωu at Kc = K cu :

Real part:

Tu :=

2

3

2

Imaginary part:

−1

min

8s + 14s + 7 + 0.5Kc s + 1 + Kc = 0

−8 ω u + 7 + 7ω u − 0.5 = 0

(e)

8

Kcu := 14min ω u − 1

−14ω u + 1 + Kcu = 0 3

7

2

−14ω u + 1 + Kcu = 0

Substitute s = iωu at Kc = K cu :

Imaginary part:

2

3

−8 ω u + 7ω u = 0

0.5s + 1 G1 ( s) = ( 4s + 1 ) ⋅ ( 2s + 1 ) ⋅ ( s + 1 )

Real part:

2

3

−0.08ω u. + 4.3ω u = 0 2

−1.22ω u + 1 + Kcu = 0 2

2

Kcu := 1.22min ω u − 1

ω u :=

4.3 0.08

−1

min

Tu :=

2π ωu

Tu = 0.857 min %CO Kcu = 64.6 %TO

(f)

− 0.6s

e G1 ( s) = 6s + 1

6s + 1 + Kc⋅ e

− 0.6s

=0

2

Padé approximation:

(

e

− 0.6s

=

1 − 0.3s 1 + 0.3s

)

1.8s + 6.3 − 0.3Kc s + 1 + Kc = 0 Substitute s = iωu at Kc = K cu :

2

(

)

−1.8ω u + 6.3 − 0.3Kcu ω u ⋅ i + 1 + Kcu = 0 + 0i 6.3 %CO 6.3 − 0.3Kcu ω u = 0 Kcu := Kcu = 21 0.3 %TO

Imaginary part:

(

)

Real part:

−1.8ω u + 1 + Kcu = 0

2

ω u :=

1 + Kcu 1.8

−1

min

Tu :=

2π ωu

Tu = 1.797 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 beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-9. Ultimate gain and period with integral controller.

D(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

KI Gc( s) = s

Integral controller:

(a)

G1 ( s) =

1

4

3

KI s

G1 ( s) = 0

2

s + 4s + 6s + 4s + s + KI = 0

4

5

5

4

3

2

ω u i + 4ω u − 6ω u i − 4ω u + ω u ⋅ i + KIu = 0 + 0i

Substitute s = iωu at KI = K Iu: 3

ω u − 6ω u + ω u = 0

Imaginary part: 2

6+

ω u :=

1+

Characteristic equation:

5

( s + 1)

C(s)

6 −4

−1

ω u = 2.414 min

2

2min

4

3

4 ⋅ ω u − 4ω u + KIu = 0

Real part:

ω u :=

6−

2

6 −4 2

2min 3

4

KIu := −4 min ⋅ ω u + 4min⋅ ω u %CO

KIu = 0.569 %TO⋅ min

−1

ω u = 0.414 min 2

Tu :=

2π ωu

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)

3

Real part:

3

2

−ω u i − 2ω u + ω u ⋅ i + KIu = 0 + 0i 3 2π −1 −ω u + ω u = 0 ω u := 1min Tu := Tu = 6.28 min ωu

Substitute s = iωu at KI = K Iu: Imaginary part:

2

s + 2s + s + KI = 0

2

2

−2 ω u + KIu = 0

KIu := 2minω u

2

%CO KIu = 2 %TO⋅ min

(c)

1 G1 ( s) = ( 4s + 1 ) ⋅ ( 2s + 1 ) ⋅ ( s + 1 )

(d)

3

−14ω u + ω u = 0

4

2

2

8ω u − 14ω u i − 7ω u + ω u ⋅ i + KIu = 0 + 0i

3

Imaginary part:

3

4

Substitute s = iωu at KI = K Iu:

Real part:

4

8s + 14s + 7s + s + KIu = 0

1

ω u :=

2

Tu :=

14

4

3

8ω u − 7ω u + KIu = 0 KIu := −8 min ⋅ ω u + 7min⋅ ω u

0.5s + 1 G1 ( s) = ( 4s + 1 ) ⋅ ( 2s + 1 ) ⋅ ( s + 1 )

4

3

Tu = 23.51 min

ωu

%CO KIu = 0.459 %TO⋅ min

2

)

3

2

(

)

8ω u − 14ω u i − 7ω u + 1 + 0.5⋅ KIu ω u ⋅ i + KIu = 0 + 0i

4

2

4

Real part:

8ω u − 7ω u + KIu = 0

Imaginary part:

−14ω u +  1 − 4ω u + 3.5ω u  3

ω u :=

KIu = −8 ω u + 7ω u

4

10.5 −

2

2

10.5 − 4 ⋅ ( −4 ) 2

2 ⋅ ( −4 ) min 3

4

1 G1 ( s) = ( 4s + 1 ) ⋅ ( 0.2s + 1 ) ⋅ ( 0.1s + 1 )

2π

Tu = 20.71 min

ωu

%CO KIu = 0.576 %TO⋅ min

2

4

2

−4 ω u − 10.5ω u + 1 = 0

3

2

0.08s + 1.22s + 4.3s + s + KI = 0 4

Substitute s = iωu at KI = K Iu:

Tu :=

2 4

ωu = 0

KIu := −8 min ⋅ ω u + 7min⋅ ω u

3

2

0.08ω u − 1.22ω u i − 4.3ω u + ω u ⋅ i + KIu = 0 + 0i 3

Imaginary part:

−1.22ω u + ω u = 0

Real part:

0.08ω u − 4.3ω u + KIu = 0

4

(

2

2π

8s + 14s + 7s + 1 + 0.5KI s + KI = 0 4

Substitute s = iωu at KI = K Iu:

(e)

−1

min

2

ω u :=

1 1.22

−1

min

Tu := 3

4

2π ωu

Tu = 6.94 min

KIu := −0.08min ⋅ ω u + 4.3min⋅ ω u

2

%CO KIu = 3.47 %TO⋅ min

(f)

− 0.6s

e G1 ( s) = 6s + 1

2

6s + s + KI⋅ e 3

2

− 0.6s

(

Padé approximation:

=0

e

− 0.6s

=

)

1 − 0.3s 1 + 0.3s

1.8s + 6.3s + 1 − 0.3KI s + KI = 0

Substitute s = iωu at KI = K Iu:

3

2

(

)

−1.8ω u i − 6.3ω u + 1 − 0.3KIu ω u ⋅ i + KIu = 0 + 0i

2

Real part:

−6.3ω u + KIu = 0

Imaginary part:

−1.8ω u +  1 − 1.89ω u  3

KIu := 6.3min⋅ ω u

2

KIu = 6.3ω u 2

ωu = 0

2

ω u :=

1 3.69

−1

min

%CO KIu = 1.707 %TO⋅ min

Tu :=

2π ωu

Tu = 12.07 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 beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-10. Open-loop unstable process and proportional controller.

D(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

Gc( s) = Kc

G1 ( s) =

K

( τs − 1) ⋅ (τ v⋅ s + 1) ⋅ (τ T⋅ s + 1)

τ := 5min

(τ ⋅ s − 1 )⋅ ( τ v⋅ s + 1 )⋅ ( τ T⋅ s + 1) + KKc = 0

Charatcteristic equation of the loop:

(a) Negligible valve and transmitter time constants: τ v := 0 τ ⋅ s − 1 + KKc = 0 %CO The loop is stable for KKc > 1 %TO

Root:

(

r=

τ T := 0

1 − KKc

τ (negative real root)

(b) Negligible valve time constant, 2

C(s)

τ T := 1min

)

τ ⋅ τ T⋅ s + τ − τ T s − 1 + KKc = 0

(

)

(τ − τ T)2 − 4τ ⋅ τ T⋅ (−1 + KKc)

− τ − τT +

Dominant root:

r1 =

The radical is:

τ − 2τ ⋅ τ T + τ T + 4τ ⋅ τ T − 4τ ⋅ τ T⋅ KKc = τ + τ T

2

The roots are real as long as:

2⋅ τ ⋅ τ T

(

2

KKc

The dominant root is negative when:

2 τ + τ T) ( ≤

4⋅ τ ⋅ τ T

)2 − 4τ ⋅ τ T⋅ KKc

(τ + τ T)2 = 1.8 %CO 4τ ⋅ τ T

(τ + τ T)2 − 4τ ⋅ τ T⋅ KKc < τ − τ T

%TO

(τ + τ T)2 − 4τ ⋅ τ T⋅ KKc < (τ − τ T)2 2

2

2

2

τ + 2τ ⋅ τ T + τ T − τ + 2τ ⋅ τ T − τ T < 4τ ⋅ τ T⋅ KKc So, the loop is stable for:

KKc > 1

%CO %TO

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

(c)

τ v := 0.1min

(

− τ − τT 2τ ⋅ τ T

) = −0.4 min− 1

τ T := 1.0min

(

3

)2 (

)

Characteristic equation: τ ⋅ τ v ⋅ τ T⋅ s + τ ⋅ τ v + τ ⋅ τ T − τ v ⋅ τ T s + τ − τ v − τ T s − 1 + KKc = 0 Substitute s = ωi at KKc = KKcu :

(

3

)

2

(

)

−τ ⋅ τ v ⋅ τ T⋅ ω u i − τ ⋅ τ v + τ ⋅ τ T − τ v ⋅ τ T ω u + τ − τ v − τ T ω u ⋅ i − 1 + KKcu = 0 + 0i 3

(

)

Imaginary part:

−τ ⋅ τ v ⋅ τ T⋅ ω u + τ − τ v − τ T ω u = 0

Real part:

− τ ⋅ τ v + τ ⋅ τ T − τ v ⋅ τ T ω u − 1 + KKcu = 0

(

)

ω u :=

2

(

)

KKcu := 1 + τ ⋅ τ v + τ ⋅ τ T − τ v ⋅ τ T ω u

2

τ − τv − τT τ ⋅ τ v⋅ τ T

Tu :=

2π ωu

Tu = 2.25 min KKcu = 43.1

%CO %TO

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

KKc := 0.99

−1 + KKc       ( τ − τ v − τ c) min− 1    −10.037    =  −0.963 polyroots    τ ⋅ τ + τ ⋅ τ ⋅ min− 2   T)  ( v   0   − 3    ( τ ⋅ τ v⋅ τ T) ⋅ min 

Root at zero, integrating response

−1 + KKc       ( τ − τ v − τ c) min− 1    −10.036  Positive root, unstable   =  −0.966  response polyroots   ( τ ⋅ τ + τ ⋅ τ ) ⋅ min− 2   −3 T  v    2.064 × 10   − 3    ( τ ⋅ τ v⋅ τ T) ⋅ min 

KKc := 1.01

−1 + KKc     Negative real roots, stable   ( τ − τ v − τ c) min− 1    −10.037  response  = −0.961 polyroots    ( τ ⋅ τ + τ ⋅ τ ) ⋅ min− 2   − 3 T  v    −2.073 × 10    τ ⋅ τ ⋅ τ ⋅ min− 3     ( v T) 

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, KK c = 0. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-11. Analyzer control loop for blender of Problem 3-18. Problem parameters: 3

V := 40m

c1b := 80 3

m

fb := 4.0 min cb := 50

c2b := 30

kg

c set(t)

kg

AC

3

m

m(t)

kg 3

m

f1(t)

3

m

Linear control valve sized for 100% overcapacity.

τ T := 3min

Transmitter: kg 3

m

cmax := 70

AT

V

f2(t) c 2(t)

∆p v := 5psi

τ v := 0.1min

cmin := 20

c 1(t)

c(t) f(t)

kg 3

m

3

Disturbance:

m ∆f1 := 0.1 min

Use subscript "b" to denote base values for linearization.

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

F1(s) Cset(s)

Ksp

+

E(s)

M(s) Gc(s)

-

Gv(s)

F2(s)

G1(s)

+ G2(s)

C(s)

+

H(s) Gc( s) = Kc

Gv ( s) =

Kv τ v⋅ s + 1

H( s) =

KT τ T⋅ s + 1 3

From the solution to Problem 3-18:

m f1b := 1.6 min

G1 ( s) =

K1 τ ⋅s + 1

3

m f2b := 2.4 min

G2 ( s) =

τ :=

V f1b + f2b

K2 τ ⋅s + 1

τ = 10 min

c1b − cb K1 := f1b + f2b

Control valve:

Cvmax :=

c2b − cb K2 := f1b + f2b

200 ⋅ %⋅ f2b⋅ gal 3.785 ⋅ 10

−3 3

⋅

m

1

K1 = 7.5

Cvmax = 567

∆p v

kg min 3

K2 = −5

3

m

m

kg min 3

3

m

m

gal 0.5

min⋅ psi

From Fig. C-10.1, p. 532, an 8-in valve is the smallest with enough capacity: Cvmax := 640

f2max := Cvmax⋅ ∆p v ⋅

3.785 ⋅ 10

−3

gal 0.5

min⋅ psi

3

⋅m

gal 3

3

f2max Kv := 100%CO

m f2max = 5.417 min

m Kv = 0.054 min⋅ %CO

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

Transmitter:

100 ⋅ %TO KT := cmax − cmin

3

Ksp := KT

Closed-loop transfer function::

C( s) =

Characteristic equation:

1+

3

(

KT = 2

Ksp ⋅ Gc( s) ⋅ Gv ( s) ⋅ G2 ( s) ⋅ C

kg

set

⋅ ( s) + G1 ( s) ⋅ F1 ( s)

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

τ T⋅ s + 1

)

%TO⋅ m

⋅ Kc⋅

Kv

τ v.⋅ s + 1 τ ⋅ s + 1

(

2

K2

=0

)

τ T⋅ τ v ⋅ τ ⋅ s + τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ⋅ s + τ T + τ v + τ s + 1 + KT⋅ Kc⋅ Kv ⋅ K2 = 0 Let

K := KT⋅ Kv ⋅ K2

K = −0.542

Substitute s = ωu i at KKc = KKcu :

(

3

)

2

(

)

−τ T⋅ τ v ⋅ τ ⋅ ω u i − τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ω u + τ T + τ v + τ ω u ⋅ i + 1 + KKcu = 0 + 0i Imaginary part:

Real part:

3

(

)

−τ T⋅ τ v ⋅ τ ⋅ ω u + τ T + τ v + τ ω u = 0

(

)

2

ω u :=

− τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ω u + 1 + KKcu = 0

Kcu :=

( τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ) ω u 2 − 1 K

τT + τv + τ τ T⋅ τ v ⋅ τ

Tu :=

2π ωu

Tu = 3.01 min %CO Kcu = −250 %TO

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

Offset:

F1 ( s) =

∆f1

C

s

E( s) = Ksp ⋅ C

set

( s) = 0

Kc :=

Kcu

%CO Kc = −125 %TO

2

set

( s) − H( s) ⋅ C( s) = −H( s) ⋅ C( s)

By the final-value theorem: K1 −KT ∆f1   ⋅ ⋅   τ T⋅ s + 1 τ ⋅ s + 1 s −KT⋅ K1 ⋅ ∆f1 Offset = lim s⋅  = KT Kv K2 s→0   1 + KT⋅ Kc⋅ Kv ⋅ K2 1+ ⋅ Kc⋅  τ T⋅ s + 1 τ v⋅ s + 1 τ s + 1   −KT K1 ⋅ ∆f1 1 + KT⋅ Kc⋅ Kv ⋅ K2

= −0.022 %TO

Although not asked in the problem, let us determine the roots of the characteristic equation: 1 + K⋅ Kc       ( τ T + τ v + τ ) min− 1    −10.226    =  −0.104 + 1.494i polyroots   ( τ ⋅ τ + τ ⋅ τ + τ ⋅ τ ) min− 2   v  T v T    −0.104 − 1.494i   −3  τ T⋅ τ v ⋅ τ ⋅ min   Decay ratio: Settling time:

Response is stable and the dominant roots are complex conjugate. The period of oscillation is: 2π

T :=

−1

1.494min T = 4.21 min e

− 0.104 min

−1

−5 −1

T

= 0.646

= 48.08 min

−0.104 min

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-12. Temperature control of non-isothermal reactor of Section 4-2.3 by manipulation of coolant flow to the jacket. Control Valve: equal-percentage, negligible time constant.

Tset(t)

α := 50

Tmin := 640R

m(t)

f(t)

Transmitter: negligible time constant.

TT

cAi(t) Ti(t)

Tmax := 700R 3

ft fcp := 0.8771 min

TC

Tc(t) V

fc(t) Tci

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

c A(t) T(t)

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

F(s) Tset(s)

Ksp

+

E(s)

M(s) Gc(s)

-

Gv(s)

GF(s)

Fc(s)

+ G(s)

T(s)

+

H(s) Ksp = KT G( s) =

Gc( s) = Kc

Gv ( s) = Kv

−35.77 ( 2.07s + 1 ) 3

2

26.27s + 36.31s + 10.14s + 1

H( s) = KT (Eq. 4-2.41)

GF( s) =

31.79 ⋅ ( 0.976 ⋅ s + 1 ) ⋅ ( 1 − 2.77⋅ s) 3

2

26.27s + 36.31s + 10.14s + 1 (Eq. 4-2.39)

−ln( α ) fcp Kv := 100%CO

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

3

ft Kv = −0.034 min⋅ %CO

The control valve fails opened (air-to-close) to prevent overheating the reactor on loss of power. This is why its gain is negative. 100%TO KT := Tmax − Tmin

Transmitter gain:

Characteristic equation of the loop:

Let

K := −35.77

R⋅ min ft

3

1 + Kc⋅ Kv ⋅ KT⋅

KT = 1.667

−35.77 ⋅ ( 2.07⋅ s + 1 ) 3

R

=0

2

26.27s + 36.31s + 10.14s + 1

⋅ Kv ⋅ KT

K = 2.046

2

%TO %CO

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

Rearrange and substitute s = ωu i at KKc = KKcu : 3

%TO

(

)

−26.27 ω u i − 36.31ω u + 10.14 + 2.07⋅ KKcu ω u ⋅ i + 1 + KKcu = 0 + 0i Real part: Imaginary part:

2

2

−36.31 ω u + 1 + KKcu = 0 3

(

KKcu = 36.31ω u − 1

)

−26.27 ω u + 10.14 + 2.07KKcu ω u = 0 ω u :=

10.14 − 2.07 26.27 − 2.07⋅ 36.31

10.14 − 2.07 = ( 26.27 − 2.07⋅ 36.31 )ω u −1

min

2

−1

ω u = 0.406i min

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-13. Temperature control of non-isothermal reactor of Section 4-2.3 by manipulation of the reactants flow. Control Valve: linear, sized for 100% overcapacity, negligible time constant. Assume ∆p v := 5psi

Tset(t)

Tmin := 640R

m(t)

f(t)

Transmitter: negligible time constant.

TT

cAi(t) Ti(t)

Tmax := 700R 3

ft fp := 1.3364 min

TC

Tc(t) V

fc(t) Tci

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

c A(t) T(t)

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

Tset(s) Ksp

+

E(s)

Gc(s)

M(s)

Gv(s)

-

F(s) Fc(s)

GF(s)

+ G(s)

T(s)

+

H(s) Ksp = KT G( s) =

Gc( s) = Kc

Gv ( s) = Kv

−35.77 ( 2.07s + 1 ) 3

2

26.27s + 36.31s + 10.14s + 1

H( s) = KT (Eq. 4-2.41)

GF( s) =

31.79 ⋅ ( 0.976 ⋅ s + 1 ) ⋅ ( 1 − 2.77⋅ s) 3

2

26.27s + 36.31s + 10.14s + 1 (Eq. 4-2.39)

Cvmax :=

Control valve size:

200%⋅ fp 7.48gal ∆p v

ft

Cvmax = 8.941

3

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

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

gal min⋅ psi gal

fmax := Cvmax⋅ ∆p v

min⋅ psi

fmax

ft

Kv := 100%CO

3

Kv = 0.033 min⋅ %CO

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

100%TO KT := Tmax − Tmin

Transmitter gain:

Characteristic equation of the loop:

Let

K := 31.79

R⋅ min ft

3

1 + Kc⋅ Kv ⋅ KT⋅

KT = 1.667

31.79 ⋅ ( 0.976 ⋅ s + 1 ) ⋅ ( 1 − 2.77s) 3

2

%TO R

=0

26.27s + 36.31s + 10.14s + 1

⋅ Kv ⋅ KT

K = 1.742

%CO

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

Rearrange and substitute s = ωu i at KKc = KKcu : 3

%TO

2

−26.27 ω u i −  36.31 + 0.976 ( −2.77) KKcu ω u +  10.14 + ( 0.976 − 2.77) ⋅ KKcu ω u ⋅ i + 1 + KKcu = 0     2

Real part:

2

− 36.31 + 0.976 ( −2.77) KKcu ω u + 1 + KKcu = 0  

3

KKcu =

−26.27 ω u +  10.14 + ( 0.976 − 2.77)KKcu ω u = 0

Imaginary part:





−26.27 ω u + 10.14 + ( −1.794 ) 4

36.31ω u − 1 1 − ( −2.704 )ω u

1 − 0.976 ( −2.77) ω u

2

0.976 − 2.77 = −1.794 0.976 ( −2.77) = −2.704

2

2

36.31ω u − 1

2

=0

2

−26.27 ⋅ ( 2.704 ) ω u + [ −26.27 + 10.14 ( 2.704 ) + ( −1.794 )36.31 ]ω u + ( 10.14 + 1.794 ) = 0 −26.27 ( 2.704 ) = −71.034 −26.27 + 10.14 ( 2.704 ) + ( −1.794 )36.31 = −63.992 4

2

−71.034ω u − 63.992ω u + 11.934 = 0

10.14 + 1.794 = 11.934

ω u :=

63.992 −

2

63.992 − 4 ⋅ ( −71.034) ( 11.934) 2

2 ( −71.034) min 2

2π ωu

Tu = 15.78 min

2

1 36.31min ω u − 1

Kcu := K

Tu :=

2

1 + 2.704min ω u

2

%CO Kcu = 1.91 %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). Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

K := 1.8R

Smith & Corripio, 3rd edition Problem 6-14. Analyzer control of three mixing tanks of Problem 4-3.

x6set(t) AC

m(t)

f1(t)

f3(t) V

AT

f4(t) V

x3(t)

V

x4(t)

f5(t)

f2(t)

f7(t)

x5(t)

x2(t)

x7(t)

f6(t)

x6(t)

Block diagram, considering only f 1 (t) anf f2 (t) as input variables:

F2(s) X6set(s)

Ksp

+

E(s)

M(s) Gc(s)

-

Gv(s)

G2(s)

F1(s)

+ G1(s)

X6(s)

+

H(s)

gal Problem parameters (Table P4-1): V := 7000gal f1 := 1900 min gal gal f5 := 500 x 5 := 80% f7 := 500 x 7 := 90% min min Assume perfectly mixed tanks, constant density. f4 := f3 + f2

x 4 :=

f3 ⋅ x 3 + f2 ⋅ x 2 f4

gal f4 = 3400 min

gal f2 := 1000 min

f3 := f1 + f5 x 3 := x 4 = 40.882 %

f5 ⋅ x 5 f3

x 2 := 99%

gal f3 = 2400 min x 3 = 16.667 %

f6 := f4 + f7

x 6 :=

f4 ⋅ x 4 + f7 ⋅ x 7

gal f6 = 3900 min

f6

x 6 = 47.179 %

Control valve: Equal-percentage valve, constant pressure drop, α = 50, negligible time constant. ln( 50) Kv := f 100%CO 1

Gv ( s) = Kv

gal Kv = 74.3 min⋅ %CO

(Eq. 5-2.24, p. 171)

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

Analyzer Transmitter: negligible lag, 30 to 70% range: 100%TO KT := ( 70 − 30)%

H( s) = KT Proportional controller:

%TO KT = 2.5 %

Gc( s) = Kc

Process Transfer Functions: From the soution to Problem 4-3:

X6 ( s) = X4 ( s) =

X3 ( s) = τ 1 := K8 :=

V

τ 2 :=

f3 x2 − x4 f4

K3 = 0.706

V f4

τ 3 :=

V f6

τ 1 = 2.917 min K5 = 7.122 × 10

f4 K6 := f6

1 τ 3⋅ s + 1 1 τ 2⋅ s + 1 −K2 τ 1⋅ s + 1 K7 :=

(K6⋅ X4(s) − K7⋅ F1( s) − K7⋅ F2(s)) (K3⋅ X3(s) − K5⋅ F1( s) + K8⋅ F2(s)) F1 ( s)

x6 − x4 f6

(where we have added F2 (s) as an input variable) f3 K3 := f4

K5 :=

τ 2 = 2.059 min τ 3 = 1.795 min K6 = 0.872

− 3 %⋅ min

gal

K2 = 6.944 × 10

Eliminate X3 (s) and X4 (s): X4 ( s) =

(

− 3 %⋅ min

−K2 ⋅ K3 − K5 ⋅ τ 1 ⋅ s + 1

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1)

gal

) F (s) + 1

x4 − x3 f4

x3 K2 := f3

K8 = 0.017

K7 = 1.615 × 10

K8 τ 2⋅ s + 1

%⋅ min

− 3 %⋅ min

F2 ( s)

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

G2 ( s) =

(

) − K7⋅ (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)⋅ (τ 3⋅ s + 1)

K6 ⋅  −K2 ⋅ K3 − K5 ⋅ τ 1 ⋅ s + 1 

(

K6 ⋅ K8 − K7 ⋅ τ 2 ⋅ s + 1

(τ 2⋅ s + 1)⋅ (τ 3⋅ s + 1)

)

gal

gal

1 + H( s) ⋅ Gc( s) ⋅ G ( s) ⋅ G1 ( s) = 0

Characteristic equation of the loop:

1 + KT⋅ Kc⋅ Kv ⋅

2

(

) − K7⋅ (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) = 0 (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)⋅ (τ 3⋅ s + 1)

K6 ⋅  −K2 ⋅ K3 − K5 ⋅ τ 1 ⋅ s + 1 

(

)2

τ 1 ⋅ τ 2 ⋅ τ 3 ⋅ s + τ 1 ⋅ τ 2 + τ 1 ⋅ τ 3 + τ 2 ⋅ τ 3 − K7 ⋅ KT⋅ Kc⋅ Kv ⋅ τ 1 ⋅ τ 2 s +

(

)

+  τ 1 + τ 2 + τ 3 − K6 ⋅ K5 ⋅ KT⋅ Kc⋅ Kv ⋅ τ 1 − K7 ⋅ KT⋅ Kc⋅ Kv ⋅ τ 1 + τ 2  s +  

(

)

+ 1 − K6 ⋅ K2 ⋅ K3 + K5 + K7 ⋅ KT⋅ Kc⋅ Kv = 0   Let τ A := τ 1 ⋅ τ 2 ⋅ τ 3

τ B := τ 1 ⋅ τ 2 + τ 1 ⋅ τ 3 + τ 2 ⋅ τ 3

(

KB := KT⋅ Kv ⋅  K6 ⋅ K5 ⋅ τ 1 + K7 ⋅ τ 1 + τ 2  3

) 2

τ A = 10.778 min

τ C := τ 1 + τ 2 + τ 3 KA := K7 ⋅ KT⋅ Kv ⋅ τ 1 ⋅ τ 2

τ B = 14.935 min

(

)

KC := K6 ⋅ K2 ⋅ K3 + K5 + K7 ⋅ KT⋅ Kv   2

τ C = 6.77 min

KA = 1.802 min

KB = 4.858 min KC = 2.248

Substitute s = ωu i at Kc = K.cu : 3

(

)

(

2

)

−τ A⋅ ω u i − τ B − KA⋅ Kcu ω u + τ C − KB⋅ Kcu ω u ⋅ i + 1 − KC⋅ Kcu = 0 + 0i Real part:

(

)

2

2

− τ B − KA⋅ Kcu ω u + 1 − KC⋅ Kcu = 0

−Kcu =

2  τ B⋅ ω u − 1  Imaginary part: −τ A⋅ ω u +  τ C + KB⋅ ωu = 0 2  − ⋅ KC KA ω u 

τ B⋅ ω u − 1 KC − KA⋅ ω u

2

3

4

(

)

2

KA⋅ τ A⋅ ω u + −τ A⋅ KC + KB⋅ τ B − KA⋅ τ C ω u + KC⋅ τ C − KB = 0 Let

a := KA⋅ τ A

b := −τ A⋅ KC + KB⋅ τ B − KA⋅ τ C 5

a = 19.419 min ω u :=

−b +

2

b − 4 ⋅ a⋅ c 2⋅ a

c := KC⋅ τ C − KB 3

b = 36.13 min

c = 10.361 min

−1

ω u = 0.595i min

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.

Test:

%CO Kc := −1 %TO

Roots of the characteristic equation:

1 − KC⋅ Kc      ( τ C − KB⋅ Kc) min− 1   −0.59       polyroots = −0.481 − 0.528i  ( τ − K ⋅ K ) min− 2   A c  B    −0.481 + 0.528i   −3  τ A⋅ min   %CO Kc := −100 %TO

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

Roots of the characteristic equation:

1 − KC⋅ Kc      ( τ C − KB⋅ Kc) min− 1   −15.183    =  −2.326 polyroots   ( τ − K ⋅ K ) min− 2   A c  B    −0.593   −3  τ A⋅ min   %CO Kc := −10000 %TO

Negative real roots. Loop is stable.

Roots of the characteristic equation:

1 − KC⋅ Kc      ( τ C − KB⋅ Kc) min− 1   −1.67 × 103   = polyroots   ( τ − K ⋅ K ) min− 2   −2.105  A c  B    −0.593   −3  τ A⋅ min  

Negative real roots. Loop is stable.

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.

Offset for an increase of 10 gal/min in flow f2: Offset = Ksp ⋅ ∆x 6 gal ∆f2 := 10 min

set

G2 ( 0 ) − KT⋅ ∆x 6 = 0 − KT⋅ ∆f 1 + KT⋅ Kc⋅ Kv ⋅ G1 ( 0 ) 2

G2 ( 0 ) = KG2

KG2 := K6 ⋅ K8 − K7

KG2 = 0.013

min⋅ % gal

(

)

KG1 := K6 ⋅ −K2 ⋅ K3 − K5 − K7 KG1 = −0.012

G1 ( 0 ) = KG1

( )

min⋅ % gal

KG2

Offset Kc := −KT⋅ ∆f 1 + KT⋅ Kc⋅ Kv ⋅ KG1 2 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. Let

X = Kc

0

Offset( X ) 0.002

0.004

10

5

0

X Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-15. Analyzer control of reactors in series of Problem 4-9.

cA2set(t) AC

m(t)

AT

f1 (t)

f0(t) cA0(t)

V1

cA1(t)

V2

cA2(t)

fR

Block diagram:

CA0(s) CA2set(s) Ksp

+

E(s) Gc(s)

-

F0(s)

M(s) Gv(s)

G2(s)

+ G1(s)

CA2(s)

+

H(s) Proportional controller:

Transmitter:

H( s) =

Control valve:

Gc( s) = Kc KT τ T⋅ s + 1

τ T := 0.5min

Gv ( s) = Kv 100 ⋅ %TO⋅ ft

KT :=

3

V2 := V1

cA0 := 7

lbmole ft

3

3

%TO⋅ ft KT = 20 lbmole

5lbmole

Problem data: V1 := 125ft

3

3

k 1 := 0.2min

ft f0 := 10 min

k 2 := k 1

f1 := f0 + fR

−1

3

ft fR := 0 min

(f1 + k1⋅ V1)⋅ cA1 = f0⋅ cAo + fR⋅ cA2

At the initial steady state:

cA1 :=

f0 ⋅ cA0

(f1 + k2⋅ V2)cA2 = f1⋅ cA1 cA2 :=

fR⋅ f1 f1 + k 1 ⋅ V1 − f1 + k 2 ⋅ V2 cA1 = 2

200%⋅ f0 ⋅ 7.48gal

ft

ft ⋅ ∆p v

Cvmax := 110

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

3

lbmole ft

3

gal min⋅ psi gal

min⋅ psi 3

f0max Kv := 100%CO

ft f0max := Cvmax⋅ ∆p v 7.48gal

cA2 = 0.571

∆p v := 5psi

Cvmax = 66.903

3

f1 + k 2 ⋅ V2

lbmole

Control valve gain: linear sized for 100% overcapacity, assume Cvmax :=

f1 ⋅ cA1

ft Kv = 0.329 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

τ 1 :=

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K2⋅ K3

f0 K1 := f1 + k 1 ⋅ V1

fR K2 := f1 + k 1 ⋅ V1

V1 f1 + k 1 ⋅ V1

f1 K3 := f1 + k 2 ⋅ V2

τ 2 :=

V2 f1 + k 2 ⋅ V2

τ 1 = 3.571 min τ 2 = 3.571 min

K1 = 0.286

K2 = 0 K3 = 0.286

CA1( s) =

CA2( s) =

1

τ1

(K1⋅ CA0(s) + K4 F0( s) + K2⋅ CA2( s) ) ⋅s + 1 K3

τ 2⋅ s + 1

CA1( s) +

K5 τ 2⋅ s + 1

cA0 − cA1 K4 := f1 + k 1 ⋅ V1

cA1 − cA2 K5 := f1 + k 2 ⋅ V2

F0 ( s)

Substitute to eliminate CA1(s): CA2( s) = K3

=

(K1⋅ CA0(s) + K4⋅ F0( s) + K2⋅ CA2( s) ) +

(τ 2⋅ s + 1)⋅ (τ 1⋅ s + 1)

(

)

K5 τ 2⋅ s + 1

K1 ⋅ K3 ⋅ CA0( s) +  K3 ⋅ K4 + K5 ⋅ τ 1 ⋅ s + 1  ⋅ F0 ( s)  

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K2⋅ K3

F0 ( s)

K4 = 0.143

K5 = 0.041

lbmole⋅ min

lbmole⋅ min ft

6

ft

6

(

)

CA2( s) K3 ⋅ K4 + K5 τ 1 ⋅ s + 1 G1 ( s) = = F0 ( s) τ 1 ⋅ s + 1 ⋅ τ 2 ⋅ s + 1 − K2 ⋅ K3

(

1+

Charactetristic equation of the loop: 3

(

KT τ T⋅ s + 1

)(

)

Kc⋅ Kv ⋅ G1 ( s) = 0

)2 (

)

τ T⋅ τ 1 ⋅ τ 2 ⋅ s + τ T⋅ τ 1 + τ T⋅ τ 2 + τ 1 ⋅ τ 2 s + τ T + τ 1 + τ 2 − K2 ⋅ K3 ⋅ τ T + KT⋅ Kc⋅ Kv ⋅ K5 ⋅ τ 1 s

(

)

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

τ A := τ 1 ⋅ τ 2 ⋅ τ T

τ B := τ T⋅ τ 1 + τ T⋅ τ 2 + τ 1 ⋅ τ 2

(

τ D := KT⋅ Kv ⋅ K5 ⋅ τ 1

)

τ A = 6.378 min

KL = 0.537

K3 ⋅ K4 = 0.041

KL := KT⋅ Kv ⋅ K3 ⋅ K4 + K5

τ C = 7.643 min

τ D = 0.959 min

τ C := τ T + τ 1 + τ 2 − K2 ⋅ K3 ⋅ τ T 3

2

τ B = 16.327 min

lbmole⋅ min ft

6

Substitute s = ωu i at Kc = K cu : 3

2

(

)

−τ A⋅ ω u i − τ B⋅ ω u + τ C + τ D⋅ Kcu ω u ⋅ i + 1 − K2 ⋅ K3 + KL⋅ Kcu = 0 + 0i 2

Real part:

2

−τ B⋅ ω u + 1 − K2 ⋅ K3 + KL⋅ Kcu = 0

Kcu =

τ B⋅ ω u − 1 + K2 ⋅ K3 KL

2  τ B⋅ ω u − 1 + K2 ⋅ K3   Imaginary part: −τ A⋅ ω u + τ C + τ D⋅ ωu = 0  KL   3

ω u :=

(

KL⋅ τ C − τ D⋅ 1 − K2 ⋅ K3

)

−1

ω u = 0.507i min

KL⋅ τ A − τ D⋅ τ B

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.

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

Ksp ⋅ ∆cA2

set

− KT⋅ ∆cA2

1 + KT⋅ Kc⋅ Kv ⋅ G1 ( 0 )

=

0 − KT G2 ( 0 ) ∆cA0 1 + KT⋅ Kc⋅ Kv ⋅ G1 ( 0 )

∆cA0 := 1

lbmole ft

3

G2 ( 0 ) = KG2

G1 ( 0 ) = KG1

K1 ⋅ K3 KG2 := 1 − K2 ⋅ K3

KG2 = 0.082

K3 ⋅ K4 + K5 KG1 := 1 − K2 ⋅ K3

KG1 = 0.082

lbmole⋅ min ft

6

KG2 Offset Kc := −KT⋅ ∆c 1 + KT⋅ Kc⋅ Kv ⋅ KG1 A0

( )

For Kc = 1 %CO/%TO:

Offset( 1 ) = −1.062 %TO

Open-loop offset (Kc = 0):

Offset( 0 ) = −1.633 %TO

For a PI controller the offset is zero. Compare these results with the simulation of this process in Problem 13-16. 0

Let

X = Kc Offset( X) 0.01

0.02 0

5

10

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Smith & Corripio, 3rd edition Problem 6-16. Temperature control of direct contact heater of Problem 4-5. Problem parameters: V := 5gal cp := 0.8

ρ := 7

gal

BTU lb⋅ degF

Design conditions: gal f1p := 25 min

T3set(t)

lb

T1p := 60degF

f1(t) TT

T1(t) Liquid

f3(t) T3(t)

T3p := 80degF Control valve:

w2(t)

∆p v := 10psi

Saturated steam

∆p v lb w2 ( t) = 1.954 vp( t) ⋅ min psi τ v := 4s⋅

min 60s

TC

Disturbance:

gal ∆f1 := 2 min

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

BTU Hs := 1145.4 lb

Block diagram, ignoring the inlet temperature as a variable:

m(t)

"p" denotes base value for linearization.

Tmin := 50degF Tmax := 100degF τ T := 10s⋅

min 60s

F1(s) T3set(s) Ksp

+

E(s)

M(s) Gc(s)

-

Gv(s)

G2(s)

+

W2(s)

G3(s)

T3(s)

+

H(s) Proportional controller:

Gc( s) = Kc

Linear control valve:

Gv ( s) =

Temperature transmitter: H( s) =

∆p v lb 1 lb Kv := 1.954 ⋅ K = 0.062 psi 100%CO v min min⋅ %CO

Kv τ v⋅ s + 1 KT

100%TO KT := Tmax − Tmin

τ T⋅ s + 1

Ksp := KT

%TO KT = 2 degF

Enthalpy balance on the tank, neglecting heat losses:

(

)

(

)

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

d ⋅ T3 ( t ) dt

d⋅ V f1 ( t) ⋅ ρ + w2 ( t) − f3 ( t) ⋅ ρ = ρ ⋅ =0 dt

Mass balance, assuming constant volume:

f3 ( t) ⋅ ρ = f1 ( t) ⋅ ρ + w2 ( t) Sustitute into enthalpy balance: V⋅ ρ ⋅ cv ⋅

d ⋅ T3 ( t ) dt

(

)

(

)

= f1 ( t) ⋅ ρ ⋅ cp ⋅ T1 ( t) − T3 ( t) + w2 ( t) ⋅  Hs − cp ⋅ T3 ( t) − 32⋅ degF 

At the initial steady state:

(

)

f1p⋅ ρ ⋅ cp ⋅ T3p − T1p w2p := Hs − cp ⋅ T3p − 32degF

(

)

lb w2p = 2.529 min

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

d⋅ Γ 3( t) dt

Γ 1 ( t) = T1 ( t) − T1p

(

)

= a1 ⋅ Γ 1 ( t) − Γ 3 ( t) + a2 ⋅ F1 ( t) + a3 ⋅ W2 ( t) − a4 ⋅ Γ 3 ( t) Γ 3 ( t) = T3 ( t) − T3p

F1 ( t) = f1 ( t) − f1p

W2 ( t) = w2 ( t) − w2p

(

a1 := f1p⋅ ρ ⋅ cp Rearrange: where

τ⋅ τ :=

a2 := ρ ⋅ cp ⋅ T1p − T3p

d ⋅ Γ 3 ( t) dt

V⋅ ρ ⋅ cp a1 + a4

)

(

a3 := Hs − cp ⋅ T3p − 32degF

()

+ Γ 3 ( t) = K1 ⋅ Γ 1 t + K2 ⋅ F1 ( t) + K3 ⋅ W2 ( t) a1 K1 := a1 + a4

a2 K2 := a1 + a4

K1 = 0.986

K2 = −0.789

gal

a4 := w2p⋅ cp

Γ 3(0) = 0

a3 K3 := a1 + a4

degF⋅ min

)

τ = 0.197 min

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. K1 K2 K3 Laplace transform: Γ 3 ( s) = Γ 1 ( s) + F1 ( s) + W2 ( s) τ ⋅s + 1 τ ⋅s + 1 τ ⋅s + 1 G2 ( s) =

K2

G3 ( s) =

τ ⋅s + 1

K3 τ ⋅s + 1

Ultimate gain and period of the loop: Characteristic equation:

1+

KT τ T⋅ s + 1

⋅ Kc⋅

Kv

⋅

K3

=0

τ v⋅ s + 1 τ ⋅ s + 1

let KKc = KT⋅ Kc⋅ Kv ⋅ K3

(

3

)

(

2

)

τ T⋅ τ v ⋅ τ ⋅ s + τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ⋅ s + τ T + τ v + τ s + 1 + KKc = 0 Substitue s = ωu i at KK c = KKcu : 3

(

)

2

(

)

−τ T⋅ τ v ⋅ τ ⋅ ω u i − τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ⋅ ω u + τ T + τ v + τ ω u ⋅ i + 1 + KKcu = 0 3

(

)

Imaginary part:

−τ T⋅ τ v ⋅ τ ⋅ ω u + τ T + τ v + τ ⋅ ω u = 0

Real part:

− τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ω u + 1 + KKcu = 0

(

)

Kcu :=

ω u :=

τT + τv + τ τ T⋅ τ v ⋅ τ

2

Tu :=

%CO Kcu = 10.61 %TO

KT⋅ Kv ⋅ K3

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

set

− KT⋅ ∆T3

1 + KT⋅ Kc⋅ Kv ⋅ K3

=

0 − KT⋅ K2 ⋅ ∆f1 1 + KT⋅ Kc⋅ Kv ⋅ K3

ωu

Tu = 0.448 min

( τ T⋅ τ v + τ T⋅ τ + τ v ⋅ τ ) ω u 2 − 1

Ksp ⋅ ∆T3

2π

Kc :=

Kcu 2

Offset :=

−KT⋅ K2 ⋅ ∆f1 1 + KT⋅ Kc⋅ Kv ⋅ K3

Offset = 0.516 %TO Offset KT

= 0.258 degF

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-17. Composition control of three isothermal reactors in series.

c3set(t) AC

m(t)

f(t)

c1(t)

c2(t)

V

ci(t)

Problem data:

V

−1

V := 1000gal

Analyzer transmitter:

k := 0.1min

cmin := 0

lb gal

AT V

gal fp := 100 min

cmax := 1.0

lb gal

cip := 4

lb gal

100%TO KT := cmax − cmin

KT = 100

Control valve: linear with constant pressure drop and sized for 100% overcapacity. 200%⋅ fp gal ∆p v := 5psi Cvmax := Cvmax = 89.443 min⋅ psi ∆p v From Fig. C-10.1, page 532, a 3-in valve is required:

Kv :=

Cvmax := 110

100%CO

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

V⋅

V⋅

%TO⋅ gal lb

τ v := 0.1min

gal

min⋅ psi The valve fails closed (air-to-open) gal to prevent overflowing the reactors Kv = 2.46 min⋅ %CO on power failure.

Cvmax⋅ ∆p v

V⋅

c3(t)

d ⋅ c1 ( t ) dt d ⋅ c2 ( t ) dt d ⋅ c3 ( t ) dt

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

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

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

Linearize:

V⋅ V⋅

V⋅

At the initial steady-state: Rearrange: τ⋅

τ⋅

τ⋅

τ :=

d ⋅ C1 ( t ) dt d ⋅ C2 ( t ) dt d ⋅ C2 ( t ) dt V

fp + k ⋅ V

τ = 5 min

d ⋅ C2 ( t )

)

(

d ⋅ C3 ( t )

(

)

(

)

(

)

(

)

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

dt

F( t) = ( t) − fp

c1p :=

fp ⋅ cip

c2p :=

fp + k ⋅ V

fp ⋅ c1p

Ki⋅ Ci( s) + K1 ⋅ F( s) τ ⋅s + 1

C3 ( 0 ) = 0

cip − c1p K1 := fp + k ⋅ V lb⋅ min gal

C2 ( s) =

c1p − c2p K2 := fp + k ⋅ V

K2 = 0.005

2

fp + k ⋅ V

C2 ( 0 ) = 0

+ C2 ( t) = Ki⋅ Ci( t) + K3 ⋅ F( t)

K1 = 0.01

fp ⋅ c2p

C1 ( 0 ) = 0

+ C2 ( t) = Ki⋅ C1 ( t) + K2 ⋅ F( t)

fp Ki := fp + k ⋅ V

c3p :=

fp + k ⋅ V

+ C1 ( t) = Ki⋅ Ci( t) + K1 ⋅ F( t)

Ki = 0.5

)

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

dt

Laplaxce transform: C1 ( s) =

(

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

dt

Cj( t) = Cj( t) − cjp

where

where

d ⋅ C1 ( t )

Ki⋅ C1 ( s) + K2 ⋅ F( s) τ ⋅s + 1

lb⋅ min gal

2

C3 ( s) =

c2p − c3p K3 := fp + k ⋅ V

K3 = 0.0025

G1 ( s) =

(

τ ⋅s + 1

2

G1 ( s) =

)2

(τ ⋅ s + 1)3 Ki ⋅ K1 + Ki⋅ K2 ⋅ ( τ ⋅ s + 1 ) + K3 ⋅ ( τ ⋅ s + 1 )

( τ ⋅ s + 1) 3

2

C3 ( s)

gal

2

Ki⋅ C2 ( s) + K3 ⋅ F( s)

Combine to obtain the transfer functions: C3 ( s)   1  Ki  Ki⋅ K1 G1 ( s) = = + K2 + K3   F( s) τ ⋅s + 1 τ s + 1  τ ⋅s + 1    Ki⋅  Ki⋅  K1 + K2 ⋅ ( τ ⋅ s + 1 )  + K3 ⋅ τ s + 1   

lb⋅ min

Ki

3

G2 ( s) = = Ci( s) (τ ⋅ s + 1)3

(a) Block diagram of the loop:

Ci(s) C3set(s) Ksp

+

E(s)

+

F(s)

M(s) Gc(s)

-

G2(s)

Gv(s)

G1(s)

C3(s)

+

H(s) Ksp := KT

G2 ( s) =

H( s) = KT Ki

Kv

Gv ( s) =

Proportional controller:

τ v⋅ s + 1

3

2

G1 ( s) =

( τ ⋅ s + 1) 3 2

KA := Ki K1 + Ki⋅ K2 + K3

KA + KB⋅ τ ⋅ s + K3 ⋅ τ ⋅ s

(τ ⋅ s + 1)3

KB := 2 ⋅ K3 + Ki⋅ K2

τ = 5 min

K3 = 0.0025

(b) Offset for a change in inlet concentration

Offset =

Gc( s) = Kc

Ksp ⋅ ∆c3

set

− KT⋅ G2 ( 0 ) ∆ci

1 + KT⋅ Kc⋅ Kv ⋅ G1 ( 0 )

KA = 0.0075

lb⋅ min gal

∆ci := 1

2

KB = 0.0075

lb gal

gal

2

lb⋅ min gal

2

%CO Kc := 1. %TO

3

Offset := −

KT⋅ Ki ∆ci 1 + KT⋅ Kc⋅ Kv ⋅ KA

Offset = −4.39 %TO For the open loop, Kc = 0:

lb⋅ min

3

Offset := −KT⋅ Ki ∆ci

Offset KT

Offset = −12.5 %TO

For a PI controller the offset is zero.

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

= −0.044

lb gal

Characteristic equation of the loop:

1 + KT⋅ Kc⋅

2 2

Kv τ v⋅ s + 1

⋅

KA + KB⋅ τ ⋅ s + K3 ⋅ τ ⋅ s

( τ ⋅ s + 1) 3

=0

Rearrange and substitute s = ωu i at Kc = K cu : 3

4

2

(

3

)

(

2

)

τ v ⋅ τ ω u − 3τ v ⋅ τ ω u i − 3τ v ⋅ τ + τ A⋅ Kcu ω u + τ v + τ B⋅ Kcu ω u ⋅ i + 1 + KL⋅ Kcu = 0 where

τ A := KT⋅ Kv ⋅ K3 ⋅ τ

2

τ B := KT⋅ Kv ⋅ KB⋅ τ

KL := KT⋅ Kv ⋅ KA 2

τ A = 15.373 min Real part:

3

4

(

)

%TO KL = 1.845 %CO τ B = 9.224 min

2

τ v ⋅ τ ω u − 3 ⋅ τ v ⋅ τ + τ A⋅ Kcu ω u + 1 + KL⋅ Kcu = 0 2

Kcu =

3

4

3⋅ τ v⋅ τ ⋅ ω u − τ v⋅ τ ω u − 1 KL − τ A⋅ ω u

2

2 3 4  3⋅ τ v⋅ τ ⋅ ω u − τ v⋅ τ ω u − 1  Imaginary part: −3 ⋅ τ v ⋅ τ ω u +  τ v + τ B⋅ ωu = 0 2  KL − τ A⋅ ω u   3 ⋅ τ ⋅ τ 2⋅ τ − τ ⋅ τ ⋅ τ 3 ω 4 +  −3⋅ τ ⋅ τ 2⋅ K − τ ⋅ τ + τ ⋅ 3 ⋅ τ ⋅ τ  ω 2 − τ = 0 A B v  u v L v A B v  u B  v  2

Let

3

2

a := 3τ v ⋅ τ ⋅ τ A − τ B⋅ τ v ⋅ τ

3

2

b := −3 ⋅ τ v ⋅ τ ⋅ KL − τ v ⋅ τ A + τ B⋅ 3 ⋅ τ v ⋅ τ − 14

a = −1.962 × 10

ω u :=

−b −

( )

2

b − 4 ⋅ a⋅ −τ B 2⋅ a

5

min

3

b = −1.537 min

−1

ω u = 2.45i min

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

%CO := %

%TO := %

kscf := 1000ft psig := psi

Problem 6-18. Compressor suction pressure control. Problem Data: Ps( s) =

Kp τ p⋅ s + 1

(Fi( s) − Fc( s) ) PC

psi⋅ min

Kp := 0.5 kscf

m(t) SC

τ p := 7.5s Fc( s) =

Ksc τ sc⋅ s + 1

Steam

M ( s)

fc(t) Discharge

PT

kscf Ksc := 0.36 min⋅ %CO

Suction

τ sc := 2.5s

ps(t)

fi(t)

Pressure transmitter, PC:

Pmin := 0psig

Pmax := 20psig

τ T := 1.2s

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

Fi(s) Psset(s) Ksp

+

E(s)

-

M(s) Gc(s)

Gsc(s)

Fc(s)

Gp(s)

+ Gp(s)

Ps(s)

-

H(s) 100%TO KT := Pmax − Pmin

Ksp := KT

Closed-loop transfer function:

H( s) =

KT τ T⋅ s + 1

%TO KT = 5 psi

Gsc( s) =

Gp ( s) =

Ksc τ sc⋅ s + 1 Kp τ p⋅ s + 1

3

−Ksp ⋅ Gc( s) ⋅ Gsc( s) ⋅ Gp ( s) Gp ( s) set Ps( s) = Ps ( s) + F ( s) 1 − H( s) ⋅ Gc( s) ⋅ Gsc( s) Gp ( s) 1 − H( s) ⋅ Gc( s) ⋅ Gsc( s) ⋅ Gp ( s) i

(

)

Characteristic equation of the loop: 1 − H( s) ⋅ Gc( s) ⋅ Gsc( s) ⋅ Gp ( s) = 1 −

KT

Gc( s)

τ T⋅ s + 1

Kp

Ksc

τ sc⋅ s + 1 τ p ⋅ s + 1

=0

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.

(b) Ultimate gain and period for a preportional controller. Gc( s) = Kc Rearrange the characteristic equation and sunbstiture s = ωu i at Kc = K cu :

(

3

)

2

(

)

−τ T⋅ τ sc⋅ τ p ⋅ ω u i − τ T⋅ τ sc + τ T⋅ τ p + τ sc⋅ τ p ω u + τ T + τ sc + τ p ω u ⋅ i + 1 + KL⋅ Kcu = 0 + 0i %TO where KL := −KT⋅ Ksc⋅ Kp KL = −0.9 %CO 3

Real part:

(

)

−τ T⋅ τ sc⋅ τ p ⋅ ω u + τ T + τ sc + τ p ω u = 0

Imaginary part:

(τ T⋅ τ sc + τ T⋅ τ p + τ sc⋅ τ p)ω u

Kcu :=

2

−1

Tu :=

KL

τ T + τ sc + τ p

ω u :=

τ T⋅ τ sc⋅ τ p 2π ωu

Tu = 8.91 s %CO Kcu = −15.9 %TO

(c) Offset caused by a change in inlet flow. ∆fi := 1 set

Offset =

Ksp ⋅ ∆p s

− KT⋅ Kp ⋅ ∆fi

1 + KL⋅ Kc

Offset :=

kscf min

0 − KT⋅ Kp ⋅ ∆fi 1 + KL⋅ Kc

Kc :=

Kcu 2

%CO Kc = −7.9 %TO

Offset = −0.307 %TO Offset KT

= −0.061 psi

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

kJ := 1000joule

degC := K

Problem 6-19. Temperature control of stirred-tank cooler of Problem 4-7. Problem Data: 3

V := 5m

Tset(t)

kJ

U := 200

TC

2

m min⋅ degC 2

f(t)

3

A := 4m

Vc := 1.1m kg

ρ := 800

cp := 3.8

3

m ρ c := 1000

kg 3

Tc(t)

kg⋅ degC

cpc := 4.2

m

Ti(t)

kJ

m(t)

TT

V

kJ

SP FC

kg⋅ degC

Design conditions:

FT

3

f := 0.1

m

min

T := 45degC

Ti := 70degC Tci := 25degC

fc(t)

T(t)

Tci

Temperature transmitter (TT): Tmin := 20degC

Tmax := 70degC

τ T := 0.6min

3

m Flow transmitter(FT): fmin := 0 min

3

m fmax := 0.8 min

τ FC := 0.1min

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

Ti(s) Tset(s) Ksp

+

E(s)

-

M(s) Gc(s)

GFC(s) H(s)

Fc(s)

G2(s) G1(s)

+ -

T(s)

100%TO KT := Tmax − Tmin

Temperature transmitter:

KFC :=

Flow contriol loop:

Ksp := KT

fmax − fmin

GFC( s) =

100%TO

KT

H( s) =

%TO KT = 2 degC

τ T⋅ s + 1

KFC

KFC = 8 × 10

τ FC⋅ s + 1

3

m

−3

min⋅ %TO

From the results of Problem 4-7: K2 ⋅ K3

G1 ( s) =

where

τ 1 :=

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K2⋅ K4 V⋅ ρ ⋅ cp

K1 :=

f ⋅ ρ ⋅ cp + U⋅ A

f ⋅ ρ ⋅ cp

K2 :=

f ⋅ ρ ⋅ cp + U⋅ A

(

)

(

)

(

)

(

)

f ⋅ ρ ⋅ cp ⋅ Ti − T − fc⋅ ρ c⋅ cpc⋅ Tc − Tci = 0

τ 2 :=

Vc⋅ ρ c⋅ cpc fc⋅ ρ c⋅ cpc + U⋅ A

K3 :=

U⋅ A

fc :=

(

τ 1 = 13.77 min

f ⋅ ρ ⋅ cp + U⋅ A

(

f ⋅ ρ ⋅ c p ⋅ Ti − T

Tc := T −

)

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K2⋅ K4

K1 = 0.275

At the initial steady state (design) conditions: f ⋅ ρ ⋅ cp ⋅ Ti − T − U⋅ A⋅ T − Tc = 0

(

K1 ⋅ τ 2 ⋅ s + 1

G2 ( s) =

K2 = 0.725

)

Tc = 35.5 degC

U⋅ A

(

f ⋅ ρ ⋅ c p ⋅ Ti − T

(

)

3

m fc = 0.172 min

)

ρ c⋅ cpc⋅ Tc − Tci

)

ρ c⋅ cpc⋅ Tc − Tci

K4 :=

fc⋅ ρ c⋅ cpc + U⋅ A

τ 2 = 3.03 min

U⋅ A fc⋅ ρ c⋅ cpc + U⋅ A

K3 = 28.94

degC ⋅ min

K4 = 0.525

3

m

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+

KT

τ T⋅ s + 1

Kc⋅

KFC

(

)(

K2 ⋅ K3

Rearrange and substitute s = ωu i at Kc = K cu : 4

3

2

)

τ FC⋅ s + 1 τ 1 ⋅ s + 1 τ 2 ⋅ s + 1 − K2 ⋅ K4

τ A⋅ ω u − τ B⋅ ω u i − τ C ⋅ ω u + τ D⋅ ω u ⋅ i + 1 − K2 ⋅ K4 + KL⋅ Kcu = 0 + 0i

=0

where

τ A := τ T⋅ τ FC⋅ τ 1 ⋅ τ 2

τ B := τ T⋅ τ FC⋅ τ 1 + τ T⋅ τ FC⋅ τ 2 + τ FC⋅ τ 1 ⋅ τ 2 + τ T⋅ τ 1 ⋅ τ 2 4

τ = 2.505 min τ C := τ T⋅ τ FC⋅ 1 − K2 ⋅ K4 + τ T⋅ τ 1 + τ T⋅ τ 2 + τ FC⋅ τ 1 + τ FC⋅ τ 2 + τ 1 ⋅ τ 2 A 3 τ B = 30.228 min

(

)

(

)(

)

τ D := τ T + τ FC 1 − K2 ⋅ K4 + τ 1 + τ 2

KL := −KT⋅ KFC⋅ K2 ⋅ K3

2

τ C = 53.54 min

%TO KL = −0.336 %CO 3

τD

Imaginary part:

−τ B⋅ ω u + τ D⋅ ω u = 0

Real part:

τ A⋅ ω u − τ C⋅ ω u + 1 − K2 ⋅ K4 + KL⋅ Kcu = 0

4

τB

Tu :=

2π

Tu = 8.32 min

ωu

2

4

Kcu :=

ω u :=

τ D = 17.234 min

2

−τ A⋅ ω u + τ C ⋅ ω u − 1 + K2 ⋅ K4

%CO Kcu = −86.7 %TO

KL

(c) Offset for change in inlet temperature with a proportional copntroller. ∆Ti := 5degC

Offset =

Ksp ⋅ ∆T

set

Kc :=

Kcu

− KT⋅ G2 ( 0 ) ∆Ti

1 + KT⋅ Kc⋅ KFC⋅ G1 ( 0 )

%CO Kc = −43.3 %TO

2

Offset :=

0 − KT⋅ K1 ⋅ ∆Ti 1 − K2 ⋅ K4 − KT⋅ Kc⋅ KFC⋅ K2 ⋅ K3 Offset = −0.18 %TO

Open-loop, Kc = 0:

Offset := 0 −

KT⋅ K1 ⋅ ∆Ti 1 − K2 ⋅ K4

Offset KT

= −0.09 degC

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

degF := R

scf := ft

V := 550000ft

MW := 50 3

lb

p2set(t)

lbmole

PC

T := 350degF

scf

f1 := 1500 p := 90psi min 1 scf

f3 := 500 min

p 3 := 30psig

p 2 := 45psig

p 4 := 15psig

psia := psi

lbmole := 453.59mole

Problem 6-20. Pressure control og gas torage tank. Problem Data:

3

p3(t)

PT

p1(t)

vp3(t) p4(t)

p2(t)

vp1(t)

vp4(t)

Pressure transmitter PT: p min := 0psig

p max := 100psig

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

Use Eqs. 5-2.3 and 5-2.5, page 160. p1 − p2 1.63 Valve 1: y 1 := ⋅ p 1 + 14.7psia Cf Cv1max :=

gal⋅ psia⋅ hr 60min

836 ⋅ C f ⋅ p 1 + 14.7psia  y 1 − 0.148y 1 

)

3

Cv3max :=

1.63

y 3 :=

Cf

p2 − p3

⋅

p 2 + 14.7psia

3

0.5

 min⋅ R scf

From initial steady state conditions: y 4 :=

1.63 Cf

⋅

p2 − p4 p 2 + 14.7psia

hr

gal min⋅ psi

gal min⋅ psi

Cv3max = 62.6

Cv3max := 110

From Fig. C-10.1, page 532, a 3-in valve is required: Valve 4:

G = 1.724

Cv1max = 90.9

Cv1max := 110

gal⋅ psia⋅ hr 60min

836 ⋅ C f ⋅ p 2 + 14.7psia  y 3 − 0.148y 3 

)

29lb

y 3 = 0.908

200%⋅ f3 ⋅ G⋅ ( T + 460R )

(

hr

0.5  min⋅ R scf

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

Valve 3:

MW⋅ lbmole

y 1 = 1.187

200%⋅ f1 ⋅ G⋅ ( T + 460R )

(

G :=

f4 := f1 − f3

y 4 = 1.284

gal min⋅ psi

gal min⋅ psi scf f4 = 1000 min

Cv4max :=

200%⋅ f4 ⋅ G⋅ ( T + 460R )

gal⋅ psia⋅ hr 60min

836 ⋅ C f ⋅ p 2 + 14.7psia  y 4 − 0.148y 4 

(

)

3

hr

0.5  min⋅ R scf

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

Cv4max = 102.9

Cv4max := 110

gal min⋅ psi

gal min⋅ psi

(c) Block diagram of the loop considering all disturbances. Transmitter PT: Process model: V⋅

Mass balance:

d ⋅ ρ 2( t) dt

ρ 2( t) =

Ideal gas law: Rg := 10.73 ⋅

100%TO KT := p max − p min

H( s) = KT

psia⋅ ft

lbmole⋅ R

gal⋅ psia⋅ hr

y1 ( t) =

1.63 Cf

⋅

1.63 Cf

⋅

1.63 Cf

⋅

Substitute ideal gas law:

ρ s = 0.132

R g ⋅ 520 R

⋅

1 ⋅ hr 60⋅ min

⋅

lb scf

(

C v1max⋅ C f ⋅ p 1 ( t ) + 14.7⋅ psia G⋅ ( T + 460R)

⋅

1 ⋅ hr 60⋅ min

) y

3  1( t) − 0.148 y 1( t)  vp1 ( t)

4 eqns. 6 unks. (y1 )

⋅

(

C v3max⋅ C f ⋅ p 2 ( t ) + 14.7⋅ psia G⋅ ( T + 460R)

) y

3  3( t) − 0.148 y 3( t)  vp3 ( t)

6 eqns. 7 unks. (y3 )

p 2 ( t ) + 14.7psia

gal⋅ psia⋅ hr

y4 ( t) =

MW⋅ 14.7psia

p2 ( t) − p3 ( t)

836 ⋅ scf ⋅ R ⋅ min

Valve 4: f4 ( t ) =

2 eqns. 5 unks. (p 2 )

Rg ⋅ ( T + 460R )

p 1 ( t ) + 14.7psia

gal⋅ psia⋅ hr

y3 ( t) =

1 eqn. 4 unks. (ρ2 , f1 , f2 , f3 )

)

p1 ( t) − p2 ( t)

836 ⋅ scf ⋅ R ⋅ min

Valve 3: f3 ( t ) =

(

ρ s :=

836 ⋅ scf ⋅ R ⋅ min

Valve 1: f1 ( t ) =

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

MW⋅ p 2 ( t ) + 14.7psia

3

%TO KT = 1 psi

Ksp := KT

⋅

1 ⋅ hr 60⋅ min

⋅

(

C v4max⋅ C f ⋅ p 2 ( t ) + 14.7⋅ psia G⋅ ( T + 460R)

p2 ( t) − p4 ( t)

) y

3  4( t) − 0.148 y 4( t)  vp4 ( t)

8 eqns. 8 unks. (y4 )

p 2 ( t ) + 14.7psia MW⋅ V Rg ⋅ ( T + 460 ⋅ R ) ρ s

⋅

d⋅ p 2( t) dt

= f1 ( t ) − f3 ( t ) − f4 ( t )

Linearize and substitute deviation variables: MW⋅ V Rg ⋅ ( T + 460 ) ⋅ ρ s

⋅

d ⋅ P2 ( t ) dt

= a1 ⋅ VP 1 ( t ) + b 1 ⋅ P1 ( t ) − c1 ⋅ P2 ( t ) − a3 ⋅ VP 3 ( t ) − b 3 ⋅ P2 ( t ) + c3 ⋅ P3 ( t )

−a4 ⋅ VP 4 ( t ) − b 4 ⋅ P2 ( t ) + c4 ⋅ P4 ( t ) k v :=

Let

a1 = a3 =

a4 =

836scf ⋅ R ⋅ min

hr

gal⋅ psia⋅ hr

60min

δ ⋅ f1 ( t )

a1 :=

δ ⋅ vp δ ⋅ f3 ( t )

a3 :=

δ ⋅ vp δ ⋅ f4 ( t )

a4 :=

δ ⋅ vp

k y11 =

b1 =

δ⋅ p1

δ ⋅ f1 ( t )

scf

⋅ C v3max⋅ p 2 + 14.7psia ⋅  y 3 − 0.148y 3

3

scf

⋅C ⋅ p + 14.7psia ⋅  y 4 − 0.148y 4  100%CO v4max 2

)

3

f4 vp4 := a4

vp1 = 41.3 %CO

100%CO kv 100%CO

)

(

kv

)

(

f3 vp3 := a3

k v ⋅ C v1max⋅ vp1 100%CO

c1 =

k y32 =

b3 =

δ ⋅ y1 ( t) δ⋅ p2

−δ ⋅ f1 ( t ) δ⋅ p2 δ ⋅ y3 ( t) δ⋅ p2

δ ⋅ f3 ( t ) δ⋅ p2

k y12 :=

c1 :=

− 0.5

⋅

p1 − p2  1.63 1  ⋅ ⋅ Cf 2 p 1 + 14.7⋅ psia  

−k v ⋅ C v1max⋅ vp1 100%CO

k v ⋅ C v3max⋅ vp3 100%CO

k y33 =

δ⋅ p3

k y33 :=

 a3 = 17.565 min⋅ %CO scf

 a4 = 21.39 min⋅ %CO vp3 = 28.5 %CO vp4 = 46.8 %CO

(p1 + 14.7⋅ psia) − (p1 − p2) (p1 + 14.7psia)2

− 0.5

⋅

−1

−1

(p1 + 14.7psia)

b 1 = 18.819

scf min⋅ psi −1

k y12 = −0.013 psi

( p + 14.7psia)  1 − 3 ⋅ 0.148 y 2 k  c = 7.878 scf 1  y12 1  1  min⋅ psi − 0.5

⋅

(p2 + 14.7⋅ psia) − (p2 − p3) (p2 + 14.7psia)2

 y − 0.148 ⋅ y 3 + ( p + 14.7psia)  1 − 3 ⋅ 0.148 y 2 k  3 2 3  y32   3  k y32 = 0.02266 psia

δ ⋅ y3 ( t)

 a1 = 36.312 min⋅ %CO

 y − 0.148 ⋅ y 3 + ( p + 14.7psia)  1 − 3 ⋅ 0.148 y 2 k  1 1 1  y11   1 

p2 − p3  1.63 1  k y32 := ⋅ ⋅ Cf 2 p 2 + 14.7⋅ psia   b 3 :=

gal⋅ psia

3

(

k y11 = 0.00752 psia k y12 =

scf

⋅ C v1max⋅ p 1 + 14.7psia ⋅  y 1 − 0.148y 1

p1 − p2  1.63 1  k y11 := ⋅ ⋅ Cf 2 p 1 + 14.7⋅ psia   b 1 :=

δ⋅ p1

k v = 0.336

G⋅ ( T + 460R)

f1 vp1 := a1

Initial valve positions:

δ ⋅ y1 ( t)

kv

Cf

p2 − p3  1.63 1  ⋅ ⋅ Cf 2 p 2 + 14.7⋅ psia  

− 0.5

⋅

−1

−1

(p2 + 14.7psia)

b 3 = 17.387

scf min⋅ psi −1

k y33 = −0.03 psi

c3 =

−δ ⋅ f3 ( t ) δ⋅ p3

k y42 =

b4 =

c3 :=

−k v ⋅ C v3max⋅ vp3 100%CO

( p + 14.7psia)  1 − 3 ⋅ 0.148 y 2 k  c = 12.036 scf 3  y33 3  2  min⋅ psi

p2 − p4  1.63 1  k y42 := ⋅ ⋅ Cf 2 p 2 + 14.7⋅ psia  

δ ⋅ y4 ( t) δ⋅ p2

δ ⋅ f4 ( t )

b 4 :=

δ⋅ p2

k v ⋅ C v4max⋅ vp4 100%CO

− 0.5

⋅

(p2 + 14.7⋅ psia) − (p2 − p4) (p2 + 14.7psia)

 y − 0.148 ⋅ y 3 + ( p + 14.7psia)  1 − 3 ⋅ 0.148 y 2 k  4 2 4  y42   4  k y42 = 0.01065 psia

k y44 =

c4 =

δ ⋅ y4 ( t)

p2 − p4  1.63 1  k y44 := ⋅ ⋅ Cf 2 p 2 + 14.7⋅ psia  

δ⋅ p4

−δ ⋅ f4 ( t )

c4 :=

δ⋅ p4

−k v ⋅ C v4max⋅ vp4 100%CO

2

− 0.5

⋅

(

−1

−1 p 2 + 14.7psia

)

b 4 = 19.691

scf min⋅ psi −1

k y44 = −0.021 psi

( p + 14.7psia)  1 − 3 ⋅ 0.148 y 2 k  c = 5.911 scf 4  y44 4  2  min⋅ psi

Rearrange equation into standard first-order form: τ⋅

d ⋅ P2 ( t ) dt

where

+ P2 ( t ) = K1 ⋅ VP 1 ( t ) − K3 ⋅ VP 3 ( t ) − K4 ⋅ VP 4 ( t ) + Kp1⋅ P1 ( t ) + Kp3⋅ P3 ( t ) + Kp4⋅ P4 ( t )

τ :=

MW⋅ V

(

Rg ⋅ ( T + 460R ) ⋅ ρ s⋅ c1 + b 3 + b 4

a4 K4 := c1 + b 3 + b 4

)

b1 Kp1 := c1 + b 3 + b 4

a1 K1 := c1 + b 3 + b 4 c3 Kp3 := c1 + b 3 + b 4

psig τ = 534.3 min K1 = 0.808 %CO Kp1 = 0.419

a3 K3 := c1 + b 3 + b 4 c4 Kp4 := c1 + b 3 + b 4

psi K3 = 0.391 %CO

psi K4 = 0.476 %CO

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:

P1(s)

Kp1

VP4(s)

K4

F1(s) VP3(s)

K3 P2set(s)

Ksp

+

E(s)

-

VP1(s)

Gc(s)

K1

Kp4

+ Kp3 - + - + 1 + Js + 1

P4(s) P3(s) P2(s)

H(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 Offset :=

Ksp ⋅ ∆p 2set 1 + KT⋅ Kc⋅ K1

Offset = 0.121 %TO

%CO Kc := 50 %TO Offset KT

= 0.121 psi

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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

wB(t) TB(t)

m2(t)

AC

vp2(t)

wA(t)

m(t) TC

vp1(t)

TT AT

T(t) x B(t)

TA(t) Problem Data:

b(t)

wA( t ) = Kv1⋅ vp1 ( t )

wB ( t ) = Kv2⋅ vp2 ( t )

q ( t ) = q max⋅ m( t )

Analyzer AT measures solution conductivity:

C( t) =

β

mho

x B( t) m

Range is C L to C H.

(a) Model of the composition control loop Assumptions: • Perfectly mixed • Constant mass and density • Negligible transportation lag • Inlet streams are pure A and B, respectively Total mass balance:

d⋅ M wA( t ) + wB ( t ) − w( t ) = =0 dt

Mass balance on component B:

wB ( t ) − w( t ) ⋅ x B( t ) = M ⋅

w( t ) = wA( t ) + wB( t )

d ⋅ xB( t) dt

2 eqns. 2 unks. (w, xB)

(b) Linearize the equations, transfer functions and block diagram: Linearize and express in terms of deviation variables:

(

)

WB( t ) − w⋅ XB ( t ) − x B ⋅ WA( t ) + WB( t ) = M ⋅

d ⋅ XB( t ) dt

XB ( 0 ) = 0

xB 1 − xB M d ⋅ XB( t ) ⋅ + XB( t ) = WB( t ) − WA( t ) dt w w w

Rearrange in the standard first-order form:

Let

τ =

xB KA = w

M w

KB =

Laplace transform and solve for output:

Transmitter:

Linearize:

Control valves:

b( t) =

B( t) =

C( t) − CL CH − CL

1 − xB

τ⋅

w XB ( s) =

100%TO =

d ⋅ XB ( t ) dt

1

τ ⋅s + 1

100%TO

(KB⋅ WB(s) − KA⋅ WA(s))

 β

CH − CL  x B( t) 

100%TO  −β  XB ( t )  CH − CL  2  xB 

+ XB = KB ⋅ WB( t ) − KA⋅ WA( t )



− CL



H( s) = KAT =

−100 %TO⋅ β

( C H − C L) x B 2

Kv2 WB( s) = Kv2⋅ VP 2 ( s) = M ( s) 100%CO 2

WA( s) = Kv1⋅ VP 1 ( s)

Block diagram of the composition loop:

VP1(s)

Kv1KA X (s) set B

Ksp

+

E(s)

-

M2(s)

GAC(s)

Kv2KB 100

+

+

1 Js + 1

XB(s)

H(s)

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 = KT

(c) Model of the temperature control loop Assumptions: • perfectly mixed • constant mass and physical properties

(negative set point scale)

•

negligible heat losses and ransportantioon lag

Energy balance: M ⋅ cv ⋅

d ⋅ T( t )

(

Substitute

)

(

)

(

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

dt

Assume

w( t ) = wA( t ) + wB( t )

M ⋅ cp ⋅

d ⋅ T( t ) dt

(

1 eqn. 1 unk. (T)

cp = cpA = cpB = cv

)

(

)

)

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

(d) Linearize the equation, transfer functions, block diagram. Linearize and express in terms of deviation variables: M ⋅ cp ⋅

d ⋅ Γ ( t) dt

(

)

(

)

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

(

)

+ wB ⋅ cp ⋅ Γ B( t ) − wA + wB ⋅ cp ⋅ Γ ( t ) Rearrange into the standard first-oder equation form:

(

)

(

)

TB − T TA − T wA wB M d⋅ Γ ( t) 1 ⋅ + Γ ( t) = WA( t ) + WB( t ) + Q( t ) + Γ A( t ) + Γ ( t) dt w w w w⋅ cp w w B

Γ ( 0) = 0 Laplace transform: Γ ( s) = τ = Electric heater:

M w

Q( s) =

1

τ ⋅s + 1

(K1⋅ WA( s) + K2⋅ WB( s) + K3⋅ Q( s) + K4⋅ Γ A(s) + K5⋅ Γ B( s) )

K1 =

TA − T

q max 100%CO

w M ( s)

K2 =

TB − T w

1 K3 = w⋅ cp

Temperature transmitter TT:

wA K4 = w

wB K5 = w

100%TO H( s) = KTT = TH − .L

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

VP1(s)

Kv1K1 VP2(s)

Kv2K2

'set(s)

Ksp

E(s)

+

-

M(s)

GTC(s)

qmaxK3 100

K4 K5

+ + + + 1 + Js + 1

'A(s) 'B(s) '(s)

H(s) (e) Characteristic equations of the control loops. Analyzer control loop: Kv2 KB 1 + KAT⋅ GAC( s) ⋅ ⋅ =1+ 100%CO τ ⋅ s + 1

−100 %TO⋅ β

( C H − C L) x B

(

)

Kv2 1 − xB GAC( s) ⋅ =0 2 100%CO w( τ ⋅ s + 1 )

Temperature control loop: q max 1 + KTT⋅ GTC ( s) ⋅ ⋅ 100%CO

For both loops:

τ =

K3

τ ⋅s + 1

=1+

q max 1 GTC ( s) ⋅ ⋅ =0 TH − TL 100%CO w⋅ cp ⋅ ( τ ⋅ s + 1 ) 100%TO

M w

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-22. Control of reactors in series. Problem Data: V := 500gal

ρ A := 2

c A2set(t)

−1

k 1 := 0.25min

lbmole gal

MWA := 25

c A1(t)

−1

k 2 := 0.50min

lb

V

lbmole

DA

∆p v := 10psi

fi(t)

cAH := 0.5

V LT

c Ai(t)

Analyzer transmitter: cAL := 0.05

AT

fA(t)

Control valve, linear, sized for 100% overcapacity.

AC

LC

c A2(t)

lbmole gal

lbmole gal

τ T := 0.5min

Design conditions:

cAi := 0.8

lbmole gal

gal fi := 50 min

gal fA := 50 min

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

ρ A⋅ MWA⋅ gal

Cvmax := 200%⋅ fA⋅

8.33lb

G

∆p v

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

Valve gain:

∆p v Cvmax Kv := ⋅ G 100%CO

Cvmax = 77.475 Cvmax := 110

gal min⋅ psi gal

min⋅ psi

gal Kv = 1.42 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. Total mass balance:

ρ ⋅ f ( t ) = ρ A⋅ fA( t ) + ρ i ⋅ fi ( t )

ρA = ρi = ρ

1 eqn. 1 unk. (f)

Reactant balances: V⋅

d ⋅ cA1( t )

= ρ A⋅ fA( t ) + fi ( t ) ⋅ cAi( t ) − V⋅ k 1 ⋅ cA1( t ) − f ( t ) ⋅ cA1( t ) 2 eqns. 2 d ⋅ cA2( t ) unks. (c A1) V⋅ = f ( t ) ⋅ cA1( t ) − f ( t ) ⋅ cA2( t ) − V⋅ k 2 ⋅ cA2( t ) dt

Reactor 1: Reactor 2:

At the initial steady state:

dt

f := fA + fi

cA1 :=

cA2 :=

3 eqns. 3 unks. (cA2)

ρ A⋅ fA + fi ⋅ cAi f + V⋅ k 1 f ⋅ cA1 f + V⋅ k 2

f = 100 cA1 = 0.622

lbmole

gal min

cA2 = 0.178

gal

lbmole gal

(c) Linearize the model equations and obtain the block diagram. Linearize and express in terms of deviation variables: V⋅ V⋅

d ⋅ CA1( t ) dt d ⋅ CA2( t ) dt

(

)

(

= ρ A⋅ FA( t ) + cAi⋅ Fi ( t ) + fi ⋅ CAi( t ) − V⋅ k 1 + f C A1( t ) − cA1⋅ FA( t ) + Fi ( t )

(

)(

)

(

)

= cA1 − cA2 ⋅ FA( t ) + Fi ( t ) + f ⋅ CA1( t ) − f + V⋅ k 2 CA2( t )

Rearrange in the standard first-order form: d ⋅ C A1( t ) τ 1⋅ + CA1( t ) = K1 ⋅ FA( t ) + K2 ⋅ Fi ( t ) + K3 ⋅ CAi( t ) dt τ 2⋅

where

d ⋅ C A2( t )

τ 1 :=

τ 2 :=

dt

)

CA1( 0 ) = 0

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

V f + V⋅ k 1 V f + V⋅ k 2

τ 1 = 2.222 min

ρ A − cA1 K1 := f + V⋅ k 1

cAi − cA1 K2 := f + V⋅ k 1

cA1 − cA2 K4 := f + V⋅ k 2

f K5 := f + V⋅ k 2

K1 = 0.006123

lbmole⋅ min gal

2

τ 2 = 1.429 min Laplace transform: CA1( s) =

CA2( 0 ) = 0 fi K3 := f + V⋅ k 1

K2 = 0.00079

K4 = 0.00127 K1 ⋅ FA( s) + K2 ⋅ Fi ( s) + K3 ⋅ C Ai( s)

τ 1⋅ s + 1

lbmole⋅ min gal

2

lbmole⋅ min gal

2

K3 = 0.222

K5 = 0.286

CA2( s) =

K4 ⋅ FA( s) + K4 ⋅ Fi ( s) + K5 ⋅ C A1( s)

τ 2⋅ s + 1

Combine to obtain over-all transfer function: CA2( s) =

K4

τ2

(FA( s) + Fi( s) ) + ⋅s + 1

K5

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) (

K1 ⋅ FA( s) + K2 ⋅ Fi ( s) + K3 ⋅ C Ai( s)

)

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

where G1 ( s) =

(

)

K4 ⋅ τ 1 ⋅ s + 1 + K5 ⋅ K1

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

G2 ( s) =

(

)

K4 ⋅ τ 1 ⋅ s + 1 + K5 ⋅ K2

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

G3 ( s) =

K5 ⋅ K3

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

Block diagram:

CAi(s) Fi(s) CA2set(s) Ksp

+

E(s)

-

M(s) Gc(s)

Gv(s)

FA(s)

G3(s) G2(s) G1(s)

+ + +

CA2(s)

H(s)

Control valve: Gv ( s) = Kv

Transmitter: H( s) =

KT

τ T⋅ s + 1

100%TO %TO⋅ gal KT := KT = 222 lbmole cAH − cAL

(d) Closed-loop transfer functions. CA2( s) CA2

set

⋅ ( s)

=

Ksp := KT

Ksp ⋅ Gc( s) ⋅ Gv ( s) ⋅ G1 ( s)

CA2( s)

1 + H( s) ⋅ Gc( s) ⋅ Gv ( s) ⋅ G1 ( s)

Fi ( s) CA2( s) C Ai( s)

=

=

G2 ( s) 1 + H( s) ⋅ Gc( s) ⋅ Gv ( s) ⋅ G1 ( s) G3 ( s) 1 + H( s) ⋅ Gc( s) ⋅ Gv ( s) ⋅ G1 ( s)

(e) Ultimate gain and period of the loop. Proportional controller: Characteristic equation: Let

τ A := τ T⋅ τ 1 ⋅ τ 2

1+

KT

τ T⋅ s + 1

⋅ Kc⋅ Kv ⋅

(

)

K4 ⋅ τ 1 ⋅ s + 1 + K5 ⋅ K1

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

(

Kp := KT⋅ Kv ⋅ K4 + K5 ⋅ K1

τ C := τ 1 + τ 2 + τ T τ A = 1.587 min

)

Substitute s = ωu i at Kc = K cu 3

2

(

=0 3

τ B := τ T⋅ τ 1 + τ T⋅ τ 2 + τ 1 ⋅ τ 2

τ D := KT⋅ Kv ⋅ K4 ⋅ τ 1

Gc( s) = Kc

2

τ B = 5 min

τ C = 4.151 min

τ D = 0.89 min

Kp = 0.953

)

−τ A⋅ ω u i − τ B ⋅ ω u + τ C + τ D⋅ Kcu ω u ⋅ i + 1 + Kp ⋅ Kcu = 0 + 0 ⋅ i 2

Real part:

Imaginary part:

2

−τ B⋅ ω u + 1 + Kp ⋅ Kcu = 0

Kcu =

τ B⋅ ω u − 1 Kp

2  τ B⋅ ω u − 1   −τ A⋅ ω u +  τ C + τ D⋅ ωu = 0 Kp   3

ω u :=

τ D − Kp ⋅ τ C τ D⋅ τ B − τ A⋅ Kp −1

ω u = 1.021i min

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-23. Temperature control of two heaters in series.

T4set(t) TC

f1

Tc1(t)

T1(t) f2

T3(t)

T2(t)

f

Steam

Cond.

TT

T4(t)

f Model the process.

Assumptions: • Constant volumes in the tanks • Each tank is perfectly mixed • Negligible heat losses • Heating coil temperature T c1 is uniform (high flow of heating fluid) • •

Constant and uniform densities and specific heats Negligible transportation lag.

Energy balances, Tank 1: V1 ⋅ ρ ⋅ cv ⋅

d ⋅ T3 ( t ) dt

)

(

−f ⋅ ρ ⋅ cp ⋅ T3 ( t ) − Tref

Tank 2: V2 ⋅ ρ ⋅ cp ⋅

(

d ⋅ T4 ( t ) dt

(

)

(

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

(

)

1 eqn. 2 unks. (f, T 3 )

)

(

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

Total mass balance on tank 1:

Substitute and simplify:

)

ρ ⋅ f1 + ρ ⋅ f2 = ρ ⋅ f

f = f1 + f2

)

2 eqns. 3 unks. (T 4 ) 3 eqns. 3 unks.

V1 ⋅ ρ ⋅ cv ⋅ V2 ⋅ ρ ⋅ cv ⋅

d ⋅ T3 ( t )

(

)

(

)

(

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

dt d ⋅ T4 ( t )

(

)

)

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

dt

Linearize and express in terms of the deviation variables: d ⋅ Γ 3 ( t) V1 ⋅ ρ ⋅ cv ⋅ = f1 ⋅ ρ ⋅ cp ⋅ Γ 1 ( t ) + f2 ⋅ ρ ⋅ cp ⋅ Γ 2 ( t ) + U⋅ A⋅ Γ c1( t ) − f ⋅ ρ ⋅ cp + U⋅ A Γ 3 ( t ) dt

(

V2 ⋅ ρ ⋅ cv ⋅

d ⋅ Γ 4 ( t) dt

)

= f ⋅ ρ ⋅ cp ⋅ Γ 3 ( t ) + λ ⋅ W( s) − f ⋅ ρ ⋅ cp ⋅ Γ 4 ( s)

Rearrange into first-order standard form: τ 1⋅ τ 2⋅ where

d⋅ Γ 3( t) dt d⋅ Γ 4( t) dt

τ1 =

τ2 =

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

Γ 3(0) = 0

+ Γ 4 ( t ) = Γ 3 ( t ) + K4 ⋅ W( s)

Γ 4(0) = 0

V1 ⋅ ρ ⋅ cv f ⋅ ρ ⋅ cp + U⋅ A V2 ⋅ ρ ⋅ cv

K1 :=

K4 =

f ⋅ ρ ⋅ cp

f1 ⋅ ρ ⋅ cp f ⋅ ρ ⋅ cp + U⋅ A

K2 =

f2 ⋅ ρ ⋅ cp f ⋅ ρ ⋅ cp + U⋅ A

K3 :=

U⋅ A f ⋅ ρ ⋅ cp + U⋅ A

λ f ⋅ ρ ⋅ cp

Laplace transform and solve for the outputs: Γ 3 ( s) =

K1 ⋅ Γ 1 ( S) + K2 ⋅ Γ 2 ( s) + K3 ⋅ Γ c1( s)

τ 1⋅ s + 1

Γ 4 ( s) =

Γ 3 ( s) + K4 ⋅ W( s) τ 2⋅ s + 1

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

G4 ( s) =

K1

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

G2 ( s) =

K2

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

K4

τ 2⋅ s + 1

Control valve:

Gv ( s) =

Kv

τ v⋅ s + 1

G3 ( s) =

K3

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

Temperature transmitter TT:

H( s) =

KT

τ T⋅ s + 1

100%TO KT = TH − TL

Ksp = KT

Block Diagram:

'c1(s) G (s) 3 '2(s) '1(s) '4set(s)

+

Ksp

E(s)

-

G1(s)

W(s)

M(s) Gc(s)

G2(s)

Gv(s)

G4(s)

++ + '4(s) +

H(s)

Characteristic equation of the loop. 1 + H( s) ⋅ Gc( s) ⋅ Gv ( s) ⋅ G4 ( s) = 1 + 3

(

KT

τ T⋅ s + 1

)

2

Gc( s)

(

Kv

K4

⋅

τ v⋅ s + 1 τ 2⋅ s + 1

=0

)

τ T⋅ τ v ⋅ τ 2 ⋅ s + τ T⋅ τ v + τ T⋅ τ 2 + τ v ⋅ τ 2 ⋅ s + τ T + τ v + τ 2 ⋅ s + 1 + KT⋅ Kv ⋅ K4 ⋅ Gc( s) = 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-24. Temperature and level control of oil heater.

T3set(t) TC N2

TT p2 = 40 psia 3 ft

Steam

fs(t)

AO

hset(t)

LT

LC

AO

h(t)

T1(t) p1(t)

5 ft

vp2(t) T3(t) p3(t)

T Condensate Problem Data:

ρ := 53

lb ft

3

cp := 0.45

BTU

T1 := 70degF

lb⋅ degF

Steam, saturated at 115 psig (130 psia). from steam tables: p 2 := 40psia

Heating coil:

p 1 := 45psig

Do := 0.5in Ao := π ⋅ Do ⋅ L

p 3 := 15psig

(20 BWG)

U := 136

Ts1 := 347degF BTU 2

hr⋅ ft degF lb wm := 0.178 ft

L := 974ft

Ao = 127.5 ft

T3 := 200degF D := 3ft

2

CM := wm⋅ L⋅ cpm

Level transmitter:

h L := 7ft

h H := 10ft

Temperature transmitter:

TL := 100degF TH := 300degF

λ s := 873

BTU

gal f1 := 100 min cpm := 0.12

BTU lb⋅ degF

CM = 20.805

τ LT := 0.01min τ TT := 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.

BTU degF

lb

At the initial steady state:

ws :=

(

f3 := f1

(

f1 ⋅ ρ ⋅ cp ⋅ T3 − T1

)

λs

(

)

−f1 ⋅ ρ ⋅ cp ⋅ T3 − T1 + ws⋅ λ s = 0 lb ws = 47.5 min

ws⋅ λ s Ts := T3 + U⋅ Ao

)

ws⋅ λ s − U⋅ Ao ⋅ Ts − T3 = 0

Ts = 343.4 degF

Size inlet oil valve: Assume initially the level is at 50% of the level transmitter range: ft  ∆p v1 := p 1 + 14.7psia − p 2 − ρ ⋅ g ⋅ ( h − 5ft) ⋅   12in  3

ρ ⋅ ft Gf := 62.4lb

h :=

2

From Fig. C-10.1, page 532, a 3-in valve is required. Size exit oil valve:



(

)

Gf

Cv2max := 150%⋅ f3 ⋅

From Fig. C-10.1, page 532, a 2-in valve is required. Size steam valve:

p s1 := 130psia y :=

Cvsmax :=

1.63 Cf

⋅

G :=

p s1 − p s

(

C f ⋅ p s1 ⋅ y − 0.148 ⋅ y

)

3

)

Assume

G = 0.621

gal⋅ hr 836 ⋅ min⋅ scf

min⋅ psi gal

min⋅ psi

gal min⋅ psi

Cv2max := 46

gal min⋅ psi

Cf := 0.9

ws 60min scf fs := 380scf ⋅ fs = 60138 18lb hr hr

y = 0.42

p s1

150 ⋅ %⋅ fs⋅ G⋅ Ts1 + 460 ⋅ R

(

29

gal

∆p v2 = 13.43 psi Cv2max = 37.7

∆p v2

18

∆p v1 = 18.41 psi

Cv1max := 46

ft  ∆p v2 := p 2 + ρ ⋅ g ⋅ h ⋅  − p 3 + 14.7psia 12in



2

Cv1max = 32.2

∆p v1

2

(steam tables) hH + hL

h = 8.5 ft

Gf

Cv1max := 150%⋅ f1 ⋅

p s := 123psia

⋅

psia R

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

Cvsmax = 50.4

Cvsmax := 110

gal min⋅ psi gal

min⋅ psi

(b) Block diagram of the level control loop. 2

Mass balance:

Valves:

π⋅D 4

⋅

7.48gal ft

3

⋅ρ⋅

d⋅ h ( t) dt

= ρ ⋅ f1 ( t ) − ρ ⋅ f3 ( t )

1 eqn. 3 unks. (h, f1 , f3 )

1  ft  f1 ( t ) = C v1max⋅ vp1 ( t ) ⋅ ⋅  p 1 ( t ) − p 2 − ρ ⋅ g ⋅ ( h ( t ) − 5 ⋅ ft) ⋅  Gf   12⋅ in 

2

 

2 eqns. 3 unks. 1  ft  f3 ( t ) = C v2max⋅ vp2 ( t ) ⋅ ⋅  p − p 3 ( t ) + ρ ⋅ g ⋅ h ( t ) ⋅  Gf  2 12  ⋅ in 

2

 

3 eqns. 3 unks. ∆p v1

f1max := Cv1max⋅

f3max := Cv2max⋅

Gf

∆p v2 Gf

gal gal f1max = 214.2 f3max = 182.9 min min

Linearize the model equations and express in terms of deviation variables: 2

π⋅D

⋅

4

where

7.48gal d ⋅ H( t ) ⋅ = f1max⋅ VP 1 ( t ) − f3max⋅ VP 2 ( t ) + a1 ⋅ P1 ( t ) − a3 ⋅ H( t ) + a2 ⋅ P3 ( t ) − a4 ⋅ H( t ) 3 dt ft

a1 =

δf1 ( t )

a2 =

δp 1

f1 vp1 := f1max

−δf3 ( t ) δp 3

f3 vp2 := f3max

1 a1 := Cv1max⋅ vp1 ⋅ ⋅ 2 1 a3 := −C v1max⋅ vp1 ⋅ ⋅ 2

a3 =

a1 = 2.716

ft

 

144 ⋅ in

2 2

gal

1 Gf ⋅ ∆p v2

⋅ ( −1 )

1 a4 := Cv2max⋅ vp2 ⋅ ⋅ 2



a2 = 3.723

min⋅ psi

δh

vp2 = 0.547





⋅  −ρ ⋅ g ⋅

δf3 ( t )

a4 =

1 a2 := −C v2max⋅ vp2 ⋅ ⋅ 2

Gf ⋅ ∆p v1

Gf ⋅ ∆p v1

δh

vp1 = 0.467

1

1

−δf1 ( t )

gal min⋅ psi

a3 = 1

1 Gf ⋅ ∆p v2 gal min⋅ ft





ft

 

144 ⋅ in

⋅  ρ ⋅ g⋅

a4 = 1.37

2 2



gal min⋅ ft

Rearrange model equation in standrd first-oder form: τ⋅

d ⋅ H( t )

+ H( t ) = K1 ⋅ VP 1 ( t ) − K2 ⋅ VP 2 ( t ) + K3 ⋅ P1 ( t ) + K4 ⋅ P3 ( t )

dt

H( 0 ) = 0

where 2

τ :=

πD 4

⋅

7.48gal

1

3

a3 + a4

ft

f1max K1 := a3 + a4

f3max K2 := a3 + a4

τ = 22.31 min K1 = 90.4 ft Laplace transform: H( s) =

1

τ ⋅s + 1

K2 = 77.2 ft

a1 K3 := a3 + a4

a2 K4 := a3 + a4

ft K3 = 1.146 psi

ft K4 = 1.571 psi

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

Level control valve:

VP 1 ( s) =

Level transmitter LT:

HLT( s) =

Proportional controller, LC:

1 100%CO

M 1 ( s)

KLT

100%TO KLT := hH − hL τ LT⋅ s + 1

KLT = 33.33

%TO ft

The controller is reverse acting: increasing level decreases the controller output closing the valve and decreasing the inlet flow.

GLc( s) = KLc

Block diagram of the level control loop:

P3(s)

K4

P1(s)

K3

VP2(s) R(s)

+

E(s) KLc

-

M1(s)

K2 K1 100

++ 1 + Js + 1

H(s)

HLT(s) (c) Block diagram and characteristic equation of the temperature control loop. Energy balance on tank: 2

πD 4

⋅

7.48gal ft

3

ρ ⋅ cv ⋅

d dt

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

−f3 ( t ) ⋅ ρ ⋅ cp ⋅ T3 ( t ) − Tref Energy balance on steam chest:

CM ⋅

d ⋅ Ts( t ) dt

)

1 eqn. 2 unks. (T 3 , Ts)

(

= λ s⋅ ws( t ) − U⋅ Ao ⋅ Ts( t ) − T3 ( t )

)

2 eqns. 3 unks. (w s) 2 eqns. 3 nks. (ws)

Steam valve: 836 ⋅ scf ⋅ min

ws( t ) =

hr⋅ gal

R

⋅

hr

⋅

18⋅ lb

⋅

psia 60⋅ min 380 ⋅ scf

⋅

(

C vsmax⋅ vps( t ) ⋅ Cf ⋅ p s1 ⋅ y − 0.148 ⋅ y

(

G⋅ Ts1 + 460R

)

3

)

3 eqns. 3 unks.

Let

836 ⋅ scf ⋅ min

wsmax :=

hr⋅ gal

R

⋅

hr

⋅

⋅

18⋅ lb

psia 60⋅ min 380 ⋅ scf

⋅

(

Cvsmax⋅ C f ⋅ p s1 ⋅ y − 0.148 ⋅ y

(

G⋅ Ts1 + 460R

)

3

)

lb wsmax = 155.3 min

Substitute mass balance into energy balance and simplify: 2 d ⋅ T3 ( t ) π ⋅ D 7.48⋅ gal ⋅ ⋅ ρ ⋅ cv ⋅ h ( t ) ⋅ = f1 ( t ) ρ ⋅ cp ⋅ T1 ( t ) − T3 ( t ) + U⋅ Ao ⋅ Ts( t ) − T3 ( t ) 4 3 dt ft

(

)

(

)

ws( t ) = wsmax⋅ vps( t ) 2

Linearize the equations:

V⋅ ρ ⋅ cv ⋅ CM ⋅

d⋅ Γ 3( t) dt

d ⋅ Γ s( t ) dt

Let

(

V :=

πD 4

⋅

7.48gal ft

3

⋅h

V = 449.4 gal

(

)

)

Assume

cv := cp

(

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

(

= λ s⋅ Ws( t ) − U⋅ Ao ⋅ Γ s( t ) − Γ 3 ( t )

Γ 3(0) = 0

)

Γ s( 0 ) = 0

Rearrange into standard first-order form: τ 1⋅ τ 2⋅

d⋅ Γ 3( t) dt d ⋅ Γ s( t ) dt

where τ 1 :=

τ 2 :=

+ Γ 3 ( t ) = −K1 ⋅ F1 ( t ) + K2 ⋅ Γ 1 ( t ) + K3 ⋅ Γ s( t ) + Γ s( t ) = K4 ⋅ Ws( t ) + Γ 3 ( t ) V⋅ ρ ⋅ cv

f1 ⋅ ρ ⋅ cp + U⋅ Ao CM U⋅ Ao

K1 :=

λs K4 := U⋅ Ao

(

−ρ ⋅ cp ⋅ T1 − T3

)

f1 ⋅ ρ ⋅ cp + U⋅ Ao

K2 :=

τ 1 = 2.357 min K2 = 0.525

Laplace transform:

Γ 3 ( s) =

f ⋅ ρ ⋅ cp f ⋅ ρ ⋅ cp + U⋅ Ao

U⋅ Ao f ⋅ ρ ⋅ cp + U⋅ Ao

τ 2 = 0.072 min

K1 = 0.682

K3 = 0.475

K4 = 3.021

−K1 ⋅ F1 ( s) + K2 ⋅ Γ 1 ( s) + K3 ⋅ Γ s( s) τ 1⋅ s + 1

K3 :=

Γ s( s) =

degF⋅ min gal degF⋅ min lb

K4 ⋅ Ws( s) + Γ 3 ( s)

τ 2⋅ s + 1

Combine:

where

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

( ) (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K3

G3 ( s) =

K1 ⋅ τ 2 ⋅ s + 1

G2 ( s) =

K3 ⋅ K4

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K3 wsmax Ws( s) = M ( s) 100%CO s

Steam control valve: Temperature transmitter TT:

HTT( s) =

KTT



wsmax 100%CO 100%TO KTT := TH − TL

τ TT⋅ s + 1

GTc( s) = KTc⋅  1 +

PID controller:

Inlet flow:

( ) (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K3 K2 ⋅ τ 2 ⋅ s + 1

= 1.553

lb min⋅ %CO

%TO KTT = 0.5 degF

s   τ DT⋅ + 1  τ IT⋅ s  α D⋅ τ DT⋅ s + 1   1

f1max F1 ( s) = M ( s) + a1 ⋅ P1 ( s) − a3 ⋅ H( s) 100%CO 1

f1max 100%CO

= 2.142

a1 = 2.716

gal min⋅ %CO

gal

a3 = 1

min⋅ psi

gal min⋅ ft

Block diagram of the temperature control loop:

M1(s) P1(s) H(s)

'3set(s) Ksp

+

E(s)

-

f1max 100 a1 a3

+ +

F1(s)

- '1(s)

G2(s)

Ms(s) w Ws(s) smax GTc(s) G3(s) 100 HTT(s)

Characteristic equation:

G1(s)

+ +

-

'3(s)

wsmax 1 + HTT( s) GTc( s) ⋅ G ( s) = 1 + 100%CO 3

KTT

τ TT⋅ s + 1

wsmax GTc( s) ⋅ 100%CO

K3 ⋅ K4

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K3

=0

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

GTc( s) = KTc

Substitute s = ωu i at KTc = K cu : where

τ A := τ TT⋅ τ 1 ⋅ τ 2

3

τ B := τ TT⋅ τ 1 + τ TT⋅ τ 2 + τ 1 ⋅ τ 2

wsmax Kp := KTT⋅ K ⋅K 100%CO 3 4 3

3

τ A = 0.085 min

Imaginary part: −τ A⋅ ω u + τ C ⋅ ω u = 0

Real part:

2

−τ A⋅ ω u ⋅ i − τ B ⋅ ω u + τ C⋅ ω u ⋅ i + 1 − K3 + Kp ⋅ Kcu = 0 + 0 ⋅ i

ω u :=

τC τA

2

−τ B⋅ ω u + 1 − K3 + Kp ⋅ Kcu = 0

Kcu :=

(

)

τ C := τ TT⋅ 1 − K3 + τ 1 + τ 2 2

τ B = 1.384 min

Tu :=

τ C = 2.692 min %TO Kp = 1.116 %CO

2π

Tu = 1.069 min

ωu

K3 − 1 + τ B⋅ ω u

2

Kp

%CO Kcu = 38.9 %TO

Compare these results with the simulation of this process in Problem 13-24. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-25. Level control of two tanks in series.

LC

.fi(t)

.f3(t) h1(t)

f1(t)

fo(t)

LT

h2(t)

.m 1(t)

f2(t)

fmax 1 Fo ( s) = M ( s) 100%CO τ p ⋅ s + 1 1

Linear pump with a time constant:

Kv

Linear control valve with constant pressure drop and time constant:

Gv ( s) =

Level transmitter LT has negligible time constant:

HLT( s) = KT =

Kv = Cv3⋅

τ v⋅ s + 1

∆p v3 Gf

100%TO h max

(a) Block diagram and transfer functions of level control loop. Assume • constant and uniform densities • constant valve positions and inlet valve pressures Mass balance, tank 1:

tank 2:

π ⋅ D1

2

⋅ρ⋅

4

π ⋅ D2 4

2

⋅ρ⋅

d⋅ h 1( t) dt d⋅ h 2( t) dt

= ρ ⋅ fi ( t ) + ρ ⋅ f3 ( t ) − ρ ⋅ fo ( t ) − ρ ⋅ f1 ( t ) 1 eqn. 2 unks. (h1 , f1 ) = ρ ⋅ f1 ( t ) − ρ ⋅ f2 ( t )

2 eqns. 4 unks. (h 2 , f2 )

2

Valves:

ft  f1 ( t ) = C v1⋅ ρ w⋅ g ⋅ h 1 ( t ) ⋅  = k v1⋅ h 1 ( t ) 12in





3 eqns. 4 unks.

2

ft  f2 ( t ) = C v2⋅ ρ w⋅ g ⋅ h 2 ( t ) ⋅  = k v2⋅ h 2 ( t ) 12in



4 eqns. 4 unks.



Linearize the equations and express in terms of deviation variables:: π ⋅ D1

2

4

π ⋅ D2

d ⋅ H1 ( t )

⋅

dt

2

4

d ⋅ H2 ( t )

⋅

dt

= Fi ( t ) + F3 ( t ) − Fo ( t ) −

=

k v1 2⋅ h1

H1 ( t ) −

k v2 2⋅ h2

k v1 2⋅ h1

H1 ( t )

H1 ( 0 ) = 0

H2 ( 0 ) = 0

H2 ( t )

Rearrangle into the standard first-order form: τ 1⋅

τ 2⋅

d ⋅ H1 ( t ) dt d ⋅ H2 ( t ) dt

2

where

τ1 =

π ⋅ D1 ⋅ h 1

K1 =

2 ⋅ k v1

Laplace transform:

H1 ( s) =

H2 ( s) =

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

+ H2 ( t ) = K2 ⋅ H1 ( t )

2⋅ h1 k v1 K1

τ 1⋅ s + 1 K2

τ 2⋅ s + 1

2

π ⋅ D2 ⋅ h 2

τ2 =

2 ⋅ k v2

k v1 h 2 K2 = ⋅ k v2 h 1

(Fi( s) + F3(s) − Fo( s) ) H1 ( s) =

K1 ⋅ K2

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) (

Fi ( s) + F3 ( s) − Fo ( s)

)

Block diagram of the level control loop:

M1(s) Fi(s) R(s)

+

E(s)

-

KLc

.fmax 100(Jps+1)

M1(s)

Gv(s)

Fo(s)

+ K1K2 + (J1s+1)(J2s+1)

F3(s) KT

H2(s)

(b) Characteristic equation and ultimate gain and period. 1 + KT⋅ KLC ⋅

Kv

(

)(

τ v⋅ s + 1 τ 1⋅ s + 1 ⋅ τ 2⋅ s + 1

Substitute s = ωu i at KLC = K.cu : 3

K1 ⋅ K2

Let

(

)

)

=0

Kp = KT⋅ Kv ⋅ K1 ⋅ K2 =

(

2

100 ⋅ %TO f3max 2 ⋅ h 2 ⋅ ⋅ h max 100 ⋅ CO k v2

)

−τ v ⋅ τ 1 ⋅ τ 2 ⋅ ω u i − τ v ⋅ τ 1 + τ v ⋅ τ 2 + τ 1 ⋅ τ 2 ω u + τ v + τ 1 + τ 2 ⋅ ω u ⋅ i + 1 + Kp ⋅ Kcu = 0 + 0 ⋅ i Imaginary part:

(

(

3

τv + τ1 + τ2

)

−τ v ⋅ τ 1 ⋅ τ 2 ⋅ ω u + τ v + τ 1 + τ 2 ω u = 0 ω u =

)

τ v⋅ τ 1⋅ τ 2

(

Tu =

)

2π

ωu 2

2 τ v⋅ τ 1 + τ v⋅ τ 2 + τ 1⋅ τ 2 ω u − 1 Real part: − τ v ⋅ τ 1 + τ v ⋅ τ 2 + τ 1 ⋅ τ 2 ω u + 1 + Kp ⋅ Kcu = 0 Kcu = Kp Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 6-26. Control of heating value of fuel stream. Problem Data:

hvset(t)

hv( t ) = c + g ⋅ x 3 ( t )

AC

G( t ) = a + b ⋅ x 3 ( t )

hv AT

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

Natural gas

Sensor transmitter, lag τT, range hvL to hvH.

Gc( s) = Kc⋅  1 +



1



τ I⋅ s



f3(t) x 3(t)

f2(t) x 2(t)

Controller is PI:

p2

p(t)

Waste gas

f1(t) x 1(t)

(a) Block diagram and transfer functions. Assume • Gas obeys ideal gas law • Temperature is constant • Perfectly mixed tank Total mole balance: Methane mole balance:

V⋅ V⋅

d ⋅ ρ ( t) dt d dt

= ρ s⋅ f1 ( t ) + ρ s⋅ f2 ( t ) − ρ s⋅ f3 ( t )

1 eqn. 2 unks. (ρ, f3 )

(ρ (t) ⋅ x3(t) ) = ρ s⋅ f1( t)⋅ x1( t) + ρ s⋅ f2( t)⋅ x2( t) − ρ s⋅ f3( t)⋅ x3( t) 2 eqns. 3 unks. (x3 ) 836scf ⋅ min

(

)

R hr p ( t) 3 C ⋅C ⋅ y ( t ) − 0.148 ⋅ y ( t ) psia 60min v3 f G( t ) ⋅ T 3 eqns. 5 unks. (y, p) 1.63 p ( t ) − p 2 y( t) = ⋅ 4 eqns. 5 unks. p( t) Cf

Exit valve:

f3 ( t ) =

Ideal gas law:

ρ ( t) =

hr⋅ gal

⋅

p ( t ) lbmole Rg ⋅ T

ft

5 eqns. 5 unks.

3

ρ s :=

14.7psia Rg ⋅ 520 R

− 3 lbmole

ρ s = 2.635 × 10

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

scf

V Rg ⋅ T

⋅

d ⋅ P( t ) dt

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

V⋅ x 3 d ⋅ P( t ) V⋅ P d ⋅ x 3 ( t ) ⋅ + ⋅ = ρ s⋅ f1 ⋅ X1 ( t ) + f2 ⋅ X2 ( t ) − f3 ⋅ X3 ( t ) + x 1 ⋅ F1 ( t ) + x 2 ⋅ F2 ( t ) − x 3 ⋅ F3 ( t ) Rg ⋅ T dt Rg ⋅ T dt δ ⋅ f3 ( t ) −δ ⋅ f3 ( t ) where F3 ( t ) = a1 ⋅ P( t ) − a2 ⋅ X3 ( t ) a1 = a2 = δp δx 3

(

)

Combine total and methane balances, and substitute valve equation: V⋅ ρ d ⋅ x 3 ( t ) ⋅ = f1 ⋅ X1 ( t ) + f2 ⋅ X2 ( t ) − f3 ⋅ X3 ( t ) + x 1 − x 3 F1 ( t ) + x 2 − x 3 ⋅ F2 ( t ) dt ρs

(

V R g ⋅ T⋅ ρ s

a1 =

a2 =

836scf ⋅ min hr⋅ gal

⋅

d ⋅ P( t ) dt

C v3⋅ C f 

hr

hr⋅ gal

R

⋅

)

(

hr

psia 60min

3

p

Cv3⋅ Cf ⋅

T

(y − 0.148 ⋅y3)

x3( 0) = 0

P( 0 ) = 0

)

 p − p2  y − 0.148 ⋅ y + p ⋅ 1 − 3 ⋅ 0.148 ⋅ y  Cf ⋅ 2  p  G⋅ T  

psia 60min

−836 scf ⋅ min

(

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

R

⋅

)

2 1.63

− 0.5

p 2  p

2



−b

(

2 a + b⋅ x3

)

1.5

Rearrange into standrad first-order form: τ 1⋅ τ 2⋅

d ⋅ X3 ( t ) dt d ⋅ P( t )

τ1 =

dt

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

+ P( t ) =

V⋅ ρ f3 ⋅ ρ s

Laplace transform:

1 a1

⋅ F1 ( t ) +

f1 K1 = f3

1

τ 1⋅ s + 1

a2 a1

K3 =

X3 ( t ) x3 − x1 f3

K4 =

x2 − x3 f3

τ2 =

V Rg ⋅ T⋅ ρ s⋅ a1

(K1⋅ X1(s) + K2⋅ X2( s) − K3⋅ F1(s) + K4⋅ F2( s) )

1 1  F1 ( s) + a F2 ( s) + τ 2 ⋅ s + 1  a1 1 1

Linearize the eating value equation: Variable speed fan:

a1

⋅ F2 ( t ) +

f2 K2 = f3

X3 ( s) =

P( s) =

1

HV( t ) = g ⋅ X3 ( t ) F2 ( s) =

KF

τ F⋅ s + 1

M ( s)

a2



X ( s) a3 3  HV( s) = g ⋅ X3 ( s) f2max KF = 100%CO

Heating value transmitter:

H( s) =

KT

100%TO KT = hvH − hvL

τ T⋅ s + 1

Ksp = KT

Block diagram of the heating value control

X1(s)

K1

X2(s)

K2

F1(s) HVset(s)

Ksp

+

E(s)

Gc(s)

-

K3

M(s) K K F 4 JFs + 1

+ + +

g J1s + 1

HV(s)

H(s)

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. 1+

KT

τ T⋅ s + 1

Gc( s) ⋅

KF

K4

τ F⋅ s + 1 τ 1 ⋅ s + 1

=0

(τ T⋅ s + 1)⋅ (τ F⋅ s + 1)⋅ (τ 1⋅ s + 1) + KT⋅ Kc⋅  1 + τ ⋅ s  ⋅ KF⋅ K4 = 0 1



I



Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

%TO := %

%CO := %

Kd := K

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

U(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

G1 ( s) =

K := 2.5

%TO

τ 1 := 5min

%CO

K

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)⋅ (τ 3⋅ s + 1) τ 2 := 0.8min 1 + Gc( s) ⋅

Characteristic equation of the loop:

Ultimate gain and period:

C(s)

Gc( s) = Kcu 3

τ 3 := 0.2min K

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)⋅ (τ 3⋅ s + 1)

=0

s = ω u⋅ i 2

−τ A⋅ ω u i − τ B⋅ ω u + τ C⋅ ω u ⋅ i + 1 + KKcu = 0 + 0 ⋅ i where

τ A := τ 1 ⋅ τ 2 ⋅ τ 3

τ B := τ 1 ⋅ τ 2 + τ 1 ⋅ τ 3 + τ 2 ⋅ τ 3

τ C := τ 1 + τ 2 + τ 3 2

τ B = 5.16 min Imaginary part:

3

−τ A⋅ ω u + τ C⋅ ω u = 0 2

−τ B⋅ ω u + 1 + KKcu = 0

Real part:

ω u :=

τC τA

Tu :=

2π ωu

2 1 Kcu :=  τ B⋅ ω u − 1  K

3

τ A = 0.8 min τ C = 6 min

Tu = 2.29 min %CO Kcu = 15.1 %TO

(a) Quarter decay ratio tuning parameters for a proportional controller. From Table 7-1.1:

Gc( s) = Kc

Kc :=

Kcu

%CO Kc = 7.5 %TO

2

(b) Quarter decay ratio tuning parameters for a proportional-integral controller. Gc( s) = Kc⋅  1 +



1



τ I⋅ s

 From Table 7-1.1:

Kcu Kc := 2.2

τ I :=

Tu 1.2

%CO Kc = 6.9 %TO

τ I = 1.9 min

(c) Quarter decay ratio tuning parameters for a series PID controller. Gc( s) = Kc⋅  1 +



 τ ⋅s + 1 (D ) τ I⋅ s  1

Kcu Kc := 1.7

From Table 7-1.1:

τ I :=

Tu 2

τ D :=

%CO Kc = 8.9 %TO Tu 8

τ I = 1.1 min τ D = 0.29 min

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Smith & Corripio, 3rd edition Problem 7-2. Feedback control loop of Problem 7-1. From the solution to Problem 7-1, the characteristic equation is: 3

2

τ A⋅ s + τ B⋅ s + τ C⋅ s + 1 + Gc( s) ⋅ K = 0 with

3

τ A = 0.8 min

2

τ B = 5.16 min

τ C = 6 min

K = 2.5

%TO %CO

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

Find the roots:

  1 + K⋅ Kc      τ C⋅ min− 1    −5.8883       polyroots = −0.2808 − 2.0282i   τ ⋅ min− 2    B     −0.2808 + 2.0282i    τ ⋅ min− 3  A  −ζ

Damping ratio:

1−ζ Decay ratio:

%CO Kc := 7.5 %TO

Gc( s) = Kc

e

2

=

2.0282 −1

⋅T

T = 3.098 min 0.2808

ζ :=

Roots of:

4

2

0.2808 + 2.0282

= 0.419

Gc( s) = Kc⋅  1 +

−

2.0282min

ζ = 0.137

The decay ratio is higher than one fourth. b) PI Controller

2π

Period: T :=

(2.0282⋅ ζ ) 2 = 0.28082(1 − ζ 2)

−0.2808

− 0.2808min

Dominant roots are the complex conjugate pair.

1





τ I⋅ s



3

2

%CO Kc := 6.9 %TO

(

)

τ A⋅ s + τ B⋅ s + τ C⋅ s + 1 + K⋅ Kc ⋅ s +

K⋅ Kc τI

τ I := 1.9min

=0

2

  K⋅ Kc  ⋅ min  τ I    −5.776    1 + K⋅ Kc    −0.555  − 1  =  polyroots  τ min ⋅  C    −0.06 + 1.881i      −0.06 − 1.881i    τ B⋅ min− 2        τ A⋅ min− 3   0.06

Damping ratio:

2

0.06 + 1.881

2

The dominant roots are the complex conjugate roots: Period:

−1

1.8814min T = 3.34 min

Decay ratio:

= 0.032

2⋅ π

T :=

e

− 0.06min

−1

⋅T

= 0.818

This is a very undamped response; extremely high decay ratio. Gc( s) = Kc⋅  1 +

c) Series PID controller:



Roots of:

(

%CO Kc := 8.9 %TO

 τ ⋅s + 1 (D ) τ I⋅ s  1

)

τ I := 1.1min τ D = 0.29 min









τ A⋅ s + τ B⋅ s + τ C + K⋅ Kc⋅ τ D ⋅ s +  1 + K⋅ Kc⋅  1 +   4

3

2

K⋅ Kc   min   τI    τ    −4.531    1 + K⋅ Kc⋅  1 + D     τ I   −0.952   polyroots    =  −0.483 − 2.372i  − 1  ( τ C + K⋅ Kc⋅ τ D) min        −0.483 + 2.372i  −2   τ B⋅ min    −3  τ A⋅ min   Damping ratio:

0.483 2

0.483 + 2.372

2

= 0.2

Decay ratio:

τ D 

K⋅ K

c  ⋅s + =0 τI  τI 

The dominant roots are the complex conjugate roots. Period:

2⋅ π

T :=

−1

2.372min T = 2.649 min

e

− 0.483min

−1

⋅T

= 0.278

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

(b) Simulate tye loop and plot responses to a unit step change in set point. 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, G c(s), is obtained from

• 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

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Smith & Corripio, 3rd edition Problem 7-3. Feedback control of a second-order plus dead-time process.

U(s) R(s) + E(s)

Gc(s)

G2(s)

M(s)

+

G1(s)

+

-

G1 ( s) =

(

K⋅ e

− t0⋅ s

)(

τ 1⋅ s + 1 ⋅ τ 2⋅ s + 1

K := 1.25

)

%TO %CO

τ 1 := 1min τ 2 := 0.6min τ2

First -order plus dead-time parameters from Fig. 7-2.8:

τ1

= 0.6

(the dead-time equivalent is added to the actual dead time) PI controller:

Gc( s) = Kc⋅  1 +



1

t0e := t´0 + t0

τ I⋅ s   (a) PI controller tuned for quarter-decay ratio response From Table 7-2.1:

0.9 − 1 Kc := P K u

C(s)

t0 := 0.20min

τ´ := 1.32⋅ τ 1 τ´ = 1.32 min t´0 := 0.39⋅ τ 1 t´0 = 0.39 min Pu :=

t0e τ´

Pu = 0.447

%CO Kc = 1.6 %TO

τ I := 3.33⋅ t0e

τ I = 2 min

(b) PI controller tuned for minimum IAE on disturbance inputs. From Table 7-2.2:

Kc :=

0.984

Pu K

− 0.986

τ I :=

τ´

P 0.608 u

0.707

%CO Kc = 1.7 %TO

τ I = 1.2 min

(c) PI controller tuned for minimum IAE on set-point inputs. From Table 7-2.3:

Kc :=

0.758 K

Pu

− 0.861

τ I :=

τ´ 1.02 − 0.323 ⋅ Pu

%CO Kc = 1.2 %TO

τ I = 1.5 min

(d) PI controller tuned by controller syntesis for 5% overshoot on a set-point chang From Table 7-4.1:

0.5 − 1 Kc := P K u

τ I := τ´

%CO Kc = 0.89 %TO

τ I = 1.3 min

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Smith & Corripio, 3rd edition Problem 7-4. Process of Problem 7-3 with PID controller. τ´ := 1.32min

From the solution to Problem 7-3: Series PID:

Gc( s) = K´c⋅  1 +



t0e := ( 0.39 + 0.2)min Pu :=

 ⋅ τ´ ⋅ s + 1 Parallel PID: ( D ) τ´I⋅ s 

t0e

Gc( s) = Kc⋅  1 +

1

Pu = 0.447

τ´



1 τ I⋅ s

+ τ D⋅ s



(a) PID controller tuned for quarter-decay ratio response From Table 7-2.1, series:

K´c :=

1.2

P K u

−1

τ´I := 2 ⋅ t0e

K´c = 2.1

%CO %TO

τ´D := 0.5⋅ t0e



τ´D 



τ´I

Parallel PID: Kc := K´c⋅  1 + 

τ I := τ´I + τ´D τ D :=



τ´D = 0.3 min τ´I⋅ τ´D

%CO Kc = 2.7 τ = 1.5 min %TO I

τ´I + τ´D

τ D = 0.24 min

(b) PID controller tuned for minimum IAE on disturbance inputs. From Table 7-2.2:

Kc :=

For parallel PID.

1.435

Pu K

− 0.921

τ I :=

τ´I = 1.2 min

τ´

P 0.878 u

0.749

τ D := 0.482 ⋅ τ´ Pu

%CO Kc = 2.4 %TO

1.137

τ I = 0.82 min τ D = 0.25 min

(c) PI controller tuned for minimum IAE on set-point inputs. From Table 7-2.3:

Kc :=

1.086

Pu K

− 0.869

τ I :=

τ´

%CO Kc = 1.7 0.740 − 0.130 ⋅ Pu %TO

For parallel PID. τ D := 0.348 ⋅ τ´ Pu

0.914

τ I = 1.9 min τ D = 0.22 min

(d) PID controller tuned by controller syntesis for 5% overshoot on a set-point change. From Table 7-4.1, series:

K´c :=

0.5

P K u

τ´D   Parallel PID: Kc := K´c⋅  1 +  τ´I 

−1

t0e τ´I := τ´ τ´D := 2

τ I := τ´I + τ´D τ D :=

K´c = 0.89 τ´I⋅ τ´D

%CO %TO

τ´I = 1.3 min τ´D = 0.3 min

%CO τ = 1.6 min τ´I + τ´D Kc = 1.1 %TO I τ D = 0.24 min

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Smith & Corripio, 3rd edition Problem 7-5. Process of Problem 7-3 with sampled-data PI controller. Sample time:

T := 0.1min

Use Eq. 7-2.18: PI controller:

From the solution to Problem 7-3:

t0e := ( 0.39 + 0.2)min + Gc( s) = Kc⋅  1 +

T 2

τ´ := 1.32min t0e Pu := τ´

Pu = 0.485



1

τ I⋅ s   (a) PI controller tuned for quarter-decay ratio response From Table 7-2.1:

0.9 − 1 Kc := P K u

%CO Kc = 1.5 %TO

τ I := 3.33⋅ t0e

τ I = 2.1 min

(b) PI controller tuned for minimum IAE on disturbance inputs. From Table 7-2.2:

Kc :=

0.984 K

Pu

− 0.986

τ I :=

τ´ 0.707 P 0.608 u

%CO Kc = 1.6 %TO

τ I = 1.3 min

(c) PI controller tuned for minimum IAE on set-point inputs. From Table 7-2.3:

Kc :=

0.758

Pu K

− 0.861

τ I :=

τ´ 1.02 − 0.323 ⋅ Pu

%CO Kc = 1.1 %TO

τ I = 1.5 min

(d) PI controller tuned by controller syntesis for 5% overshoot on a set-point chang From Table 7-4.1:

0.5 − 1 Kc := P K u

τ I := τ´

%CO Kc = 0.82 %TO

τ I = 1.3 min

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Smith & Corripio, 3rd edition Problem 7-6. Controller Synthesis for the process of Problem 7-3. (a) Assuming no dead time. G( s) =

K

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1)

Dahlin syntheis formula:

(

)(

)

τ 1⋅ s + 1 τ 2⋅ s + 1 τ1 1 1 1 1 + 1  τ ⋅s + 1 Gc( s) = = ⋅ = 2 K G( s) τ c⋅ s τ c⋅ s K⋅ τ c  τ 1⋅ s   Compare with the series PID controller:

(

Gc( s) = K´c⋅  1 +



K´c =

τ1

 τ´ ⋅ s + 1 ( D ) τ´I⋅ s  1

τ´I = τ 1

K⋅ τ c

)

τ´D = τ 2

(b) Second-order plus dead time. G( s) =

K⋅ e

− t0⋅ s

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1)

C( s )

Dahlin synthesis formula:

1

Gc( s) = G( s)

C( s)

R( s )

C( s)

1−

R( s)

(τ 1⋅ s + 1) (τ 2⋅ s + 1) G ( s) =

Substitute:

c

K⋅ e

=

R( s)

− t0⋅ s

e

e

− t0⋅ s

τ c⋅ s + 1

− t0⋅ s

τ c⋅ s + 1 − e

− t0⋅ s

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

e

− t0⋅ s

1− = 1+

Substitute:

t0 2 t0 2

s 1−e s

− t0⋅ s

1+ =

t0 2

⋅s − 1 +

1+

t0 2

s

t0 2

s =

t0⋅ s 1+

t0 2

⋅s

Gc( s) =

(

)(

τ 1⋅ s + 1 ⋅ τ 2⋅ s + 1 K

)⋅

1+

t0 2

t

⋅s

t0   τ c⋅ s⋅  1 + s + t0 ⋅ s 2  

=



(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)⋅  20 s + 1 

τ c⋅ t 0



2

K⋅ s⋅  τ c + t0 +

τe =

 t0

  ⋅s + 1 τ1 1  2  Gc( s) = ⋅  1 + ⋅ (τ 2⋅ s + 1)⋅ τ e⋅ s + 1 K⋅ ( τ c + t0 ) τ 1⋅ s  

(



s



τ c⋅ t 0

2 ⋅ τ c + t0

)

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

τ1 K⋅ τ c

τ´I = τ 1

τ´D = τ 2

t0 and a second derivative with τ´D2 = 2

In practice astandard PID controller is used with the tuning parameters of Problem 7-4. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-7. Simulation of the control loop of Problem 7-3. To simulate the loop use • a 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:

(τ 1⋅ s + 1)(τ 2⋅ 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: %CO Kc := 1.8 %TO τ I := 1.9min τ D := 0.22min

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Smith & Corripio, 3rd edition Problem 7-8. Quarter decay tuning of PI controller for the blender of Problem 6-11. From the solution to Problem 6-11:

%CO Kcu := −250 %TO

PI controller quarter-decay tuning from Table 7-1.1:

Tu := 3.01min

Kcu Kc := 2.2

The negative gain means the controller is direct acting.

τ I :=

Tu 1.2

%CO Kc = −114 %TO τ I = 2.5 min

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Smith & Corripio, 3rd edition Problem 7-9. Quarter decay tuning of PID controller for the reactor of Problem 6-12. 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: K :=

Gain:

10.1%TO 5%CO K = 2.02

%TO %CO

By fit 3 (two-point method): 0.283 ⋅ 10.1%TO = 2.858 %TO t1 := ( 6 − 1 )min 0.632 ⋅ 10.1%TO = 6.383 %TO t2 := ( 10 − 1 )min

(

τ := 1.5 t2 − t1

)

t0 := t2 − τ

τ = 6 min t0 = 3 min

From Table 7-2.1, the quarter-decay ratio tuning parameters for a series PID controller are: 1.2  t0 

Kc := K

 τ

−1

τ I := 2 ⋅ t0

τ D :=

t0 2

%CO Kc = 1.2 %TO

τ I = 6 min

τ D = 1.5 min

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Smith & Corripio, 3rd edition Problem 7-10. Quarter decay tuning of a PI controller for the three-tank process of Problem 6-14. 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: −11.2%TO

K :=

Gain:

5%CO K = −2.24

%TO %CO

Fit 3 (two-point method): 0.283 ( −11.2%TO) = −3.17 %TO t1 := ( 3 − 1 )min 0.632 ( −11.2%TO) = −7.078 %TO t2 := ( 5.8 − 1 )min

(

τ := 1.5 t2 − t1 t0 := t2 − τ

From Table 7-2.1:

0.9  t0 

Kc := K

 τ

−1

τ I := 3.33t0

%CO Kc = −2.8 %TO

)

τ = 4.2 min t0 = 0.6 min

τ I = 2 min

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Smith & Corripio, 3rd edition Problem 7-11. Quarter decay tuning of a PI controller for the reactors of Problem 6-17. 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:

From this response we get: ∆c := ( 74 − 50)%TO K :=

∆c 5%CO

K = 4.8

%TO %CO

Two-point method: 0.283 ⋅ ∆c = 6.792 %TO t1 := ( 7.5 − 1 )min 0.632 ⋅ ∆c = 15.168 %TO t2 := ( 15 − 1 )min

(

τ := 1.5⋅ t2 − t1 t0 := t2 − τ

From Table 7-2.1:

0.9  t0 

Kc := K

−1

 τ

%CO Kc = 0.77 %TO

τ I := 3.33⋅ t0

1 + KT⋅ Kc⋅  1 +



where

1 τ I⋅ s

⋅

Kv

 τ v⋅ s + 1

2 2

⋅

KA + KB⋅ τ ⋅ s + K3 ⋅ τ ⋅ s

τ := 5min τ v := 0.1min KT := 100

(τ ⋅ s + 1)3 %TO⋅ gal lb

=0

gal Kv := 2.46 min⋅ %CO

τ = 11.25 m t0 = 2.75 mi

τ I = 9.2 min

Roots of the characteristic equation, damping ratio, and decay ratio. From the solution to Problem 3-17, the characteristic equation is:

)

KA := 0.0075

)(

lb⋅ min gal

KB := 0.0075

2

)

s⋅ τ v ⋅ s + 1 ⋅ τ ⋅ s + 3 ⋅ τ ⋅ s + 3 ⋅ τ ⋅ s + 1 + KT⋅ Kc⋅ Kv ⋅  s +

(

3 3

2 2



3 5

4

3

lb⋅ min gal

K3 := 0.0025

2

2

τ A := 3 ⋅ τ v ⋅ τ + τ

3

KL

2

2

3

4

τ v ⋅ τ = 12.5 min

τ E := 1 + KL + 3

τ A = 132.5 min

2

 ⋅  K + K ⋅ τ ⋅ s + K ⋅ τ 2⋅ s2 = 0 A B 3  τI   =0

τI

τ B := 3 ⋅ τ v ⋅ τ + 3τ + KT⋅ Kc⋅ Kv ⋅ K3 ⋅ τ

τ C := τ v + 3τ + KT⋅ Kc⋅ Kv ⋅ KB⋅ τ

gal

1

τ v ⋅ τ ⋅ s + τ A⋅ s + τ B⋅ s + τ C⋅ s + τ E⋅ s + where

lb⋅ min

2

KL := KT⋅ Kc⋅ Kv ⋅ KA

KT⋅ Kc⋅ Kv ⋅ KB⋅ τ

KL = 1.415

τI 2

τ B = 88.293 min

τ C = 22.176 min

τ E = 3.188

Roots of the characteristic equaton:

  KL   ⋅ min   τ I   τE −9.905        − 0.39  −1 τ min   =  −0.112 + 0.164i  polyroots  C     −2   − 0.112 − 0.164i    τ B min     −0.082    τ A min− 3       τ ⋅ τ 3 min− 4  v 

The dominant root is a real root with time constant: −1 = 12.2 min −1 −0.082 min There is also a comlex conjugate root with period: 2π

T :=

−1

T = 38.3 min

0.164min Damping ratio:

0.112 2

0.112 + 0.164 Decay ratio:

e

2

− 0.112 min

= 0.564

−1

T

= 0.014

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

kscf := 1000ft

3

Problem 7-12. Control of suction pressure for compressor of Problem 6-18. From the solution to Problem 6-18: KT %TO H( s) = KT := 5 psi τ T⋅ s + 1

PC

τ T := 1.2s

Gsc( s) =

m(t) SC

Ksc τ sc⋅ s + 1

τ sc := 2.5s

Steam

kscf

PT

Ksc := 0.36 min⋅ %CO Gp ( s) =

Kp

Suction

τ p := 7.5s

τ p⋅ s + 1

fc(t) Discharge ps(t)

fi(t)

psi⋅ min Kp := 0.5 kscf Ultimate gain and period:

%CO Kcu := −15.9 %TO

Tu := 8.91s

Quarter-decay ratio tuning parametes for a PI controller: Kcu Kc := 2.2

From Table 7-1.1:

τ I :=

Tu

%CO Kc = −7.2 %TO

1.2

τ I = 7.4 s

Roots of the characteristic equation, damping ratio, and decay ratio. From the solution to Problem 6-18: 4

(

1−

KT τ T⋅ s + 1

)

3

Kc⋅  1 +



(

1



Ksc

Kp

τ I⋅ s τ sc⋅ s + 1 τ p ⋅ s + 1 

)2 (

)

τ T⋅ τ sc⋅ τ p ⋅ s + τ T⋅ τ sc + τ T⋅ τ p + τ sc⋅ τ P ⋅ s + τ T + τ sc + τ p s + 1 − KL s − where

KL := KT⋅ Kc⋅ Ksc⋅ Kp

Find roots:

KL = −6.505

=0 KL τI

=0

−KL     sec τI      −1.16  1 − KL    −0.135  =   −1 polyroots  τ + τ + τ sec  ( T sc p)    −0.036 − 0.498i      −0.036 + 0.498i   ( τ T⋅ τ sc + τ T⋅ τ p + τ sc⋅ τ p) sec− 2     −3   τ T⋅ τ sc⋅ τ p ⋅ sec   Damping ratio:

0.036 2

0.036 + 0.498

2

= 0.072

Decay ratio:

The dominant roots are the complex conjugate pair. The period of the oscillations is: 2π

T :=

0.498sec

−1

T = 12.62 s

−1

e

− 0.036sec

⋅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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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: GFC( s) =

Tset(t)

KFC

τ FC := 0.1min

τ FC⋅ s + 1

f(t)

3

H( s) =

KT

Ti(t)

Tc(t)

τ T := 0.6min

τ T⋅ s + 1

m(t)

TT

m

KFC := 0.008 min⋅ %CO

TC

V

SP FC

%TO KT := 2 degC G1 ( s) =

FT

K2 ⋅ K3

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1) − K2⋅ K4

τ 1 := 13.77min τ 2 := 3.03min K2 := 0.725

K3 := 28.94

Tci

fc(t)

T(t)

degC ⋅ min 3

m

K4 := 0.525

%CO Kcu := −86.7 %TO

Ultimate gain and period:

Tu := 8.32min

Quarter-decay ratio tuning of a PI controller. Kcu Kc := 2.2

τ I :=

Tu

%CO Kc = −39 %TO

1.2

τ I = 6.9 min

Roots of the characteristic equation, damping ratio, and decay ratio. From the solution to Problem 6-19:

1−

KT τ T⋅ s + 1 5

Kc⋅  1 +



4

1



KFC

(

τ A := τ T⋅ τ FC⋅ τ 1 ⋅ τ 2

)

τ I⋅ s τ FC⋅ s + 1 τ 1 ⋅ s + 1 ⋅ τ 2 ⋅ s + 1 − K2 ⋅ K4  3

2

τ A⋅ s + τ B⋅ s + τ C⋅ s + τ D⋅ s + τ E s − where

)(

K2 ⋅ K3

KL τI

=0

=0

τ B := τ T⋅ τ FC⋅ τ 1 + τ T⋅ τ FC⋅ τ 2 + τ FC⋅ τ 1 ⋅ τ 2 + τ T⋅ τ 1 ⋅ τ 2

4

τ = 2.503 min τ C := τ T⋅ τ FC⋅ 1 − K2 ⋅ K4 + τ T⋅ τ 1 + τ T⋅ τ 2 + τ FC⋅ τ 1 + τ FC⋅ τ 2 + τ 1 ⋅ τ 2 A 3 τ B = 30.214 min

(

(

)

)(

)

τ D := τ T + τ FC 1 − K2 ⋅ K4 + τ 1 + τ 2 τ E := 1 − K2 ⋅ K4 − KL

KL := KT⋅ Kc⋅ KFC⋅ K2 ⋅ K3 2 τ C = 53.52 min

τ E = 13.849

KL = −13.23

τ D = 17.234 min

Find the roots:

   min  τ I      τ E    −9.993   −1.882 − 1   τ D⋅ min      polyroots = −0.154     −2   τ C⋅ min    −0.02 − 0.512i      −0.02 + 0.512i    τ B⋅ min− 3       τ ⋅ min− 4  A  −KL

The dominant roots are the complex conjugate pair. The period of oscillation is: 2π

T :=

Tu = 8.32 min

−1

0.512min

0.02

Damping ratio:

2

0.02 + 0.512

Decay ratio:

e

− 0.02min

−1

⋅T

2

= 0.039

= 0.782

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

lbmole := 433.59mole

Problem 7-14. Composition control of reactors in series of Problem 6-22. From the solution to Problem 6-22: gal Kv := 1.42 min⋅ %CO H( s) =

KT

τ T⋅ s + 1

cA2set(t)

cA1(t)

τ T := 0.5min

%TO⋅ gal KT := 222 lbmole G1 ( s) =

(

)

K4 ⋅ τ 1 ⋅ s + 1 + K5 ⋅ K1

V fA(t)

DA

(τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)

τ 1 := 2.222min τ 2 := 1.429min K1 := 0.006123

K4 := 0.00127

fi(t) cAi(t)

AC

AT

V LT

LC

c A2(t)

lbmole⋅ min 2

gal

lbmole⋅ min gal

2

K5 := 0.286

Quarter-decay tuning parameters for a PI controller. 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:

Gain: K :=

4.7

K = 0.94

5

%TO %CO

Two-point method: 0.283 ⋅ 4.7%TO = 1.33 %TO t 1 := ( 2.5 − 1 )min 0.632 ⋅ 4.7⋅ %TO = 2.97 %TO t 2 := ( 4 − 1 )min

(

τ := 1.5⋅ t 2 − t 1

)

τ = 2.3 min

t 0 := t 2 − τ

t 0 = 0.8 min

From Table 7-2.1: 0.9  t 0 

Kc := K

−1

 τ

τ I := 3.33⋅ t 0

%CO Kc = 2.9 %TO

τ I = 2.5 min

Roots of the characteristic equation, damping ratio, and decay ratio. From the solution to Problem 2-22: 4

(

1+

KT

τ T⋅ s + 1

Kc⋅  1 +



s  K ⋅ K4 ⋅ ( τ 1⋅ + 1) + K5 ⋅ K1 = 0 v τ I⋅ s (τ 1⋅ s + 1)⋅ (τ 2⋅ s + 1)  1

)3 (

)2 (

)

τ T⋅ τ 1 ⋅ τ 2 ⋅ s + τ T⋅ τ 1 + τ T⋅ τ 2 + τ 1 ⋅ τ 2 s + τ T + τ 1 + τ 2 + τ A s + 1 + KL s + where

τ A := KT⋅ Kc⋅ Kv ⋅ K4 ⋅ τ 1

Find the roots:

(

KL := KT⋅ Kc⋅ Kv ⋅ K4 + K5 ⋅ K1

)

KL

τI

=0

τ A = 2.555 min KL = 2.736

KL     min τI       −1.204 1 + KL    −1.204   − 1  =  polyroots τ + τ + τ + τ min   ( T 1 2 A)    −0.371     −0.371  ( τ T⋅ τ 1 + τ T⋅ τ 2 + τ 1⋅ τ 2) min− 2     −3   τ T⋅ τ 1 ⋅ τ 2 ⋅ min  

Damping ratio:

0.371 2

0.371 + 0.435

2

= 0.649

− 0.813i 

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

+ 0.813i 

T :=

− 0.435i 

−1

0.435min

+ 0.435i 

Decay ratio:

2π

T = 14.44 min

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

%TO := %

%CO := %

Problem 7-15. Solid moisture control of a vacuum filter.

AC M AT To incinerator Ferric Chloride Sludge

 0  1  ∆m := 12.5%CO  1.5  2.5 • First column is time in  minutes  3.5 • Second column is outlet %  4.5 moisture in the solids  5.5   6.5  7.5  8.5  txm :=  9.5 Problem data:  10.5 x in := 95% x o := 75%   11.5  13.5 Transmitter (AT):  15.5  x min := 55% x max := 95%  17.5  19.5 100%TO %TO KT := KT = 2.5  21.5 x max − x min %   25.5  29.5   33.5

Response to a step change of:

Filtrate

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 

∆x in := 0.5%

0 1  2 3 4  5 6  7 8 9  10 txi :=   11  12   13  14   15  17  19   21  25  29   33

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 75.0

77.0 





(a) Block diagram of the moisture control loop.

Xin(s) Xoset(s) KT

R(s) + E(s) -

Gc(s)

G2(s)

M(s)

C(s)

G1(s)

+

Xo(s)

+

KT

(b) Transfer functions by fit 3, two-point method. ∆x m :=  txm − txm  % 0 , 1  20 , 1

∆x m = −8 %

K1 :=

∆x m

%TO KT⋅ K1 = −1.6 %CO

∆m

72.736 − txm 8, 1

txm % + 0.283 ⋅ ∆x m = 72.736 % 0, 1

t1 := txm +  tx − txm  ⋅ 8 , 0  m9 , 0 8 , 0 txm

txm % + 0.632 ⋅ ∆x m = 69.944 % 0, 1

t2 := txm +  tx − txm ⋅ 12 , 0  m13 , 0 12 , 0 txm

t1 = 7.877

t2 = 12.212

69.944 − txm 12 , 1 13 , 1

(

)

τ 1 := 1.5 t2 − t1 min

t01 := t2 ⋅ min − τ 1 τ 1 = 6.5 min KT⋅ G1 ( s) =

∆x i :=  txi − txi  % 0 , 1  21 , 1

∆x i = 2 %

t01 = 5.71 min

− 5.71s

6.5⋅ s + 1

∆x i

%TO %CO

%TO KT⋅ K2 = 10 %

∆x in

t1 := txi +  tx − txi  ⋅ 9 , 0  i 10 , 0 9 , 0 txi

10 , 1

− txi 9, 1

76.264 − txi 14 , 1

txi % + 0.632 ⋅ ∆x i = 76.264 % 0, 1 t2 = 13.64

K2 :=

−1.6e

− txm 12 , 1

75.566 − txi 9, 1

txi % + 0.283 ⋅ ∆x i = 75.566 % 0, 1

t1 = 8.66

− txm 8, 1

9, 1

t2 := txi +  tx − txi ⋅ 14 , 0  i 15 , 0 14 , 0 txi

(

)

τ 2 := 1.5 t2 − t1 min

15 , 1

− txi 14 , 1

t02 := t2 ⋅ min − τ 2 τ 2 = 7.47 min t02 = 6.17 min − 6.17s

%TO 10e KT⋅ G2 ( s) = 7.47⋅ s + 1 %

(c) Discuss the controllability of the loop and the controller action. The loop is difficult to control by feedback control because its ratio of dead time to time constant is high: t01 = 0.878 τ1 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.

(d) Gain of a proportional controller for minimum IAE response and offset to a 1% increase in inlet moisture. From Table 7-2.2:

Offset =

KT⋅ ∆x o

set

 t01  Kc :=  KT⋅ K1 τ 1   0.902

− K2 ⋅ ∆x in

−1

Offset :=

1 + KT⋅ Kc⋅ K1

%CO Kc = −0.64 %TO 0 − K2 ⋅ 1 % 1 + KT⋅ Kc⋅ K1

(e) Quarter-decay response tuning of a PI controller. From Table 7-2.1:

 t01  Kc :=  KT⋅ K1 τ 1   0.9

−1

τ I := 3.33⋅ t01

%CO Gc( s) = −0.64 %TO

Offset = −2 %TO Offset KT

= −0.789 %

%CO Kc = −0.64 %TO

τ I = 19 min

%CO  1  Gc( s) = −0.64 1 + %TO  19⋅ s  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

−6

Smith & Corripio, 3rd edition

ppm := 10

Problem 7-16. Composition control of an absorber. (a) Design a control loop to control the air outlet composition. Transmitter (AT): y min := 0ppm

SP

y max := 200ppm

Air Out

Negligible lag.

AC

AT

Control valve, assumed linear.

Water In

gal fmax := 500 min Negligible lag.

Air In NH3 solution

(b) Block diagram of the loop and transfer function of each block. Use fit-3 on response data (two-point method).

Yin(s) set o

Y (s) KT

R(s) + E(s)

Transmitter (AT):

Control valve:

-

Gc(s)

M(s)

Kv

C(s)

F(s)

G2(s) G1(s)

+

Yo(s)

+

KT

100%TO KT := y max − y min

%TO KT = 0.5 ppm

fmax Kv := 100%CO

gal Kv = 5 min⋅ %CO

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

Response to a step change in inlet water flow of

 0  20  30  40   50  60   70  80 ty :=  90   100  110  120   130  140   160  180  250

∆f := −50

50.00 

gal min

50.00  50.12 

• •

  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  50.30

First column is time in seconds Second column is outlet ammonia ppm

(

∆y o := ty K1 :=

16 , 1

)ppm

− ty

∆y o = 1.77 ppm

0, 1

∆y o

K1 = −0.035

∆f

y o1 := ty

0, 1

ppm + 0.283 ⋅ ∆y o

3, 0

y o2 := ty

(

+ ty

0, 1

4, 0

− ty

)

3, 0

⋅

y o1⋅ ppm

3, 1

4, 1

7, 0

Fit 3

(

(

+ ty

)

τ 1 := 1.5 t2 − t1 sec

8, 0

− ty

)

7, 0

⋅

t1 = 46.7

3, 1

y o2 = 51.12 ppm −1

t2 := ty

− ty

− ty

ty

ppm + 0.632 ⋅ ∆y o

gal

y o1 = 50.5 ppm −1

t1 := ty

ppm⋅ min

y o2⋅ ppm ty

t01 := t2 ⋅ sec − τ 1

8, 1

− ty

7, 1

− ty

t2 = 84.58

τ 1 = 0.95 min

t01 = 0.46 min

7, 1

G1 ( s) =

−0.035 e

0.46s

0.95⋅ s + 1

ppm⋅ min gal

(c) Quarter decay ratio tuning for proportional controller and offset to a set-point change of ( 60 − 50)ppm = 10 ppm From Table 7-2.1:

 t01  Kc :=  KT⋅ Kv ⋅ K1 τ 1   1

−1

%CO Kc = −23 %TO

%CO Gc( s) = −23 %TO

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.

Offset :=

KT⋅ 10ppm

Offset = 1.641 %TO

1 + KT⋅ Kc⋅ Kv ⋅ K1

Offset KT

= 3.3 ppm

(d) Quarter decay ratio tuning of series PID controller and offset. From Table 7-2.1:

K´c = −28

%CO %TO

 t01  K´c :=  KT⋅ Kv ⋅ K1 τ 1   1.2

τ´I = 0.93 min

−1

τ´I := 2 ⋅ t01

τ´D = 0.23 min

t01 τ´D := 2

%CO  1  Gc( s) = −28 ( 0.23s + 1 ) 1 + %TO  0.93s 

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

degF := R

Problem 7-17. Temperature control of a furnace. Transmitter (TT): Tmin := 300degF Tmax := 500degF

SP

Response to step change of: ∆m := 5%CO

 0  0.5  1.0  2.0   2.5  3.0   3.5  4.0  4.5  5.0 tt :=   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 

  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  425.0

445.0 

• •

First column is time in minutes Second column is temperature in ºF

TT

Fuel

Air

(a) Block diagram of the loop, fail-safe position of the valve, and controller action. The control valve must fail closed (air-to-open) to prevent overheating the furnace on instrument power failure. 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.

Block diagram of the loop:

Fin(s) Toset(s)TC KT EF

TC

Process air

R(s) + E(s) %TO

-

TC

scfh

Gc(s)

C(s) %TO

G2(s)

M(s) Furnace %CO

KT

G1(s)

TT

-

To(s)

+

EF

(b) Transfer functions using fit 3 (two-point method). Transmitter (TT):

100%TO KT := Tmax − Tmin

(

∆T := tt T1 := tt

0, 1

19 , 1

degF + 0.283∆T

)degF

− tt

degF + 0.632∆T

(

+ tt

∆m

7, 0

)

− tt

6, 0

⋅

T1 ⋅ degF tt

7, 1

−1

− tt

6, 1

t1 = 3.517

− tt

6, 1

T2 = 437.64 degF t2 := tt

11 , 0

Process by fit 3:

%TO KT⋅ K1 = 2 %CO

T1 = 430.66 degF

6, 0

0, 1

∆T

K1 :=

0, 1

t1 := tt T2 := tt

%TO KT = 0.5 degF

(

)

τ 1 := 1.5⋅ t2 − t1 min

(

+ tt

12 , 0

− tt

)

11 , 0

⋅

T2 ⋅ degF

t01 := t2 ⋅ min − τ 1

tt

−1

− tt

11 , 1

12 , 1

− tt

11 , 1

t2 = 6.022

τ 1 = 3.76 min t01 = 2.26 min − 2.26s

%TO 2e KT⋅ G1 ( s) = 3.76s + 1 %CO

(c) Quarter decay ratio tuning of series PID controller. From Table 7-2.1:

K´c = 1

%CO %TO

 t01  K´c :=  KT⋅ K1 τ 1   1.2

τ´I = 4.5 min

−1

τ´I := 2 ⋅ t01

τ´D = 1.1 min

t01 τ´D := 2

%CO  1  Gc( s) = 1 ( 1.1s + 1 ) 1 + %TO  4.5s 

(d) Synthesis tuning of series PID controller for 5% overshoot. From Table 7-4.1:

K´c = 0.42

%CO %TO

 t01  K´c :=  KT⋅ K1 τ 1   0.5

τ´I = 3.8 min

−1

τ´D = 1.1 min

τ´I := τ 1

t01 τ´D := 2

%CO  1  Gc( s) = 0.42⋅ ( 1.1s + 1 ) 1 + %TO  3.8s 

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Smith & Corripio, 3rd edition Problem 7-18. Temperature control of oil heater of Problem 6-24.

T3set(t) TC N2

TT p2 = 40 psia 3 ft

Steam

fs(t)

AO

hset(t)

LT

LC

AO

h(t) 5 ft

T Condensate

T1(t) p1(t)

vp2(t) T3(t) p3(t)

Quarter decay ratio tuning parameters of series PID temperature controller TC. From the solution to Problem 6-24, the ultimate gain and period are:

%CO Kcu := 38.9 %TO Tu := 1.116min

From Table 7-1.1:

K´c = 23

%CO %TO

K´c :=

Kcu 1.7

τ´I = 0.56 min

Tu τ´I := 2

Tu τ´D := 8

τ´D = 0.14 min

%CO  1  Gc( s) = 23 ( 0.14s + 1 ) 1 + %TO  0.56s 

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-19. Temperature control of exothermic catalytic reactor.

SP TC

Reactants

Steam

SP LT LC TT Water Products

Problem data:

α := 50

Control valve, equal percentage:

Tmin := 150degF

Temperature transmitter (TT): Oil density:

ρ := 55

lb ft

Design conditions:

3

∆p v := 10psi

gal fd := 200 min

Tmax := 350degF

3

ρ ⋅ ft Gf := 62.4lb

gal fpump := 400 min

Gf = 0.881 T := 275degF

(constant)

Open-loop test on temperature loop:

∆vp := 5%

∆T := 4.4degF

Closed-loop test on temperature loop:

%CO Kcu := 16 %TO

Tu := 24min

(a) Size control valve for 100% overcapacity. Valve fail-safe position and controlle action. Cvmax := 200%⋅ fd ⋅

Gf ∆p v

Cvmax = 119

gal min⋅ psi

From Fig. C-10.1, p. 532, a 4-in valve is required. Cvmax := 195

gal min⋅ psi

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.

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.

(b) Valve position at design conditions and maximum flow through the valve when fully opened. 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.

fv = Cv ⋅

∆p va

 fpump − ∆p va = ∆p v ⋅   fpump −

Gf

fv 

2

 fv  = Gf ⋅  fd   Cv 

2

Cv = Cvmax⋅ α

vp− 1

 Cv 

At design conditions:

Cv := fd ⋅

ln

Gf ∆p v

∆p v  fpump − fvmax 

Flow when fully opened:

 Cvmax  C = 59.4 v ln( α )

vp := 1 +

fvmax

vp = 69.6 %

min⋅ psi

Let k v := Cvmax⋅

=  Gf  fpump − fd  Cvmax

kv fv := f 1 + k v pump

gal

∆p v Gf

⋅

1 fpump − fd

gal fv = 306.6 min

k v = 3.284

(c) Block diagram of the loop.

Fin(s) Tset(s) TC KT EF

R(s) + E(s) %TO

-

TC

lb/hr

Gc(s)

M(s) Furnace %CO

C(s)

KT

%TO 100%TO KT := Tmax − Tmin ln( α ) ⋅ fd Kv := 100%CO

%TO KT = 0.5 degF

G2(s) G1(s)

+

To(s)

+

EF

TT

Valve fails closed. Controller is direct acting. (See part (a).)

gal Kv = 7.824 min⋅ %CO

(d) Calculate the process gain atb design conditions, including the control valve an

( ) p g the temperature transmitter. K1 :=

∆T ∆vp

g

g

%TO KT⋅ K1 = 0.44 %CO

K1 = 88 degF

The gain of the valve is included in K1, because the step change is in valve position.

(e) Quarter decay tuning parameters for series PID temperature controller. From Table 7-1.1:

K´c = 9.4

Kcu

K´c :=

1.7

%CO

1

%TO

τ´I

Tu τ´I := 2

= 0.083

repeats min

Tu τ´D := 8 τ´D = 3 min

repeats := 1 %CO  0.083  Gc( s) = 9.4 ( 3s + 1 1 + %TO  s 

(f) Quarter-decay ratio tuning of proportional temperature controller and offset for a set point change of -10ºF. From Table 7-1.1:

Offset :=

Kc :=

KT⋅ ( −10degF) 1 + KT⋅ Kc⋅ K1

Kcu 2

%CO Kc = 8 %TO Offset = −1.11 %TO

%CO Gc( s) = 8 %TO Offset KT

= −2.21 degF

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

weight% := %

Problem 7-20. Composition control of a double-effect evaporator SP AC 13

Vapors

Vapors

Steam SP

FT

SP

FC 12

LT

SP

LC

Cond.

13

AT

Feed Problem data:

LC

LT

12

Product Feed := 50000

lb hr

x F := 5weight%

Open loop step response in feed composition:

x min := 10weight%

∆x F := 0.75weight%

x max := 35weight%

25

24.5

Product composition, wt%

24

23.5

23

22.5

22

21.5

21 0

100

200

300

400

500

600

700

Time, sec

Open-loop step response to change in controller output:

∆m := 2.5%CO

25

24.5

Product composition, wt%

24

23.5

23

22.5

22

21.5

21 0

100

200

300

400

500 Time, sec

600

700

800

900

1000

(a) Block diagram of the composition control loop, transfer functions, cntrol valve fa safe position, and controller action.

XF(s) Xset(s) AC wt% KT

AC

R(s) + E(s) %TO

-

Gc(s)

M(s) %CO

C(s)

KT

%TO Analyzer transmitter:

G2(s)

wt%

100%TO KT := x max − x min

X(s)

+

G1(s)

+

wt%

AT

%TO KT = 4 weight%

Determine process transfer functions by fit 3: Change in controller output:

K1 :=

( 24.7 − 21.5)weight% ∆m

weight% K1 = 1.28 %CO

21.5weight% + 0.283 ⋅ ( 24.7 − 21.5) weight% = 22.41 weight%

t1 := 256sec

21.5weight% + 0.632 ⋅ ( 24.7 − 21.5) weight% = 23.52 weight%

t2 := 419sec

(

τ 1 := 1.5 t2 − t1

(

min

) 60sec

t01 := t2 − τ 1

min ) 60sec

τ 1 = 4.08 min

t01 = 2.91 min

KT⋅ G1 ( s) = Change in feed composition:

K2 :=

( 24.7 − 21.5)weight% ∆x F

K1 = 1.28

t1 := 143sec

21.5weight% + 0.632 ⋅ ( 24.7 − 21.5) weight% = 23.52 weight%

t2 := 237sec

(

min

) 60sec

(

t02 := t2 − τ 2

min ) 60sec

5.12e

4.08s + 1

τ 2 = 2.35 min KT⋅ G2 ( s) =

− 2.91s

%TO %CO

%TO KT⋅ K2 = 17.07 weight%

21.5weight% + 0.283 ⋅ ( 24.7 − 21.5) weight% = 22.41 weight%

τ 2 := 1.5 t2 − t1

%TO KT⋅ K1 = 5.12 %CO

17.07e

t02 = 1.6 min − 1.6s

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

the controller output closing the steam control valve; this decreases the rate of evaporation reducing the product composition.

(b) Quarter-decay ratio tuning of PI composition controller. From Table 7-2.1:

 t01  Kc :=  KT⋅ K1 τ 1   0.9

−1

τ I := 3.33⋅ t01

%CO Kc = 0.25 %TO

Quarter decay ratio tuning is based on fit 1 parameters. From the figure above: min t01 := 130sec τ 1 := 580sec − t01 60sec

(

From Table 7-2.1:

 t01  Kc :=  KT⋅ K1 τ 1   0.9

)

−1

τ I := 3.33⋅ t01

%CO Kc = 0.61 %TO

τ I = 9.7 min

τ 1 = 7.5 min τ I = 7.2 min

This is over twice the gain and 25% faster reset than with fit 3 parameters.

(c) Controller synthesis tuning for 5% overshoot of PI composition controller. From Table 7-4.1:

 t01  Kc :=  KT⋅ K1 τ 1   0.5

−1

τ I := τ 1

%CO Kc = 0.34 %TO

τ I = 7.5 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 beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

psia := psi

Smith & Corripio, 3rd edition Problem 7-21. Temperature control of stirred-tank reactor. Design conditions: p 1 := 30psia p 2 := 15psia gal TR := 210degF fcw := 350 min

TC 11

Coolant

Coil pressure drop: ∆p L := 10psi Temperature transmitter: Tmin := 190degF

Product

P1

Tmax := 230degF

P2

α := 50

Equal-percenage valve: Open-loop test:

TT

gal ∆fcw := 10 min

Feed

∆TR := −5.2degF %CO Closed-loop test: Kcu := 8.0 %TO

Tu := 14min

(a) Size control valve for 100% overcapacity, valve gain at design flow, valve fail-safe position. ∆p v := p 1 − p 2 − ∆p L

Cvmax := 200%⋅ fcw⋅

Gf := 1 Gf ∆p v

k L :=

∆p L Gf ⋅ fcw

Cvmax = 313.05

gal min⋅ psi

From Fig. C-10.1, page 532, a 6-in valve is required:

Valve gain, Eq. 5-2.27, page 171:

∆p v = 5 psi

2

−ln( α ) Kv := 100%CO

−5

k L = 8.163 × 10

Cv := fcw⋅ Cvmax := 400 fcw

1 + k L⋅ C v

2

Gf ∆p v

psi⋅ 

Cv = 156.5

min 

 gal  gal

min⋅ psi

gal min⋅ psi gal Kv = −4.564 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.

2

Fin(s) lb/hr

TC Tset(s) TC R(s) E(s) +

M(s)

EF

%CO

KT

Gc(s)

%TO -

KT

%TO

K1 :=

∆TR ∆fcw

Fcw(s)

Gv(s)

C(s)

100%TO KT := Tmax − Tmin

G2(s) gpm

TR(s)

+

EF

TT

%TO KT = 2.5 degF

%TO KT⋅ K1 ⋅ Kv = 5.933 %CO

Total process gain:

G1(s)

+

K1 = −0.52

degF⋅ min gal

repeat := 1

(c) Quarter-decay ratio tuning of PID temperature controller and controller action. From Table 7-1.1:

K´c := 1 τ´I

Kcu 1.7

= 0.14

Tu τ´I := 2 repeat min

Tu τ´D := 8

100%CO K´c

= 21 %TO

100%CO  0.14  τ´D = 1.75 min Gc( s) = ( 1.75s + 1 ) 1 + 21%TO  s 

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Smith & Corripio, 3rd edition Problem 7-22. Solids moisture control of a phosphates pebbles drier

Stack AC Feed AT

Dry phospahates

Fuel Air

Design conditions:

x F := 15weight%

x := 3weight%

Transmitter AT: x min := 1weight% x max := 5weight%

(a) Block diagram of the moisture control loop.

XF(s) Xset(s) AC wt% KT

R(s) + E(s) %TO

-

AC

Gc(s)

C(s) %TO

Transmitter AT:

100%TO KT := x max − x min

G2(s)

wt%

M(s) %CO

KT

G1(s)

+

X(s)

+

AT

%TO KT = 25 weight%

(b) Process transfer functions from open-loop step responses by fit 2. Open-loop step response to change in controller output:

∆m := 8%CO

wt%

5

Product moisture,wt %

4.5

4

3.5

3

2.5 0

50

100

150

200

250

300

350

400

Time, sec

∆x := ( 4.5 − 3 )weight% From the graph:

K1 :=

t01 := 70sec ⋅

min τ 1 := t2 ⋅ − t01 60sec

∆x

∆m min

60sec

weight% K1 = 0.188 %CO

%TO KT⋅ K1 = 4.688 %CO

3.0weight% + 0.632∆x = 3.95 weight%

t01 = 1.17 min τ 1 = 2 min

Open-loop step response to change in inlet moisture:

KT⋅ G1 ( s) =

∆x F := 3weight%

4.688e

t2 := 190sec

− 1.17s

2.0s + 1

%CO %TO

5.5

5

Product moisture, wt%

4.5

4

3.5

3

2.5 0

50

100

150

200

250

300

350

400

Time, sec

∆x := ( 5.0 − 3.0)weight% From the graph:

K2 :=

t02 := 65sec ⋅

min τ 2 := t2 ⋅ − t02 60sec

∆x

weight% K2 = 0.667 weight%

∆x F

min

%TO KT⋅ K2 = 16.67 weight%

3.0weight% + 0.632∆x = 4.26 weight%

60sec

t02 = 1.08 min τ 2 = 1.75 min

KT⋅ G2 ( s) =

16.67e

t2 := 170sec

− 1.08s

1.75s + 1

%CO weight%

(c) Minimum IAE tuning of parallel PID moisture controller on disturbance inputs a controller action.  t01  From Table 7-2.2: Kc :=  KT⋅ K1 τ 1   1.435

100%CO Kc

= 199 %TO

− 0.921

τ I = 1.5 min

τ1

 t01  τ I :=  0.878 τ 1  

τ D = 0.52 min

0.749

 t01  τ D := 0.482 ⋅ τ 1 ⋅   τ1 

1.137

100%CO  1 Gc( s) = + 0.52s 1 + 199%TO  1.5s 

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.

(d) New moisture content of the product when the feed moisture content decrease by 2 weight%. Controller is proportional only tuned for quarter decay ratio respons From Table 7-2.1:

Offset :=

 t01  Kc :=  KT⋅ K1 τ 1   1

KT⋅ 0 − KT⋅ K2 ⋅ ( −2 weight%) 1 + KT⋅ Kc⋅ K1

−1

%CO Kc = 0.37 %TO Offset

Offset = 12.28 %TO

Final steady state moisture control of the product:

KT

3weight% −

= 0.49 weight%

Offset KT

= 2.51 weight%

(e) Controller output required to avoid offset for the disturbance of part (d). For the change in outlet moisture to be zero:

∆m⋅ K1 + K2 ⋅ ( −2 weight%) = 0 ∆m :=

−K2 K1

( −2 weight%)

If the initial controller output is 50%CO, the final steady state is:

∆m = 7.11 %CO

50%CO + ∆m = 57.11 %CO

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-23. Level control by manipulatiion of inlet flow. Control valve fails closed.

SP

LC

fi (t) LT

h(t) fo(t)

Block diagram of the level control loop and required controller action.

Fo(s) Cset(s)

+

%TO

ft3/min

E(s)LC -

Gc(s)

M(s) %CO

C(s) %TO

Gv ( s) =

Kv τ v⋅ s + 1

Gv(s) KT

1/As Fi(s) ft3/min

1/As

-

H(s)

+

ft

LT

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.

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) =

where

K=

(

K⋅ Gc( s)

)

s⋅ τ v ⋅ s + 1 + K⋅ Gc( s)

Kv ⋅ KT

%TO

A

%CO⋅ min

C

set

( s) −

(

(

Ku ⋅ τ v ⋅ s + 1

)

)

s⋅ τ v ⋅ s + 1 + K⋅ Gc( s)

Fo ( s)

KT %TO Ku = A 3 ft

The formulas of Section 7-3 apply to this case also. Only the action of the controller is different. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-24. Comparison of tank level dynamic models of Section 4-1.1 Problem data: D := 3m

h d := 2m

Level transmitter: h min := 1m

fi(t)

h max := 3m

SP

Control valve sized for twice the design flow of

LC

LT

3

m f1d := 0.003 s

h(t)

τ v := 5s

fo(t)

f1(t)

Proportional controller: Gc( s) = Kc

(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. The model of Section 4-1.1 must be modified to account for the variable valve position. 2

Mass balance: Valve equation:

π⋅D 4

⋅ρ⋅

d ⋅ h( t) dt

= ρ ⋅ fi( t) − ρ ⋅ fo ( t) − ρ ⋅ f1 ( t)

f1 ( t) = Cvmax⋅ vp( t) ⋅

∆p v ( t)

3

1 eqn. 2 unks. (h, f1 )

min

m

2 eqns. 3 unks. (∆pv)

264.2gal 60sec

Gf

Assume valve exit is at the level of the bottom of the tank and at the same pressure. ∆p v ( t) = ρ ⋅ g ⋅ h ( t)

1.45⋅ 10

−4

psi

3

ρ⋅m Gf = 1000kg

Size the valve:

3 eqns. 3 unks.

Pa g = 9.807

Cvmax := 200%⋅ f1d⋅

m

3

m fd = 0.013 s

2

s 1Pa

3

m

264.2gal 60sec ⋅ −4 1000kg⋅ g ⋅ h d 3 min 1.45⋅ 10 psi m ⋅

gal

Cvmax = 56.4

From Fig. C-10.1, p. 532, a 3-in valve is required.

min⋅ psi

Cvmax := 110 Let

where

gal min⋅ psi

f1 ( t) = k v ⋅ vp( t) ⋅ h ( t) 1000kg

k v := Cvmax⋅

3

⋅ g⋅

1.45⋅ 10

Valve position at design conditions:

2

π⋅D 4

⋅

vpd :=

d ⋅ h( t) dt

psi

Pa

m

Substitute and simplify:

−4

3

⋅

m

min

k v = 8.275 × 10

264.2gal 60s f1d

3 −3 m

s⋅ m

vpd = 25.6 %

kv⋅ hd

= fi( t) − fo ( t) − k v ⋅ vp( t) ⋅ h ( t)

Linearize and express in terms of deviation variables: 2

π⋅D 4

⋅

d ⋅ H( t) dt

= Fi( t) − Fo ( t) − k v ⋅ h d VP ( t ) −

Rearrange into standard firts-order form:

where

τ :=

2 2⋅ h d

π⋅D 4

⋅

k v ⋅ vpd

τ⋅

d ⋅ H( t) dt

2⋅ hd

H( t)

τ = 9425 s

K1 = 1.3 × 10

τ ⋅ min

(

)

H( s) = G1 ( s) ⋅ Fi( s) − Fo ( s) − G2 ( s) VP ( s)

Block diagram of the loop:

3 s

G1 ( s) =

K1 τ ⋅s + 1

G2 ( s) =

K2 τ ⋅s + 1

2

m

m = 157.08 min K2 = 0.156 %CO 60s

Laplace transform and rearrange:

where

H( 0 ) = 0

+ H( t) = K1 ⋅ Fi( t) − K1 ⋅ Fo ( t) − K2 ⋅ VP ( t ) 2⋅ hd K2 := vpd

2⋅ hd K1 := k v ⋅ vpd

k v ⋅ vpd

Fi(s) - Fo(s) Cset(s) %TO

+

E(s)LC

Gc (s)

-

m3/sec M(s)

Gv(s)

%CO

C(s)

KT

%TO Proportional controller:

%VP

Valve positioner:

Gc( s) = Kc

Kp := KT⋅ K2

VP(s)

Gv ( s) =

KT = 50

H(s)

-

m

1 τ v⋅ s + 1

%CO m

%TO Kp = 7.801 %CO

Ku := KT⋅ K1

G2(s)

+

LT

100%TO KT := h max − h min

Level transmitter LT:

Let

G1(s)

Ku = 6.667 × 10

4 %TO⋅ s 3

m

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. Closed-loop transfer function:

C( s) =

−Kp ⋅ Kc

(τ v⋅ s + 1)⋅ (τ ⋅ s + 1) − Kp⋅ Kc

Characteristic equation of the loop:

Roots of the caracteristic equation:

C

set

( s) +

2

(

Ku ⋅ τ v ⋅ s + 1

)

(τ v⋅ s + 1)⋅ (τ ⋅ s + 1) − Kp⋅ Kc (

(

⋅ Fi( s) − Fo ( s)

)

)

τ v ⋅ τ ⋅ s + τ v + τ s + 1 − Kp ⋅ Kc = 0

r1 =

(

)

− τv + τ +

(τ v + τ )2 − 4⋅ τ v⋅ τ ⋅ (1 − Kp⋅ Kc) 2⋅ τ v⋅ τ

The term inside the radical: 2

2

(

τ v + 2 ⋅ τ v ⋅ τ + τ − 4 ⋅ τ v ⋅ τ + 4τ v ⋅ τ ⋅ Kp ⋅ Kc = τ v − τ

)2 + 4⋅ τ v⋅ τ ⋅ Kp⋅ Kc

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:

Kcmax :=

(

− τv − τ

)2

%CO Kcmax = −60.3 %TO

4 ⋅ τ v ⋅ τ ⋅ Kp

The equivalent time constants of the closed loop at this gain are two identical roots at: τ e :=

2⋅ τ v⋅ τ

τ e = 9.995 s

τv + τ

Offset for change in inlet flow:

3

Ku

Offset :=

0.001

1 − Kp ⋅ Kcmax

m

Offset = 0.141 %TO

s

Offset KT

= 2.826 × 10

−3

m

(b) Repeat part (a) modeling the tank as in section 7-3.1 (integrating process). The model now neglects the effect of the level on the flow out of the tank:

f1 ( t) = Cvmax⋅ vp( t) ⋅

3

∆p v

m

2

Substitute into mass balance:

min

= k v ⋅ h d ⋅ vp( t)

Gf 264.2gal 60sec π⋅D

⋅

4

d ⋅ h( t) dt

= fi( t) − fo ( t) − k v ⋅ h d ⋅ vp( t)

Express this linear equation in terms of deviation variables: 2

π⋅D 4 Laplace transform:

⋅

H( s) =

d ⋅ H( t) dt 4

= Fi( t) − Fo ( t) − k v ⋅ h d VP ( t ) 1

2 s

πD

(Fi(s) − Fo(s) − kv⋅

(

h d VP ( s)

H( 0 ) = 0

)

)

= G1 ( s) ⋅ Fi( s) − Fo ( s) − G2 ( s) VP ( s) where

G1 ( s) =

4

1

G2 ( s) =

2 s

πD

4 2

⋅

kv⋅ hd

π⋅D

s

The block diagram is the same as in part (a) with these transfer functions. Let

Kp :=

4 2

πD

k v ⋅ h d ⋅ KT

Ku :=

4 2

πD

KT

Kp = 8.278 × 10

Ku = 0.071

%TO

−4

%CO⋅ sec

%TO 3

%CO⋅ m

Closed-loop transfer function: C( s) =

−Kp ⋅ Kc

(

)

s⋅ τ v ⋅ s + 1 − Kp ⋅ Kc

C

set

( s) +

(

(

Ku ⋅ τ v ⋅ s + 1

)

(Fi( s) − Fo(s))

s⋅ τ v ⋅ s + 1 − Kp ⋅ Kc

2

Characteristic equation of the loop:

τ v ⋅ s + s − Kp ⋅ Kc = 0

Roots of the characteristic equation:

r1 =

Term in the radical:

)

−1 +

(

1 − 4 ⋅ τ v ⋅ −Kp ⋅ Kc

)

2⋅ τ v

1 + 4 ⋅ τ v ⋅ Kp ⋅ Kc

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: Kcmax :=

−1

%CO Kcmax = −60.4 %TO

4 ⋅ τ v ⋅ Kp

The equivalent time constants of the closed loop are identical for this gain (the radical is zero): τ e := 2 ⋅ τ v

The offset for the change in inlet flow is:

τ e = 10 s

Offset :=

Ku −Kp ⋅ Kcmax

3

0.001

m s

Offset = 0.141 %TO Offset KT

= 2.829 × 10

−3

m

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-25. Proportional-integral level control.

Fi(s) Cset(s)

+

%TO

E(s)LC

Gc(s)

-

Gc( s) = Kc⋅  1 +

PI controller:



1



τ I⋅ s



G1(s)

M(s)

G2(s)

%CO

G2 ( s) =

C(s)

+ -

Kv ⋅ KT

(

A⋅ s⋅ τ v ⋅ s + 1

%TO

KT G1 ( s) = A⋅ s

)

(a) Closed-loop transfer function, characteristic equation, and offset. Closed-loop transfer function:

C( s) =

−Gc( s) ⋅ G2 ( s) 1 − Gc( s) ⋅ G2 ( s)

C

set

( s) +

G1 ( s)

F ( s) 1 − Gc( s) ⋅ G2 ( s) i

Substitute and rearrange:

(

)

−Kc⋅ τ I⋅ s + 1 ⋅ Kv ⋅ KT

C( s) =

2

(

)

(

⋅C

)

set

⋅ ( s)

A⋅ τ I⋅ s ⋅ τ v ⋅ s + 1 − Kc⋅ τ I⋅ s + 1 ⋅ Kv ⋅ KT

(

KT⋅ τ I⋅ s⋅ τ v ⋅ s + 1

+

(

2

)

)

(

)

A⋅ τ I⋅ s ⋅ τ v ⋅ s + 1 − Kc⋅ τ I⋅ s + 1 ⋅ Kv ⋅ KT

⋅ Fi( s)

To obtain the steady-state transfer functions, set s = 0: ∆c =

−Kc⋅ Kv ⋅ KT 0 − Kc⋅ Kv ⋅ KT

∆c

set

+

0 0 − Kc⋅ Kv ⋅ KT

∆fi = ∆c

set

Offset = ∆c

set

− ∆c = 0

This means there is no offset for either set-point changes or disturbances. Characteristic equation of the loop:

3

2

A⋅ τ I⋅ τ v ⋅ s + A⋅ τ I⋅ s − Kc⋅ Kv ⋅ KT⋅ τ I⋅ s − Kc⋅ Kv ⋅ KT = 0

(b) Roots of the characteristic equation when the integral time is set equal to the valve time constant. Level response under these conditions. For

(

)

(

) (

)

A⋅ τ v ⋅ s τ v ⋅ s + 1 − Kv ⋅ KT⋅ Kc⋅ τ v ⋅ s + 1 = τ v ⋅ s + 1  A⋅ τ v ⋅ s − Kv ⋅ KT⋅ Kc = 0   2

τI = τv

r1 =

−1

−Kv ⋅ KT⋅ Kc

r2 = i⋅

τv

2

r3 = −i⋅

A⋅ τ v

−Kv ⋅ KT⋅ Kc A⋅ τ v

Root r 1 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: C( t) = A⋅ sin( ω ⋅ t + φ)

where

ω=

−Kc⋅ Kv ⋅ KT A⋅ τ v

(c) For negligible valve time constant, determine the limits of the controller gain for which the loop is oscillatory. Dominant time constant at high controller gain. Characteristic equation:

Roots:

r1 =

2

A⋅ τ I⋅ s − Kc⋅ Kv ⋅ KT⋅ τ I⋅ s − Kv ⋅ KT⋅ Kc = 0

Kc⋅ Kv ⋅ KT⋅ τ I +

(Kc⋅ Kv⋅ KT⋅ τ I)2 − 4A⋅ τ I⋅ (−Kv⋅ KT⋅ Kc) 2Aτ I

The response i oscillatory when the term inside the radical is negative (complex conjugate roots).

(Kc⋅ Kv⋅ KT⋅ τ I)2 + 4⋅ Kc⋅ Kv⋅ KT⋅ A⋅ τ I = Kc⋅ Kv⋅ KT⋅ τ I⋅ (Kc⋅ Kv⋅ KT⋅ τ I + 4A) 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: −Kc⋅ Kv ⋅ KT A A⋅ τ I For very high controller gains:

−Kc⋅ Kv ⋅ KT

<

4 τI

2

s + τ I⋅ s + 1 = τ I⋅ s + 1

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: C( s) = C

set

( s) +

1

τ I⋅ s

F ( s) −Kc⋅ Kv τ I⋅ s + 1 i

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-26. Level control of an evaporator. Problem data: lb fd := 800 min A := 10ft

ρ := 98

lb ft

2

Vapors

3

∆h T := 4ft

Valve sized for 100% overcapacity, linear.

Steam

τ v := 2sec

SP

LC

LT

Condensate

Disturbance, feed flow: ∆f := 10%⋅ fd

∆f = 80

lb

Feed

min

AT

Product

Assume ∆p v := 5psi

Solution: Block diagram of the level control loop:

Ws(s) lb/min

Fp(s) Cset(s) %TO

+

E(s)LC -

Gc(s)

M(s) %CO

C(s) %TO

ft3/min

Gv(s) KT

G2(s) G1(s)

Fi(s) ft3/min

- -

G1(s)

+

H(s) ft

LT

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.

100%TO

Level transmitter LT:

KT :=

Size the control valve:

Cvmax := 200%⋅

KT = 25

∆h T

fd 7.48gal ρ

ft

3

%CO

ρ ⋅ ft

⋅

Gc( s) = Kc

ft 3

Cvmax = 68.4

62.4lb⋅ ∆p v

Cvmax := 110

From Fig. C-10.1, page 532, a 3-in valve is required. ∆p v ⋅ 62.4⋅ lb ft3 Cvmax Kv := ⋅ 3 7.48gal 100%CO ρ ⋅ ft

Valve gain:

Gv ( s) =

Kv

1 G1 ( s) = A⋅ s

τ v⋅ s + 1

G2 ( s) =

E 1 ρ A⋅ s

gal min⋅ psi gal

min⋅ psi 3

ft Kv = 0.262 min⋅ %CO

E = evaporator economy, lb vapors/lb steam.

Closed-loop transfer function: C( s) =

(

Kc⋅ Kv ⋅ KT

)

A⋅ s⋅ τ v ⋅ s + 1 + Kc⋅ Kv ⋅ KT

C

Characteristic equation of the loop:

Roots of the characteristic equation:

set

( s) −

KT

 F ( s) + E W ( s)  p s A⋅ s⋅ ( τ v ⋅ s + 1 ) + Kc⋅ Kv ⋅ KT  ρ 

(

)

2

A⋅ s⋅ τ v ⋅ s + 1 + Kc⋅ Kv = A⋅ τ v ⋅ s + A⋅ s + Kc⋅ Kv ⋅ KT = 0 r1 =

r2 =

−A +

2

A − 4 ⋅ A⋅ τ v ⋅ Kc⋅ Kv ⋅ KT 2 ⋅ A⋅ τ v

−A −

2

A − 4 ⋅ A⋅ τ v ⋅ Kc⋅ Kv ⋅ KT 2 ⋅ A⋅ τ v

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: Kcmax :=

A 4 ⋅ τ v ⋅ Kv ⋅ KT

%CO Kcmax = 11.4 %TO

At this gain the term in the radical is zero and the effective time constants are identical to each other and equal to: τ e := 2 ⋅ τ v

τe = 4 s

Offset for a 10% change in the product and vapor flow: Offset KT

= 0.011 ft

Offset := 0 −

−KT   ∆f   Kcmax⋅ Kv ⋅ KT  ρ

Offset = 0.272 %TO

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-27. Sythesize controller for process with inverse response. G( s) =

Solution:

(

)

K⋅ 1 − τ 3 ⋅ s

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1)

C( s )

1 Gc( s) = G( s)

Synthesis formula, Eq. 7-4.2, page 258:

C( s)

For teh standard closed-loop response:

The controller transfer function is:

R( s)

Gc( s) =

=

R( s )

1−

C( s) R( s)

1 τ c⋅ s + 1

(τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1) ⋅ 1 τ c⋅ s K⋅ ( 1 − τ 3 ⋅ s)

This contains a positive pole which results in an unstable response when it doesn't exactly match the time constant of the transfer function. To avoid a postive pole in the controller, define:

Substitue and reaarrange:

Gc( s) =

Gc( s) =

C( s) R( s)

=

1 − τ 3⋅ s τ c⋅ s + 1

1 − τ 3⋅ s (τ 1⋅ s + 1) ⋅ (τ 2⋅ s + 1) K⋅ ( 1 − τ 3 ⋅ s) τ c⋅ s + 1 − 1 + τ 3 ⋅ s τ1

1 + 1  τ ⋅s + 1 (2 ) K⋅ ( τ c + τ 3 )  τ 1⋅ s  

This is a series PID controller with the following tuning formulas: K´c =

(

τ1

K⋅ τ c + τ 3

)

τ´I = τ 1

τ´D = τ 2

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 7-28. Simulation of temperature control loop for the non-isothermal reactor of Section 4-2.3. 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. Open-loop response to a 2%CO step change in the signal to the valve at 1 minute: ∆c := ( 68.9 − 64.8)%TO ∆c = 4.1 %TO ∆c Kp := 2%CO %TO Kp = 2.05 %CO

Fit 3 model:

64.8%TO + 0.283∆c = 65.96 %TO

t1 := ( 6.5 − 1 )min

64.8%TO + 0.632∆c = 67.391 %TO

t2 := ( 12 − 1 )min

(

τ := 1.5⋅ t2 − t1

)

t0 := t2 − τ

τ = 8.25 min

G( s) =

2.05e

t0 = 2.75 min − 2.75s

8.25s + 1

%TO %CO

(a) Quater-decay ratio tuning of series PID temperature controller. From Table 7-2.1:

K´c :=

1.2  t0 

 Kp  τ 

τ´I = 5.5 min

−1

τ´I := 2 ⋅ t0 τ´D = 1.4 min

t0 τ´D := 2

K´c = 1.8

%CO %TO

%CO  1  Gc( s) = 1.8 ( 1.4s + 1 ) 1 + %TO  5.5s 

Response to a -0.2 ft3 /min change in rectants flow at 1 min:

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).

(b) Synthesis tuning with τc = 0 of series PID temperature controller. From Table 7-4.1: K´c :=

τ Kp ⋅ t0

τ´I = 8.3 min

τ´I := τ τ´D = 1.4 min

t0 τ´D := 2

K´c = 1.5

%CO %TO

%CO  1  Gc( s) = 1.5 ( 1.4s + 1 ) 1 + %TO  8.3s 

By comparison, the gain is slightly smaller and the integral is 60% slower than for quarter-decay ratio response.

The responses to a -0.2 ft 3 /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.

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Smith & Corripio, 3rd edition Problem 12-1. Loop interaction for the processes of Fig. 12-1.1. (a) Blending tank, Fig. 12-1.1(a). Manipulated variables: flows of each of the two inlet streams. Controlled variables: product flow and composition. Increasing the flow of either inlet stream increases the product flow. Increasing the flow of the more concentrated stream increases the product composition, while increasing the flow of the more dilute stream decreases the product composition. This means the interaction is positive (the two loops help each other): when the flow of the stream controlling the product composition increases, the flow controller decreases the flow of the other stream; as the two inlet streams have opposite effects on the product composition, the change caused by the product flow controller changes the composition in the same direction.

(b) Chemical reactor, Fig. 12-1.1(b). Manipulated variables: coolant flow and reactants flow. Assume the product flow is manipulated to control the level in the reactor. Controlled variables: flow and reactant composition in the product stream. As the reactor is cooled, the reaction must be exothermic. • •

The coolant flow has a negative effect on the temperature and no direct effect on the composition. The reactants flow has a positive effect on the reactant composition and, if the feed is at a lower temperature than the reactor, a negative effect on temperature.

So, at first glance it appears that there is no intercation between the loops, but the interaction comes through the effect or reactant composition and temperature on each other through the reaction rate. When the reactants composition is not controlled, an increase in coolant flow causes a decrease in reactor temperature; this decreases the reaction rate and increases the reactants concentration, resulting in a higher reaction rate and temperature than if the reactants composition were kept constant. So, controlling the composition constant results in a smaller change in reactor temperature when the coolant flow is changed. Similarly, if the temperature is not controlled, an increase in reactants flow increases in reactants concentration and the reaction rate resulting in an increase in temperature, so that the increase in reaction rate is higher than if the temperature is kept constant. So, controlling the temperature constant results in a larger increase in composition when the temperature is allowed vto increase. So, for this particular case, it appears that the interaction is negative with respect to the temperature loop, and positive with respect to the composition loop. This is unusual.

(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.

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mass% := %

Smith & Corripio, 3rd edition

klb := 1000lb

Problem 12-2. Control of caustic dilution process. Problem data:

SP

klb

wP := 40 hr

AC 3

SP

x P := 30mass%

FC

x C := 50mass%

SP FC

1

3

FT

Caustic

Assumptions: • Perfect mixing • Constant mass

AT

SP

FC

xP

2

FT

wP

FT

Manipulated variables: Flows of water and caustic wC

and

w Water

ww

Controlled variables: Product flow and mass fraction:

Steady state model:

wP

and

Total mass balance:

xP wC + ww = wP

Caustic balance:

wC⋅ x C = wP⋅ x P

Solve for x P:

xP =

wC⋅ x C wP

=

(pure water diluent) wC⋅ x C wC + ww

Open-loop gain matrix: wC

ww wP

 δwP   δwC KOL =   δxP  δw  C

Relative gain matrix:

δwP



1 1 1    1   w ⋅ x − w ⋅ x −w ⋅ x  w ⋅ x −w ⋅ x C C C C= w C C C = P C δx P   2 2 2  w 2 wP wP wP P     δww 

δww 

wC

ww

xP

−ww⋅ x C ww −wC⋅ x C    wC   −wC⋅ xC − ww⋅ xC −wC⋅ xC − ww⋅ xC   wC + ww wC + ww  µ = =  −ww⋅ x C −wC⋅ x C wC    ww   −wC⋅ xC − ww⋅ xC −wC⋅ xC − ww⋅ xC  wC + ww wC + ww    

wP

xP

So, the result is as for the regular blender: the pairing that minimizes the interaction is the one in which the largest of the two inlet flows is used to control the product flow, and the smaller inlet flow is used to control the product mass fraction. The interaction is positive (the loops help each other) as the relative gains are positive.

Numerical results:

wC :=

wP⋅ x P

klb ww = 16 hr

ww := wP − wC

xC

Open loop gains:

wC

   w ⋅x − w ⋅ x C C klb klb KOL :=  w C  2 hr⋅ mass%  w 2 hr⋅ mass% wP  P  1

Relative gains:

1

 wC  wP µ :=   ww  wP 

  wC  wP

wP

ww

1  1 KOL =   0.5 −0.75 

wC

ww  µ =

klb wC = 24 hr

wP xP

ww

 0.6 0.4    0.4 0.6 

wP xP



So, in this case, to minimize interaction, • the caustic stream flow, FC-1, must be manipulated to control the product flow • the water inlet flow, FC-2, must be used to control the product composition.

When the flow control loop is closed, the gain of the composition loop increases by 1 µ 1, 1

− 1 = 66.7 %

The gain is then:

KOL 1, 1 µ 1, 1

= −1.25

mass%⋅ hr klb

Notice that the gain of the composition loop is negative, so a direct acting controller is required for the composition loop. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

degF := R

Smith & Corripio, 3rd edition

gal

gpm :=

min

Problem 12-3. Automatic control of a household shower. Problem data: fo := 3gpm T1 := 170degF ρ := 8.33

lb gal

SP

To := 110degF

SP

T2 := 80degF

TC TE

cp := 1

fo To

FE

FC

BTU lb⋅ degF

Assume • Negligible time delay • Constant properties • Reference temperature of 0ºF

S

S T1 f1 Hot water

T2 f2 Cold water

(a) Develop the model of the process Total mass balance:

f1 ⋅ ρ + f2 ⋅ ρ = fo ⋅ ρ

fo = f1 + f2

Enthalpy balance:

f1 ⋅ ρ ⋅ cp ⋅ T1 + f2 ⋅ ρ ⋅ cp ⋅ T2 = fo ⋅ ρ ⋅ cp ⋅ To To =

f1 ⋅ T1 + f2 ⋅ T2 f1 + f2

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

(

)

f1 ⋅ T1 + fo − f1 T2 = fo ⋅ To To − T2 f1 := fo ⋅ T1 − T2

Open-loop gains:

f2 := fo − f1

f1 = 1 gpm

f2 = 2 gpm

 δfo   δf1 KOL =   δTo  δf  1

δfo



1    f + f ⋅T − f ⋅T − f ⋅T  =  ( 1 2) 1 1 1 2 2 δTo   2 f1 + f2 ) (  δf1  δf2

(

f1 + f2 ⋅ T2 − f1 ⋅ T1 − f2 ⋅ T2  

)

(f1 + f2)

f1

1 1   f ⋅ T − T  2 ) gpm −f1 ⋅ ( T1 − T2 ) gpm  KOL :=  2 ( 1 2 2 degF degF   fo fo  



1

 

2

f2

1 1  KOL =   20 −10 

fo To

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

(c) Relative gains and pairing that minimize interaction. f1

f2

−f2 ⋅ ( T1 − T2 ) f2    f1  −f1⋅ ( T1 − T2 )   f  −( f1 + f2) ( T1 − T2) −( f1 + f2) ( T1 − T2)   f1 + f2 f1 + f2  o µ = =  −f1 ⋅ ( T1 − T2 ) f1  −f2⋅ ( T1 − T2 )   f2  T o  −( f1 + f2) ( T1 − T2) −( f1 + f2) ( T1 − T2)   f1 + f2 f1 + f2     The pairing that minimizes interaction is the total flow with the larger of the two inlet streams (in this case the cold water, f2 ) and the tempearture with the smaller stream (in this case the hot water, f 1 ).

 f1  fo µ :=   f2  fo 

 fo   f1  f2

fo

f1 µ =

f2

 0.333 0.667    0.667 0.333 

fo To



The interaction is positive (the loops help each other), as the relative gains are positive. When one loop is closed, the gain of the other loop increases by: 1 µ 1, 0

− 1 = 50 %

The closed-loop gains are:

KOL 0, 1 µ 0, 1

= 1.5

KOL 1, 0 µ 1, 0

= 30

degF gpm

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 12-4. Control of an evaporator. Problem data: Manipulated variables: Product and steam valve signals. mP

and

wv

mS mS

Controlled variables: Feed flow (throughput) Product composition wF

and

Vapors

wS

SP FC

Steam

SP

xP

LC

Assume the evaporator economy E is constant.

wF

SP

LT

AC

FT

Feed

mP

xP

xF

AT

wP

Product

(a) Steady-state model of the evaporator. Total mass balance:

wF = wP + wv

Solute balance:

wF⋅ x F = wP⋅ x P

Combine and rearrange:

wF = wP + E⋅ wS

Control valve gains:

wP = KvP⋅ mP

wv = E⋅ wS

Economy:

x P = x F⋅

(wP + E⋅ wS) wP

E⋅ wS   = x F⋅  1 + wP  

wS = KvS⋅ mS

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

mP

mS

Open-loop gains:

 δwF   δmP KOL =   δx P  δm  P

δwF



KvS⋅ E  KvP    − E ⋅ w  E  S = KvS⋅ x F⋅  δx P  KvP⋅ x F⋅  wP   w 2    P   δmS 

δmS 

wF

xP

Relative gains:

2  KvP⋅ KvS⋅ x F⋅ E ⋅ wS KvP⋅ KvS⋅ x F⋅ E   E⋅ wS  E⋅ wS   1 2  wP⋅ KvP⋅ Kvs⋅ xF⋅ E⋅  1 + + wP ⋅ KvP⋅ Kvs ⋅ x F⋅ E⋅    wP w 2  wP w 2 P  P     µ =  2 KvP⋅ KvS⋅ x F⋅ E ⋅ wS KvP⋅ KvS⋅ x F⋅ E   2 E⋅ wS  E⋅ wS   1  1 wP⋅ KvP⋅ Kvs ⋅ x F⋅ E⋅   wP ⋅ KvP⋅ Kvs⋅ x F⋅ E⋅  + +  wP w 2  wP w 2  P  P    

mP

           

mS

wP E⋅ wS    wP + E⋅ wS wP + E⋅ wS  µ =  wP  E⋅ wS   wP + E⋅ wS wP + E⋅ wS  

wF

xP

The interaction is positive (the loops help each other), as the realative gains are positive.

(c) General pairing strategy. The pairing the minimizes interaction is: • if the product flow is larger than the vapor flow (E*W S), control throughput with the signal to the valve on the product line and the composition with the signal to the steam valve • if the product flow is smaller than the vapor flow (E*WS), control throughput with the signal to the steam valve and the composition with the signal to the product valve. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 12-5. Multivariable control of evaporator of Problem 12-4. x F := 30mass%

Problem data:

x P := 50mass%

klb wF := 80 hr

E := 0.9

klb klb

Combine balance equations to solve for design conditions: xF wF − wP klb klb wP := wF⋅ wS := wP = 48 wS = 35.556 xP E hr hr From the solution to Problem 12-4, ignoring the valve gains: Open-loop gains: 1 E    −x ⋅ E⋅ w x F⋅ E S klb klb KOL :=  F  2 w hr ⋅ mass% hr ⋅ mass%  w P P   Relative gains: wP E⋅ wS    wP + E⋅ wS wP + E⋅ wS  µ :=   wP  E⋅ wS   wP + E⋅ wS wP + E⋅ wS  

wP

wS wF xP

0.9   1 KOL =   −0.417 0.563 

The first row is in klb/klb, and the second is in mass%/(klb/hr). wP µ =

wS

 0.6 0.4  wF   0.4 0.6  xP

Pairing that minimizes interaction is • control throughput with the product rate • control composition with steam rate

When the feed flow loop is closed, the product composition loop gain changes by 1 µ 1, 1

− 1 = 66.7 %

The closed-loop gain of the composition loop is: KOL 1, 1 µ 1, 1

mass%⋅ hr = 0.938

klb

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 12-6. Distillation product composition control. Manipulated variables: Steam and reflux flows

AT

Controlled variables: Bottoms and distillate compositions

PC

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

LC 1 Condenser

LT

PT AC 1

Feed

Reflux

Distillate

AC 2 LC 2

LT

Steam Bottoms

AT

Results of tests performed on a smlation of the column: Base Test 1 Test 2

Case

wS := ( 24.0 25.0 24.0 )

Steam flow, klb/hr Reflux flow, lbmole/hr

wR := ( 70.0 70.0 75.0 )

Butane mole %

x B := ( 6.22 3.08 8.77 ) y D := ( 93.50 91.79 96.88 )

Open-loop steady state gains:  x B0 , 1 −   wS − 0, 1 KOL :=   yD −  0, 1  wS −  0, 1 Relative gains:

xB

0, 0

xB

0, 2

− xB 0, 0



wS 0, 0

wR − wR  0, 2 0, 0 

yD

yD

wS 0, 0

wR − wR 0, 2 0, 0 

0, 0

0, 2

− yD 0, 0

 

wS

wR

 −3.14 0.51  KOL =   −1.71 0.676 

xB yD

The first column is in mole%/(klb/hr) and the second in mole%/(lbmole/hr).

− KOL ⋅ KOL    KOL0 , 0⋅ KOL1 , 1 0, 1 1 , 0    K K   OL OL µ :=   KOL ⋅ KOL  − KOL0 , 1⋅ KOL1 , 0 0, 0 1, 1    KOL KOL  

wS µ =

wR

 1.697 −0.697  xB   −0.697 1.697  yD

The interaction is negative (loops fight each other): relative gains are either negative or greater than unity. Pairing the minimizes interaction: • control distillate composition with reflux rate • control bottoms composition with steam rate

Closed-loop gain of distillate control loop:

KOL 1, 1 µ 1, 1

= 0.398

mole%⋅ hr lbmole

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition

kbl := 42000gal

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

Octane

1. Alkylate 2. Straight run x := 3. Reformate

RVP

 97.0   80.0 y :=   92.0 

 5.00   11.00   3.00 

Gasoline x p := 89.0

SP

f1

FT

SP

SP

FC

AC

AC

1

2

3

AT

AT

FT

x

y

f

SP

Alkylate

y p := 7.00

FC

kbl fp := 60.0 day

f2

FT

SP

Straight run

Assume • gasoline properties are a linear combination of the feed properties weighted by volume • Constant standar density

FC f3

FT

Reformate

Required inlet flows: xP =

 x0 x1 x2   f :=  y y y  0 1 2  1 1 1

SP

FC

−1

f1 ⋅ x 1 + f2 ⋅ x 2 + f3 ⋅ x 3 f1 + f2 + f3

 xp   fp ⋅ day  yp  kbl  1

yp =

 22.5  f =  24.375   13.125 

f1 ⋅ y 1 + f2 ⋅ y 2 + f3 ⋅ y 3 f1 + f2 + f3

kbl day

Steady-state open-loop gains:  f 1⋅ ( x0 − x 1) + f 2⋅ ( x0 − x 2) f 0⋅ ( x 1 − x0) + f 2⋅ ( x 1 − x2) f 0⋅ ( x2 − x 0) + f 1⋅ ( x2 − x 1)    2 2 2   2 day  2 day  2 day  fp ⋅  fp ⋅  fp ⋅     kbl   kbl   kbl    KOL :=  f ⋅ ( y − y ) + f ⋅ ( y − y ) f ⋅ ( y − y ) + f ⋅ ( y − y ) f ⋅ ( y − y ) + f ⋅ ( y − y )  1 0 1 2 0 2 0 1 0 2 1 2 0 2 0 1 2 1   2 2 2   day day day 2 2 2    fp ⋅  fp ⋅  fp ⋅     kbl   kbl   kbl    1 1 1  

Alk St run f1 f2

Ref f3

 0.133 −0.15 0.05  KOL =  −0.033 0.067 −0.067  1 1   1

day kbl

xp

Oct

yp

RVP

fp

Flow

Relative gains: 11.25 0.375   7.5 kbl  B = −1.875 4.688 0.406  day  −5.625 −15.938 0.219 

−1

B := KOL

Inverse of the open-loop gains:

 KOL0 , 0⋅ B0 , 0 KOL0 , 1⋅ B1 , 0 KOL0 , 2⋅ B2 , 0   µ :=  KOL1 , 0⋅ B0 , 1 KOL1 , 1⋅ B1 , 1 KOL1 , 2⋅ B2 , 1     KOL ⋅ B ⋅B ⋅B K K 2 , 0 0 , 2 OL2 , 1 1 , 2 OL2 , 2 2 , 2  

Alk f1

St run f2

Ref f3

0.281 −0.281   1  µ = −0.375 0.313 1.062   0.375 0.406 0.219 

x p Oct y p RVP fp Flow

Pairing to minimize interaction: • Use alkylate (stream 1) to control the gasoline octane • Use the reformate (stream 3) to control the gasoline Reed vapor pressure • Use the straight run (stream 2) to control the product flow. These are the same pairings as in Example 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

 B0 , 0  B0 , 0   B2 , 0 D :=   B0 , 0 B  1, 0  B0 , 0 

B B B B B B

0, 1 2, 1 2, 1 2, 1 1, 1 2, 1

0, 2  B  1, 2  B 2, 2   B 1, 2   B 1, 2 

RVP

Flow

m2

m3

B

B

1, 2



m1

 1 −0.706 0.923  D =  −0.75 1 0.538   −0.25 −0.294 1 

f1 f3

Alkylate

f2

Straight run

Reformate

The instrumentation diagram is the same as Fig. 12-3.6.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 12-8. Control of a wet grinding circuit. Manipulated variables: SF = solids flow MW = mill water flow SW = slurry tank water Controlled variables: TOR = mill torque FML = flow from mill DCF = cyclone feed density

Cyclone

TOR

XT

XC FC FML

SW

FT

Mill

SF

AC AT DCF

MW

Models from open-loop tests by Hubert and Woodburn (1983): SF, kg/s Time is in sec

MW, kg/s

SW, kg/s

153 −21   119 337 ⋅ s + 1 10⋅ s + 1   217 ⋅ s + 1  0.37 0.767 −0.050  G( s) =  500 ⋅ s + 1 33⋅ s + 1 10⋅ s + 1    − 320s  0.930 −0.667 e −1.033   500 ⋅ s + 1 166s + 1 47s + 1 

TOR, N-m FML, liters/s DCF, kg/liter

(a) Relative gains and pairings that minimize interaction. Obtain the open-loop steady-state gains by setting s = 0 in the transfer functions: −21   119 153  KOL := 0.37 0.767 −0.050   0.930 −0.667 −1.033 

−1

B := KOL

Obtain the relative gains by Eq. 12-2.13, page 422:

 0.03089 −6.43731 −0.31641  B =  −0.01256 3.8685 0.06809   0.03592 −8.29331 −1.29688 

 KOL0 , 0⋅ B0 , 0 KOL0 , 1⋅ B1 , 0 KOL0 , 2⋅ B2 , 0   µ :=  KOL1 , 0⋅ B0 , 1 KOL1 , 1⋅ B1 , 1 KOL1 , 2⋅ B2 , 1     KOL ⋅ B KOL ⋅ B KOL ⋅ B 2, 0 0, 2 2, 1 1, 2 2, 2 2, 2  

SF kg/s

MW kg/s

SW kg/s

 3.676 −1.922 −0.754  TOR, N-m FML, liters/s µ =  −2.382 2.967 0.415   −0.294 −0.045 1.34  DCF, kg/liter

Pairing to minimize interaction: • Control the mill torque TOR with the solids flow SF • Control the flow from the mill FML with the water flow to the mill MW • Control the density of the cyclone feed DCF with the slurry water flow SW

(b) Design a decoupler and draw the block diagram and the instrumentation diagram of the decoupled system. Because of the complexity of the transfer functions, and the fact that five of the six interaction terms are negative, a full dynamic decoupler will probably create unstable poles. So, a static decoupler is designed. If dynamic compensation is required, 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:

 1    B1 , 0 D :=   B0 , 0 B  2, 0  B0 , 0 

B B

0, 1 1, 1

1 B B

2, 1

  B 2, 2  B 1, 2   B 2, 2   1  B

0, 2

1, 1

Decoupled system gains:

mTOR mFML mDCF SF −1.664 0.244   1  D = −0.407 1 −0.053  1   1.163 −2.144

set

MW SW

set

set

All variables in kg/sec

 mTOR mFML mDCF kg/sec kg/sec kg/sec

−0  TOR, N-m  32.372 0 FML, liters/sec KOL⋅ D =  0 0.258 0  0 −0.771  DCF, kg/liter  0 Block diagram of decoupled system:

R1(s)

-

+

GC1(s)

M1(s)

SF

+ -

+

1.664

R2(s)

-

+

GC2(s)

G13(s)

0.407

G21(s)

M2(s)

+ -

MW

-

-

GC3(s)

+ + G22(s) + FML G31(s) G32(s)

2.144 - + M3(s) +

Instrumentation diagram of decoupler:

+

TOR

G23(s)

1.163

+

+ +

G12(s)

0.244

0.053

R3(s)

G11(s)

SW

+ G33(s) +

+

DCF

FMLSP

TORSP

XC

-1.664

3

XY

DCFSP

FC

-0.407

0..244

3 FY

AC

-2.144 -0.053

1.163

AY 3

TOR

XT

FML

SW

FT SF MW

Mill

AT DCF

The summer devices, XY, FY, and AY, are the decouplers. Each signal into the summer is multiplied by the factor shown by it (unity if no factor is shown). The factors are not corrected for the valve gains. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Problem 12-9. Decoupler design for distillation product composition control. Manipulated variables: Reflux flow Steam flow to reboiler Controlled variables: Distillate purity Bottoms product purity

mR mS yD

AT

PC

Condenser

xB

xB

Reflux

AC 1

Distillate

AC 2

= composition of heavy key in the distillate = composition of the light key in the bottoms product

LT

PT

Feed yD

LC 1

LC 2

LT

Steam Bottoms

AT

Transfer functions from open-loop step tests (time parameters in hours): 0.75 1 − 0.11s  YD( s) = −3.0 M R( s) + M ( s) 1 + 0.35s S  1 + 0.35s  2.5 1 + 0.25s  XB( s) = M R( s) − 3.5 M S( s) 1 + 0.35s  1 + 0.35s  %TO := %

(a) Relative gains and pairing that minimizes interaction.

%CO := %

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

mS

 −3.0 0.75  %TO yD   2.5 −3.5  %CO xB

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

 KOL0 , 0⋅ KOL1 , 1  KOL  µ :=   ( −KOL) 0 , 1⋅ KOL1 , 0  KOL 

(−KOL)0 , 1⋅ KOL1 , 0  KOL KOL ⋅ KOL 0, 0 1, 1 KOL mR

µ =

     

mS

 1.217 −0.217  yD   −0.217 1.217  xB

The interaction is negative (loops fight each other): relative gains are negative or greater than one. Pairing that minimizes interaction: • Control distillate purity with reflux flow • Control bottoms purity with steam flow

(b) Design the block diagram and draw the block diagram for this system. 1 − 0.11⋅ s  0.75 YD( s) = −3.0⋅  + ⋅ D1 ( s) ⋅ M S( s) = 0 1 0.35 s 1 + 0.35 ⋅ s + ⋅     XB( s) =

 2.5 ⋅ D ( s) − 3.5⋅  1 + 0.25⋅ s   ⋅ M ( s) = 0  1 + 0.35⋅ s 2  1 + 0.35⋅ s  R   

3.0 D1 ( s) = ( 1 − 0.11s) 0.75 3.5 D2 ( s) = ( 1 + 0.25s) 2.5

The first decoupler required a negative lead that will produce an undersirable result in the reflux flow, so we will use a simple gain. The second decoupler can be implemented, but a small lag must be included. %CO D1 ( s) = 4.0 %CO

%CO  1 + 0.28s  D2 ( s) = 1.4  %CO  1 + 0.03s 

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

R1(s)

-

+

MR(s) D1(s)

G11(s) + + YD(s) G12(s)

D2(s)

G21(s)

+ +

R2(s) +

Gc1(s)

-

Gc2(s)

MS(s)

+ +

G22(s)

+ XB(s) +

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Smith & Corripio, 3rd edition Problem 12-10. Decoupler design for a 2x2 process. C( s) = GP( s) M ( s) + GPU ( s) ⋅ U( s)

 0.81e− 0.6s 1.2e− 1.1s   1.4s + 1 2.4s + 1  GP( s) =   − 0.3s −s  0.9e  1.1e 2s + 1   1.5s + 1

 0.5   2.2s + 1 GU( s) =    −1.5  1.8s + 1 

(a) Relative gains and pairing that minimizes interaction. Obtain open-loop steady-state gains by setting s = 0 in the transfer functions:

KOL :=

 0.81 1.2    1.1 0.9 

Relative gains by Eq. 12-2.7, page 418:

 KOL0 , 0⋅ KOL1 , 1 −KOL0 , 1⋅ KOL1 , 0   KOL KOL   µ :=    −KOL0 , 1⋅ KOL1 , 0 KOL0 , 0⋅ KOL1 , 1   KOL KOL  

m1

µ =

m2

 −1.234 2.234  c1   2.234 −1.234  c2

The interaction is negative (loops fight each other): relative gains are either negative or greater than unity. Pairing that minimizes interaction: • Control c1 with m2 •

Control c2 with m1

(b) Design decoupler and show in the block diagram.  GP1 , 2( s) ⋅ D1 ( s) + GP1 , 1( s)  M1 ( s) = 0    GP2 , 2( s) + GP2 , 1( s) ⋅ D2( s)  M2 ( s) = 0  

−GP ( s) 1, 1

D1 ( s) = = GP ( s) 1, 2

−GP ( s) 2, 2

−0.81 2.4s + 1 − ( 0.6− 1.1)s e 1.2 1.4s + 1

−0.9 1.5s + 1 − ( 1− 0.3)s D2 ( s) = = e 1.1 2.0s + 1 GP ( s) 2, 1

The first decoupler term cannot be implemented because it requires a negative delay which would require knowledge of the signal in the future. So, we will add the negative delay of 0.5 min to the lead in the lead lag unit, making the new lead: ( 2.4 − 0.6 + 1.1)min = 2.9 min

The decouplers are then: 2.9s + 1 D1 ( s) = −0.675 1.4s + 1 1.5s + 1 − 0.7s D2 ( s) = −0.818 e 2.0s + 1

Block diagram of the complete system:

R1(s) +

-

Gc1(s)

M2(s)

+

GP12(s)

+ GP11(s)

D1(s) U(s)

+ + C1(s) +

GU1(s) GU2(s)

D2(s) R2(s) +

-

Gc2(s)

M1(s)

GP22(s) + +

GP21(s)

+ +

+C2(s)

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Smith & Corripio, 3rd edition Problem 12-11. Decoupling of 2x2 process of Fig. 12-3.2--show decoupled closed-loop transfer functions The decouplers are given in Eq. 12-3.2, page 426: −Gv2( s) ⋅ G12( s) D12( s) = Gv1( s) ⋅ G11( s)

−Gv1( s) ⋅ G21( s) D21( s) = Gv2( s) ⋅ G22( s)

From the block diagram, the transfer functions are:

(

)

C1 ( s) = Gv1( s) ⋅ G11( s) + D21( s) ⋅ Gv2( s) ⋅ G12( s) ⋅ M 1 ( s)

(

)

+ Gv2( s) ⋅ G12( s) + D12( s) ⋅ Gv1( s) ⋅ G11( s) ⋅ M 2 ( s) If the decouoplers can be implemeted exactly, then: −Gv1( s) ⋅ G21( s)   C1 ( s) =  Gv1( s) ⋅ G11( s) + ⋅ Gv2( s) ⋅ G12( s) ⋅ M 1 ( s) Gv2( s) ⋅ G22( s)   −Gv2( s) ⋅ G12( s)   +  Gv2( s) ⋅ G12( s) + ⋅ Gv1( s) ⋅ G11( s) ⋅ M 2 ( s) Gv1( s) ⋅ G11( s)   Simplify:

G21( s)   C1 ( s) = Gv1( s) ⋅  G11( s) − ⋅ G12( s) ⋅ M 1 ( s) G22( s)  

So, the system would be decoupled.

Similarly for the other variable: From the block diagram, the transfer functions are:

(

)

C2 ( s) = Gv2( s) ⋅ G22( s) + D12( s) ⋅ Gv1( s) ⋅ G21( s) ⋅ M 2 ( s)

(

)

+ Gv1( s) ⋅ G21( s) + D21( s) ⋅ Gv2( s) ⋅ G22( s) ⋅ M 1 ( s) If the decouoplers can be implemeted exactly, then: −Gv2( s) ⋅ G12( s)   C2 ( s) =  Gv2( s) ⋅ G22( s) + ⋅ Gv1( s) ⋅ G21( s) ⋅ M 2 ( s) Gv1( s) ⋅ G11( s)   −Gv1( s) ⋅ G21( s)   +  Gv1( s) ⋅ G21( s) + ⋅ Gv2( s) ⋅ G22( s) ⋅ M 1 ( s) Gv2( s) ⋅ G22( s)  

Simplify:

G12( s)   C2 ( s) = Gv2( s) ⋅  G22( s) − ⋅ G21( s) ⋅ M 2 ( s) G11( s)  

So, the system would be decoupled.

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Smith & Corripio, 3rd edition Problem 12-12. Decoupler design for blending tank of Example 12-5.1. Design conditions: x 1 := 10mass% w := 100

SP

AC

x 2 := 30mass%

lb

SP

FC

x := 20mass%

min

m1

Valves are linear with constant pressure drop. lb wmax := 150 min

w1 x1

V

m

2

AT

FT

x

w

Analyzer transmitter AT: x min := 5mass%

x max := 35mass%

At the design conditions:

w2 x2

x − x1 w1 := w⋅ x2 − x1

w2 := w − w1

lb w1 = 50 min

lb w2 = 50 min

So, there is no best pairing and the relative gains are 0.5 for both pairings. As there are no flow transmitters on the inlet flows, ratio control cannot be implemented. We design then a linear decoupler. −Kv2 w = w1 + w2 = Kv1⋅ m1 + Kv2⋅ m2 ∆w = Kv1⋅ ∆m1 + Kv2⋅ ∆m2 = 0 ∆m1 = ∆m2 Kv1 w1 ⋅ x 1 + w2 ⋅ x 2 x= w1 + w2

(w1 + w2)x1 − w1⋅ x1 − w2⋅ x2 K ⋅ ∆m + (w1 + w2)x2 − w1⋅ x1 − w2⋅ x2 K ⋅ ∆m = 0 v1 1 v2 2 (w1 + w2)2 (w1 + w2)2

∆x =

(

)

(

)

w2 ⋅ x 1 − x 2 ⋅ Kv1⋅ ∆m1 + w1 ⋅ x 2 − x 1 ⋅ Kv2⋅ ∆m2 = 0 wmax Kv1 := 100%CO

The gains of the decouplers are:

wmax Kv2 := 100%CO

−Kv1 Kv2

= −1

∆m2 =

lb Kv1 = 1.5 min⋅ %CO Kv1⋅ w2 Kv2⋅ w1

=1

Kv1⋅ w2 Kv2⋅ w1

∆m1

lb Kv2 = 1.5 min⋅ %CO

Instrumentation diagram: FY: summer, keeps flow constant when the analyzer controller varies m2 AY: summer, keeps composition constant when the flow controller varies m1

-

Kv1 w2 Kv2

= 33.33 %CO

AC SP

m1

3 FY +

FC

+

3 AY

w1 x1

Both require a bias of: w1

SP

m2

+ V

AT

FT

x

w

w2 x2

= 33.33 %CO

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Smith & Corripio, 3rd edition Problem 12-13. Non-linear static decoupler for evaporator of Problem 12-4. From the solution of Problem 12-4:

wv

wF = wP + E⋅ wS

xP =

Vapors

x F⋅ wF

mS

wP

wS

SP FC

wP = KvP⋅ mP

SP

LC

wS = KvS⋅ mS

Assume the vapor rate is greater than the product rate, so that the steam is used to control the feed rate and the product to control the composition.

wF

Steam SP

LT

AC

FT

Feed

mP

xP

xF

AT

wP

Product

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

= wP + E⋅ wS

wF

So

set

wS =

wF

− wP

E

set

KvP mS = − m E⋅ KvS P E⋅ KvS wF

Now, the feed flow setpoint is obtained using a scale factor on the controller output in %CO. This scale factor is obtained from the maximum expected feed rate over 100%CO: set

wF

=

wFmax 100%CO

mS =

wFmax    100%CO⋅ E⋅ K mFC − vS  

 KvP   E⋅ K mP  vS 

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

xP

wF = x F⋅ wP

wP =

xF set

xP

⋅ wF

This is a ratio controller with the ratio adjustable by the output of the composition controller AC: xF where, at design conditions: wP = Ratio⋅ wF Ratio = set xP

This would work best if a flow controller is installed on the product line, but this is probably too much of an expense for an evaporator. So, the ratio of the signals are used:

mP = Ratio⋅ mFC

Ratio =

where

xF

wFmax

set 100%CO⋅ KvP

xP

Instrumentation diagram: FY: summer, subtracts the scaled signal to the product valve from the signal to the seam valve to maintain the feed flow constant.

SP AC SP FC

Ratio

Vapors

mFC

AY

AY: multiplier, makes the signal to the product valve proportional to the total signal from the feed controller, scaled, to make the product flow proportional to the feed flow and mintain the product com position constant.

X

FY 3 mS

SP

LC wF

mP

wS Steam

LT

FT

Feed

xF

xP wP

AT Product

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Chapter 13. Simulation of Process Control Systems Solutions to Problems 13-1 to 13-17

13-1. Simulation of Bird Mobile of Problem 2-9. The differential equation representing the motion of the bird mobile is, from Problem 2-9: 2

M⋅

d y ( t) 2

= −M ⋅ g − k ⋅ y ( t) + f ( t)

dt

Solve for the highest derivative:

2

d y ( t) 2

= −g +

dt

Integration of this second derivative results in: From the solution of problem 2-9:

f ( t) M

d⋅ y ( t)

M := 50gm

dt

−

k M

y( t)

and a second integartion gives y ( t)

g = 980.7

cm 2

s For this problem:

y ( 0 ) = 100cm

f ( t) = 0dynes

k :=

M⋅ g 27cm

k = 1816

gm 2

s

The period of oscillation is, as in the solution to problem 2-9:

Period := 2 ⋅ π ⋅

M k

Period = 1.043 s The number of complete cycles in 10 seconds is: 10s Period

= 9.592

The simulation plot shows the same result.

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13-2. Simulation of a Pendulum Application of Newton's Second Law of Motion: −M ⋅ g ⋅ tan( θ ) = M ⋅

2

d x ( t)

2

2

dt

L

For small angles θ, from the geometry: tan( θ ) = sin( θ ) =

x ( t) L

Mg sin2

Substitute and simplify to obtain: 2

d x ( t)

g

 L

2

dt g = 9.807

= −

m 2

x ( t)

M := 0.5kg

x(t) L := 0.8m

s

g r1. = i⋅ L

g r2 = −i⋅ L

The frequency of oscillation is:

x ( 0 ) = 0.1m

 s2 + g  X( s) = 0  L 

The solution of the differential equation:

Roots:

Mg

g L

 g t  L 

x ( t) = x ( 0 ) cos

= 3.501 Hz

Period :=

2⋅ π g L

(Table 2-1.1)

Period = 1.795 s

The number of oscillations in 10 s is: 10s Period

= 5.572

The simulation plot shows the same result.

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13-3. Simulation of Punctured Air Tank of Problem 2-23. In Problem 2-23 the mass balance on the tank produced the following equation: V Flow through the orifice:

dρ( t) dt

= wi( t) − wo ( t)

(

wo ( t) = Ao ⋅ 2 ρ ( t) p ( t) − p o

d⋅ p ( t) dt 3

V := 1.5m

Problem parameters:

=

Rg ⋅ T V⋅ M

Ao := 0.785cm

2

3

Rg := 8.314 ⋅

M

ρ ( t) =

Ideal gas law, assuming constant temperature:

Substitute and solve for dp(t)/dt:

)

Pa⋅ m

mole⋅ K

Rg ⋅ T

p( t)

(wi(t) − wo(t) ) M := 29

gm mole

T := ( 70 + 273.16)K

p o := 101300Pa

p ( 0 ) = ( 500000 + 101300)Pa

From the linearization of Problem 2-23, we know that the time constant is: V⋅ 2 ⋅ τ :=

M Rg⋅ T

⋅ 601300Pa⋅ 500000Pa

(

Ao ⋅ 2 ⋅ 601300Pa − p o

)

τ = 42.895 s

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: wi := Ao ⋅ 2 ⋅

M ⋅ 601300Pa Rg ⋅ T

(601300Pa − po)

kg wi = 0.194 s

As predicted by the linearized model, the pressure reaches steady state in about 200 s.

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13-4. Simulation of the turkey temperature response of Problem 2-24. From the solution to Problem 2-24, the differential equation obtained from an energy balace on the turkey is: M ⋅ cv ⋅

dT ( t) dt

= σ ⋅ ε ⋅ A Ts( t) − T( t) M := 12lb

The parameters, given in this problem are: cv := 0.95

4



BTU



A := 3.5ft

T( 0 ) = 535R

lb⋅ R

4 2

Ts := 800R −8

σ := 0.1718⋅ 10

ε := 0.6 BTU 2 4

hr⋅ ft R

By the linearization done in Problem 2-24, the time constant of the turkey is: τ :=

M ⋅ cv 4 ⋅ σ ⋅ ε ⋅ A⋅ ( 535R)

Integrate the differential equation :

τ = 5.16 hr

3

d ⋅ T( t ) dt

=

σ ⋅ε ⋅A 

4 4 Ts( t) − T( t)    M ⋅ cv

Run the simulation for 25 hrs (five time constants). Simulate the oven as a step function from an inital temperature of 535ºR to 800ºR.

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: M ⋅ cv τ := τ = 1.54 hr 3 4 ⋅ σ ⋅ ε ⋅ A⋅ ( 800R) 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.

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13-5. Non-isothermal Chemical Reactor of Section 4-2.3 lbmole := 453.59mole

Rearranging the model equations from Section 4-2.3: d ⋅ cA( t) f ( t) = c ( t) − cA( t) − rA( t) dt V Ai

(

)

cA( 0 ) = 0.2068

lbmole ft

−E

rA( t) = k o ⋅ e d ⋅ T( t )

=

dt d ⋅ Tc ( t ) dt

=

Design conditions: Parameters:

f ( t) V

2

⋅ cA ( t )

3

lb ft

∆Hr

( Ti ( t ) − T( t ) ) −

fc( t) Vc

cAi := 0.5975

Rg := 1.987

⋅ rA( t) −

ρ ⋅ Cp

(Tci( t) − Tc( t)) +

V := 13.46ft

ft fc := 0.8771 min ρ c := 62.4

Rg⋅ T( t)

V⋅ ρ ⋅ Cp

U⋅ A

lbmole ft

U⋅ A

Vc⋅ ρ c⋅ Cpc

( T( t ) − Tc ( t ) )

3

3

ft

8

ρ := 55

lbmole⋅ R

U := 75

ft

3

Tci := 540R

min

3

E := 27820

lbmole⋅ min

lb ft

Tc( 0 ) = 602.7R

f := 1.3364

k o := 8.33⋅ 10 BTU

T( 0 ) = 678.9R

( T( t ) − Tc ( t ) )

Ti := 633.5R

BTU ∆Hr := −12000 lbmole

3

3

Cp := 0.88

3

BTU 2

hr⋅ ft ⋅ R

BTU lb⋅ R

Vc := 1.56ft

3

A := 36ft

Cpc := 1

BTU lbmole

2

BTU lb⋅ R

Check that initial conditions are at steady state (derivatives = 0): −E

rA := k o ⋅ e f V

Rg⋅ 678.9R

⋅  cAi − 0.2068

f

 

Vc

 

ft

3

∆Hr ρ ⋅ Cp

(Tci − 602.7R) +

lbmole  ft

lbmole 

(T − 678.9R) − V i fc

⋅  0.2068



3

2

lbmole rA = 0.039 3 ft ⋅ min

 − 4 lbmole

− rA = −5.87 × 10

⋅ rA −

U⋅ A V⋅ ρ ⋅ Cp

U⋅ A Vc⋅ ρ c⋅ Cpc

3

ft ⋅ min

( 678.9 − 602.7 )R = −7.915 × 10

( 678.9 − 602.7 )R = −0.027

The following is the Simulink diagram for the reactor:

R min

−3 R

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

13-6. Mixing Process of Problem 3-1 The model equation, from the solution to Problem 3-1: d ⋅ cA( t) dt Problem parameters:

Total mass balance:

At the initial steady state:

=

f1 ( t) ⋅ cA1( t) + f2 ( t) ⋅ cA2( t) − f ( t) ⋅ cA( t) A⋅ h

cA1 := 0.01 f1 + f2 = f

mole cm

3

cA2 := 0.05

mole cm

3

cA( 0 ) = 0.025 f := 100

gal min

mole cm

Ah := 200gal

(assuming constant volume)

f1 ⋅ cA1 + f2 ⋅ cA2 − f ⋅ cA = 0

 c − 0.025mole⋅ cm− 3  A2  f1 := f ⋅ c −c

gal f1 = 62.5 min

f2 := f − f1

gal f2 = 37.5 min

A2

The folowing is the Simulink diagram for the mixer:

A1

3

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.

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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.

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13-8. Isothermal reactor of Problem 3-2 The model equations, from the solution to Problem 3-2, are: d ⋅ cA2( t) dt

=

f c ( t) − cA2( t) − kcA2( t) V A1

(

)

(

cA3( t) = cA2 t − to Design conditions:

Problem parameters:

f := 50

) ft

3

cA1 := 2

min

V := 150ft

Initial conditions at steady state:

cA2( 0 ) = 0.5

k :=

3

lbmole ft

3

Lp := 400ft

f ⋅  cA1 − 0.5lbmole⋅ ft



V⋅ 0.5lbmole⋅ ft

−3

The Simulink diagram is given by:

The responses to a 5 ft3/min step increase in process flow are:

Di := 5.5in − 3



−1

k = 1 min

lbmole ft

3

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: τ :=

V f + k⋅ V

τ = 0.75 min

2

to :=

π ⋅ Di ⋅ Lp 4f

to = 1.32 min

 c − 0.5 lbmole   A1 3 3 ft  ft  ∆cA2 := ⋅5 f + k⋅ V min ∆cA2 = 0.038

lbmole ft

3

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13-9. Flash drum of Problem 3-11 kmole := 1000mole

Rearranging the model equation developed in Problem 3-11: d⋅ x ( t)

=

dt

1 M

( F⋅ z( t) − V⋅ y ( t) − L⋅ x ( t) )

x ( 0 ) = 0.4

F=V+ L y( t) =

α ⋅ x( t)

1 + ( α − 1) x( t) kmole

Design conditions:

F := 10

Problem parameters:

M := 500kmole

At initial steady state:

y 0 :=

s α ⋅ 0.4

L := 5

kmole s

V := F − L

V= 5

kmole s

α := 2.5

1 + ( α − 1 ) 0.4

z0 :=

V⋅ y 0 + L⋅ 0.4 F

y 0 = 0.625 z0 = 0.513

The Simulink diagram for this problem is:

The responses to a 0.1 step increase in feed composition are:

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.

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13-10. Distillation tray of Problem 3-12 The model equations from the solution to Problem 3-12 are: d⋅ h ( t) dt

=

1

(

f ( t) − fo ( t) S i

fo ( t) = 0.415 ⋅ w⋅ h ( t) Problem parameters:

1.5

w := 3ft

Initial steady state conditions:

)

⋅ 2⋅ g

S := 11.2ft ft

2

1

3

fi := 30 min

fo := fi

fo   h 0 :=   0.415 ⋅ w⋅ 2⋅ g 

1.5

h 0 = 0.136 ft As the response is fast, convert time units to s by multiplying the derivative by 60 s/min. The Simulink diagram for the tray is:

The responses to a 20 ft3/min increase in inlet flow are:

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: −1

3

20ft ⋅ min 2

−1

= 0.06 ft

close!

331.4ft ⋅ min

The students can check the results for the change in 10 ft3/min.

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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: 3 %CO := % kg kg kg m 3 c1 := 80 c2 := 30 c0 := 50 f := 4 V := 40m %TO := % 3 3 3 min m m m f = f1 + f2

At the intial steady state:

f ⋅ c0 = f1 ⋅ c1 + f2 ⋅ c2

c0 − c1 f2 := f ⋅ c2 − c1

3

m f1 = 1.6 min

f1 := f − f2

3

m f2 = 2.4 min

From the results of Problem 6-11, the ultimate gain and period are: %CO Kcu := −250 %TO

Tu := 3.01min

From Table 7-1.1, for a series PID controller tuned for quarter decay ratio response: Kcu Kc := 1.7

τ I :=

Tu 2

τ D :=

Tu 8

%CO Kc = −147 %TO

τ I = 1.5 min

τ D = 0.38 min

3

Control valve (from F401Vlv1):

m Kv := 0.0542 min⋅ %CO Initial position:

Transmitter (from F407Trmr):

τ T := 3min

f2 Kv

τ v := 0.1min

(Solution of Problem 6-11)

= 44.28 %CO

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:

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.

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13-12. Non-interacting tanks in series of Fig. 4-1.1 The model equations developed in Section 4-1.1 are: d⋅ h 1( t) 1 = f ( t) − fo ( t) − f1 ( t) dt A1 i

(

d⋅ h 2( t) dt

=

1

(

f ( t) − f2 ( t) A2 1

h 10 := 2.5m

)

h 20 := 2.5m

f1 ( t) = Cv1⋅ h 1 ( t) 3

)

f2 ( t) = Cv2⋅ h 2 ( t) 3

Design conditions:

m fi := 5 min

m fo := 2 min

Problem parameters:

A1 := 9m

A2 := 9m

At initial steady state:

f1 := fi − fo

f2 := f1

2

2 3

Cv1 :=

f1

Cv2 :=

h 10

m f1 = 3 min 2.5

f2

Cv1 = 1.897

h 20

m

min 2.5

Cv2 = 1.897

m

The linearized gains and time constants are: K1 :=

2 ⋅ h 10 Cv1

Cv1 h 20 K2 := ⋅ Cv2 h 10

τ 1 :=

τ 2 :=

The Simulink diagram for this problem is:

2 ⋅ A1 ⋅ h 10 Cv1 2 ⋅ A2 ⋅ h 20 Cv2

K1 = 1.667

K2 = 1

min 2

m

τ 1 = 15 min

τ 2 = 15 min

min

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: 3

m K1 ⋅ 0.2 = 0.333 m min The responses for the second tank are second order with the same steady-state change in level, meaning that the gain K 2 is unity as predicted by the linear model.

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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⋅ h 1 ( t) − h 2 ( t) Design conditions are the same except for the initial condition in tank 1: At the initial steady state:

Cv1 :=

f1 h 10 − h 20

h 10 := 5m 2.5

Cv1 = 1.897

m

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. min K1 = 1.667 2 m The following is the Simulink diagram for the interacting tanks is series:

The responses to a 0.2 m3/min step increase in inlet flow are: 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 h 10 = 2.6 m, Cv1 :=

f1 2.6m − h 20 2.5

Cv1 = 9.487

m

min

The response of the second tank becomes first-order and the two tanks behave as a single tank.

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13-14. Non-interacting thermal tanks in series of Fig. 4-1.5 The model equations developed in Section 4-1.2 are: d ⋅ T2 ( t ) dt d ⋅ T4 ( t ) dt

=

= 3

fA V1

( T1 ( t ) − T2 ( t ) )

1

(

)

 f ⋅ T ( t) + fB⋅ T3( t) − fA + fB ⋅ T4 ( t)  V2  A 2 3

Design conditions:

m fA := 1 min

m 3 3 V1 := 5m V2 := 5m fB := 0 min

At initial steady state:

T2 := T1

T4 :=

fA⋅ T2 + fB⋅ T3 fA + fB

This problem is linear with a gains and time constants: fB K2 := fA + fB

τ 1 :=

V1 fA

τ 2 :=

The Simulink diagram for this problem is:

V2 fA + fB

T1 := 500K T3 := 500K

T2 = 500 K

fA K1 := fA + fB K2 = 0

T4 = 500 K K1 = 1

τ 1 = 5 min

τ 2 = 5 min

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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 ) dt d ⋅ T4 ( t ) dt

=

=

1

(

)

 f ⋅ T ( t) + fR⋅ T4( t) − fA + fR ⋅ T2 ( t)  V1  A 1 1

(

)(

f + fR ⋅ T1 ( t) − T4 ( t) V2 A

)

The design conditions and problem parameters are the same as in Problem 13-14, plus the recycle flow: 3

m fR := 1 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 Simulink diagram for this problem is:

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.

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13-16. Reactors with recycle of Problems 4-9 and 6-15 The model equatios are developed in the solution to Problem 4-9: d ⋅ cA1( t)

1 f ⋅ c ( t) + fR⋅ cA2( t) − f1 ⋅ cA1( t) − k 1 ⋅ cA1( t) V1 o Ao d ⋅ cA2( t) f1 = c ( t) − cA2( t) − k 2 ⋅ cA2( t) dt V2 A1 dt

=

( (

)

cA1( 0 ) = cA10

)

Design conditions from Problem 6-15:

cAo := 7

lbmole ft

3

cA2( 0 ) = cA20 −1

k 1 := 0.2min

3

ft fo := 10 min

V1 := 125ft

3

−1

k 2 := 0.2min

V2 := 125ft

3

3

For the base case let:

ft fR := 10 min

At the initial steady state:

f1 := fo + fR

3

cA20 =

ft f1 = 20 min

f1 ⋅ cA10 f1 + k 2 ⋅ V2 fo ⋅ cAo

cA10 :=

f1

cA10 = 1.726

f1 + k 1 ⋅ V1 − fR⋅ f1 + k 2 ⋅ V2 cA20 := The Simulink diagram for the reactors is:

f1 ⋅ cA10 f1 + k 2 ⋅ V2

cA20 = 0.767

lbmole ft

3

lbmole ft

3

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.

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13-17. Extraction process of Problem 4-6 The model equations developed in the solution to Problem 4-6 are: d ⋅ c1 ( t ) 2 ⋅ f1 ( t) = ci( t) − c1 ( t) − Ka⋅ c1 ( t) − me⋅ c2 ( t) dt V

(

d ⋅ c2 ( t ) dt

)

(

(

)

= Ka⋅ c1 ( t) − me⋅ c2 ( t) −

2 ⋅ f2 ( t) V

3

Design conditions:

Problem parameters:

)

ci := 0.4

me := 3.95

Ka := 3.646min

kmole m

c10 := ( 1 − Rec) ⋅ ci

f2 :=

Rec := 90%

3 −1

c20 :=

c2 ( 0 ) = c20

c2 ( t )

m f1 := 5 min

At the initial steady state:

(

c10 = 0.04

)

)

Ka⋅ V⋅ c10 − me⋅ c20 2 ⋅ c20

kmole 3

m

− 4 kmole

c20 = 1.2776 × 10

Ka⋅ me⋅ V

(

3

V := 25m

−2 ⋅ f1 ⋅ ci − c10 + Ka⋅ V⋅ c10

The Simulnk block diagram for the extractor is:

c1 ( 0 ) = c10

3

m f2 = 14088.6 min

3

m

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

−1

Ka := 209min

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Smith & Corripio, 3rd edition Chapter 13. Simulation of Process Control Systems (continued) 13-18. Temperature control of stirred tank cooler of Problems 4-7 and 6-19 degC := K

The model equations, from the solution to Problem 4-7, are: d ⋅ T( t ) dt d ⋅ Tc ( t ) dt Design conditions:

Problem parameters:

=

=

f ( t) V

( Ti ( t ) − T( t ) ) −

fc( t) Vc

Ti := 70degC 3

V := 5m

U⋅ A V⋅ ρ ⋅ cp

(Tci − Tc( t)) +

( T( t ) − Tc ( t ) )

U⋅ A Vc⋅ ρ c⋅ cpc

At the initial steady state:

Tc( 0 ) = Tc0

3

f := 0.1

m

min

U := 200

kJ kg⋅ degC

T0 := 45degC

Tci := 25degC

kJ

2

A := 4m

2

ρ c := 1000

kg

(

(T0 − Tc0) ρ c⋅ cpc ( Tc0 − Tci) U⋅ A

The Simulink diagram for the cooler is:

3

m

f ⋅ ρ ⋅ cp Tc0 := T0 − Ti − T0 U⋅ A fc0 :=

T( 0 ) = T0

( T( t ) − Tc ( t ) )

min⋅ m degC cp := 3.8

kJ := 1000joule

)

cpc := 4.2

3

Vc := 1.1m

ρ := 800

kg 3

m

kJ kg⋅ degC

Tc0 = 35.5 degC 3

m fc0 = 0.172 min

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 %CO := % %TO := % From the solution to Problem 6-19: %CO Kcu := −75.4 Tu := 7.59min %TO From Table 7-1.1:

Kcu Kc := 1.7

τ I :=

Tu 2

Tu

τ D :=

8

%CO Kc = −44 τ I = 3.8 min %TO τ D = 0.95 min

Initial conditions:

Transmitter output:

45 − 20 70 − 20

= 50 %TO

fc0

Controller output:

3

0.008

m

min⋅ %CO

= 21.5 %CO

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.

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gal

gpm :=

13-19. Direct contact heater of Problem 4-5

degF := R

min

The model equations developed in the solution to Problem 4-5 are: d ⋅ T3 ( t ) dt

=

1

(

)

⋅ f ( t) ⋅ T1 ( t) − f3 ( t) T3 ( t) + V 1

hs V⋅ ρ ⋅ cp

⋅ w2 ( t)

T3 ( 0 ) = T30

w2 ( t) = 1.954 10⋅ vp( t) = k v ⋅ vp( t) w2 ( t)

f3 ( t) = f1 ( t) +

ρ

d ⋅ T3 ( t )

Substitute to eliminate f.3(t):

dt

Design conditions:

f1 := 25gpm

Problem parameters:

ρ := 7 k v :=

=

f1 ( t) V

( T1 ( t ) − T3 ( t ) ) +

T1 := 60degF

lb

cp := 0.8

gal

V⋅ ρ ⋅ cp

( h s − c p ⋅ T3 ( t ) )

T30 := 80degF

BTU lb⋅ degF

1.954 ⋅ 10 lb

w2 ( t)

V := 5gal k v = 0.0618

100%CO min

lb min⋅ %CO

Steam enthalpy, saturated at 1 atm., referenced to 32 F, is: 1150.4 BTU/lb h s := 1150.4

At the initial steady state:

w2 :=

BTU lb

+ cp ⋅ 32degF

(

f1 ⋅ ρ ⋅ cp ⋅ T30 − T1 h s − cp ⋅ T30

w2 vp0 := kv

)

h s = 1176

BTU lb

lb w2 = 2.518 min

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 τ v := 4s The Simulink block diagram is:

τ v = 0.067 min

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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.

13-20. Gas flow control valve of Example 5-2.2 The flow through a gas control valve can be calculated from Eq. 5-2.3: p1 3 fs = 836 ⋅ Cv ⋅ Cf ⋅ y − 0.148 ⋅ y G⋅ T

(

1.63

y=

Cf

⋅

∆p

G=

p1

Problem parameters from Example 5-2.2: Cv := 440

gpm

p 1 := 34.7psia

psi

Functions for valve flow:

)

Mw 29

psia := psi G := 0.621

scfh := Cf := 0.8

ft

3

hr

T := ( 259 + 460 )R

1.63 ∆p y p 1 , ∆p := ⋅ Cf p1

(

)

p1 R 3 scfh fs p 1 , ∆p := 836 ⋅ ⋅ Cv ⋅ Cf ⋅ ⋅  y p 1 , ∆p − 0.148 y p 1 , ∆p    gpm psi G⋅ T

(

)

(

)

(

)

5

6 .10

5

4 .10

(

)

fs p1 , ∆p⋅ psi scfh

5

2 .10

0

0

5

10

15

20

25

∆p Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

This is the plot also obtained with the following Simulink diagram:

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.

13-21. Temperature control of reactor of Section -2.3 and Problem 6-12 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-close • f405PIDs: 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 Valve maximum flow:

2 ⋅ 0.8771

ft

3

min

= 1.754

678.9 − 640

Transmitter initial condition:

700 − 640

ft

3

Initial position:

min

−

ln( 0.5) ln( 50)

= 17.72 %CO

= 64.8 %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: %CO Kcu := 10 %TO

Tu := 6.7min

From Table 7-1.1, the quarter decay rato tuning parameters are: Kcu Kc := 1.7

τ I :=

Tu 2

τ D :=

Tu 8

%CO Kc = 5.9 %TO

The Simulink block diagram for the temperature control loop is:

τ I = 3.4 min

τ D = 0.84 min

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.

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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. 18.3036−

p2 ( t) =

14.7psia 760mmHg

e

3816.44K

( Ts( t) +460R) K 1.8R

− 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.

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13-23. Concentration control of isothermal reactors in series of Problem 6-17 The model equation for each reactor is developed in the soution to Problem 6-17: d⋅ c ( t) j f ( t) For j = 1,2,3 = c ( 0 ) = cj0  c ( t) − c j( t) − k⋅ c j( t) j dt V  j−1 and

c ( t ) = ci ( t ) 0

Design conditions:

f := 100gpm

c := 4 0

f ⋅c At the initial steady state:

c := 1

0

f + k⋅ V

lb gal

−1

k := 0.1min f ⋅c

c := 2

V := 1000gal f ⋅c

1

c :=

f + k⋅ V

3

2

f + k⋅ V

 4  2  lb c=  1  gal  0.5  

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. From the solution to Problem 6-17:

Process gain:

Initial conditions:

gpm Kv := 2.46 %CO Kv ⋅ Ko ⋅

( 100 − 0 )%TO ( 1 − 0 )lb⋅ gal

f m0 := Kv

The Simulink diagram for the control loop is:

lb Ko := 0.0075 gal⋅ gpm = 1.845

−1

m0 = 40.65 %CO

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

τ I = τ = 9min

τ D = 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.

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13-24. Temperature and level control of oil heater of Problem 6-24 The model equations from the solution to Problem 6-24 are: d⋅ h ( t) dt

ft

=

d ⋅ T3 ( t ) dt

3

4

7.48gal

=

f1 ( t)

(

f ( t) − f3 ( t) 2 1

πD

)

h( 0) = h0

U⋅ A ⋅ T1 ( t ) − T3 ( t ) + Ts( t) − T3 ( t) V( t) V( t) ⋅ ρ ⋅ cp

(

)

(

)

T3 ( 0 ) = T30

2

V( t) = 7.48⋅

gal π ⋅ D ⋅ h ( t) 3 4 ft

d ⋅ Ts( t)

1

dt

=

 λ ⋅ ws( t) − U⋅ A⋅ ( Ts( t) − T3( t) )

Ts( 0 ) = Ts0

CM 

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:

(

)

∆p v1 = p 1 + 14.7psi − p 2 − 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:

p 1 := 45psig

p 2 := 40psia

BTU

Problem parameters: U := 136

f1 := 100gpm

D := 3ft

2

T1 := 70degF

A := 974ft⋅ π ⋅ 0.5in

T30 := 200degF A = 127.5 ft

2

hr⋅ ft degF ρ := 53

lb ft

3

cp := 0.45 At the initial steady state:

CM := 974ft⋅ 0.178 BTU

p s := 115psig

lb⋅ degF

f3 := f1 ws :=

Ts0 := T30 +

(

)

U⋅ A⋅ Ts0 − T30

The Simulink diagram for the heater is:

λ

ρ ⋅ cp ⋅ f1 U⋅ A

lb ft

⋅ 0.12

BTU lb⋅ degF

λ := 872.9

(T30 − T1)

CM = 20.8

BTU degF

BTU lb Ts0 = 343.4 degF lb ws = 47.48 min

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. Intial conditions, from the results of Problem 6-24:: ws 46 wsmax := ⋅ 1.5⋅ ws ms0 := 41.8 wsmax

lb wsmax = 78.4 ms0 = 60.58 %CO min

32.2 m10 := 1.5⋅ 46

m10 = 46.67 %CO

Ultimate gain and period of temperature controller: Quarter decay tuning parameters from Table 7-1.1:

50% for both transmitters

%CO Kcu := 77.1 %TO Kcu Kc := 1.7 %CO Kc = 45 %TO

Tu := 1.12min

τ I :=

Tu 2

τ D :=

τ I = 0.56 min

Tu 8

τ D = 0.14 min

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.

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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:

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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: %CO Kc := −0.42 %TO

Cascade: −0.56

%CO %TO

τ I := 3.5min

1.0min

τ D := 0.86min

0.26min

13-26. Feedback vs. cascade control of Problem 9-3 Simple Feedback Loop. From the solution to Problem 9-3, the ultimate gain and period are: %CO Kcu := 2.38 %TO

Tu := 27.8min

From Table 7-1.1, the quaerte decay tuning parameters for a PI controller are: Kcu Kc := 2.2

τ I :=

Tu 1.2

Cascade Control. From the solution to Problem 9-3, with %CO Kcu := 13.5 %TO

Tu := 12.5min

Kcu Kc := 2.2

τ I :=

Tu 1.2

%CO Kc = 1.1 %TO

τ I = 23 min

%CO Kc2 := 2.2 %TO

%CO Kc = 6.1 %TO

τ I = 10 min

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: 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.

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13-27. Jacketed reactor of Problem 9-4. Simple feedback vs. cascade control 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: Simple Feedback Loop: %TO Kp := 0.285 %CO

τ := 13.5min

t0 := 3.5min

The tuning parameters for minimum IAE response, from Table 7-2.2:

 t0  Kc :=  Kp  τ  1.435

− 0.921

%CO Kc = 17 %TO

 t0  τ I :=  0.878  τ  τ

0.749

τ I = 5.6 min

 t0  τ D := 0.482 ⋅ τ ⋅  τ

1.137

τ D = 1.4 min

Cascade Control (with the secondary proportional controller gain set at 2%CO/%TO): %TO Kp := 0.385 %CO

 t0  Kc := ⋅ Kp  τ  1.435

%CO Kc = 19 %TO

t0 := 2.2min

τ := 12.8min − 0.921

 t0  τ I := ⋅ 0.878  τ  τ

0.749

τ I = 3.9 min

 t0  τ D := 0.482 ⋅ τ ⋅  τ

1.137

τ D = 0.83 min

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The responses to a 5ºC step increase in set point are:

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.

13-28. Feedforward temperature control of heater of Example 13-4.2 If the variation in the inlet temperature diturbance is neglected, the feedforward controller calculation becomes: set

w ⋅ ( t) =

ρ ⋅ cp ⋅ T



⋅ ( t) − Ti 

set

f ( t) = Ratio⋅ f ( t)

λ

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: ρ := 68

lb ft

3

cp := 0.8

The ratio for set point of

BTU

λ := 966

lb⋅ degF

BTU lb

Ti := 100degF

lb fmax := 75 min

Tset := 150degF Ratio :=

(

ρ ⋅ cp ⋅ Tset − Ti λ

) ⋅ 100%TO fmax

Ratio = 3.75

min⋅ %TO ft

3

The Simulink diagram for the ratio control scheme is obtained by a simple modification of the diagram of Fig. 13-4.11:

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 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.

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13-29. Ratio control of blending tank of Problems 3-18 and 13-11 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. Design conditions: 3 3 f2 m m f1 := 1.6 f2 := 2.4 Ratio := Ratio = 1.5 min min f1 3

Transmitter ranges:

m f1max := 4 min

3

m f2max := 5.42 Rmax := 3 min

(at 100%CO)

The signal from the flow transmitter must then be scaled by the factor: Rmax⋅ f1max 100%CO⋅ f2max The controller gain must be adjusted by the factor the output is multiplied by: %CO 100%CO f2max Kc := −147 ⋅ ⋅ %TO Rmax f1 The Simulink diagram for the ratio control scheme is:

%CO Kc = −166 %TO

= 0.0221

1 %CO

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.

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13-30. Cascade temperature control of reactor of Section 4-2.3 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. An open-loop step test in the slave controller results in the following parameters: Slave loop:

%TO K2 := 1.95 %CO

Synthesis tuning from Table 7-4.1: Master loop:

%TO Kcu := 10.5 %CO

τ := 3min τ I := τ

t0 := 0min Set Kc to 2%CO/%TO to obtain fast response without much oscillation.

Tu := 3.75min

Quarter decay ratio tuning from Table 7-1.1:

Kcu Kc := 1.7 τ I :=

Tu 2

The Simulink diagram for the cascade control loop is:

%CO Kc = 6 %TO τ D :=

Tu 8

τ I = 1.9 min τ D = 0.47 min

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.

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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: −GD FFC = 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: KDT 100%TO %TO HD = KDT := KDT = 50 3 −1 3 −1 0.1⋅ s + 1 2ft ⋅ min ft ⋅ min From step tests on the disturbance, the folowing parameters are obtained: − 5.5s

e %TO GD = 61.5 ⋅ 3 − 1 9s + 1 ft ⋅ min

−61.5%TO

FFC =

−1

3

− 3.25s

KDT⋅ ft min

%TO e GM = −0.98 ⋅ %CO 6.75s + 1

⋅

⋅ ( −0.98)

6.75s + 1 − ( 5.5−3.25)s e 9s + 1

−61.5%TO 3

Feedforward controller gain:

−1

KDT⋅ ft min FFC = 1.225

= 1.255

⋅ ( −0.98)

%CO %TO

%CO 6.75s + 1 − 2.25s+ 1 ⋅ ⋅e %TO 9.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

delay = 4.5 min

To start with a correct initial set point of the coolant flow controller, a bias is needed: Initial reactant transmitter output: Bias required for the same set point:

1.3364

ft

3

min

⋅ KDT = 66.82 %TO

m( 0 ) = m( 0 ) + 1.255 ⋅ 66.82 %TO + Bias = m( 0 ) Bias := −1.255 ⋅ 66.82 %TO

Bias = −83.86 %TO

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: 1.2 Kc := −0.98

 3.25    6.75 

−1

τ I := 2 ⋅ 3.25min

τ D :=

3.25min 2

τ I = 6.5 min The Simulink diagram for the linear feedforward controller is:

%CO Kc = −2.5 %TO τ D = 1.6 min

The responses to a 0.2 ft3/min step decrease in reactants flow at 1 min are

The feedforward responses are in magenta and the simple feedback responses are in yellow. The feedforward controller required some adjustments: Delay := 3.5min %CO Kc := −5 %TO τ I := 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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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: 1. Control objective: T(t) = Tset(t) (reactor temperature) 2. Manipulated variable: fc(t) (coolant flow to the jacket) 3. Disturbances: f(t) (reactants flow) 4. Steady state balances: 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) ⋅ ρ c⋅ cpc⋅ Tc − Tci Combine to eliminate rA(t): f ( t ) ⋅  ρ ⋅ c p ⋅  Ti − T





(

)(

( t)  + −∆Hr ⋅ cAi − cA 

set

Solve for the manipulated variable:

) = fc(t) ⋅ ρ c⋅ cpc⋅ (Tc − Tci)

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 changes 7. 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 := 55

lb ft

cpc := 1

3

cp := 0.88

BTU

Ratio at design conditions:

Ti := 635R

lb⋅ degF

cAi := 0.5975

BTU lb⋅ degF

lbmole ft

Ratio :=

Check:

BTU ∆Hr := −12000 lbmole T := 678.9R

cA := 0.2068

3

(

Tci := 540R

ft

)

3

min

ft

3

Tc := 602.7R

3

) ( )( ρ c⋅ cpc⋅ ( Tc − Tci) ft

lb

lbmole

ρ ⋅ cp ⋅ Ti − T + −∆Hr ⋅ cAi − cA

f := 1.3364

ρ c := 62.4

Ratio = 0.655 3

fc := Ratio⋅ f

ft fc = 0.876 min

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.

3

3

ft fmax := 2 min

Rmax := 2 The scaled variables are:

x=

ft fcmax := 1.75 min

f fmax

⋅ 100 %TO

y=

fcmax

Substitute into the feedforward equation:

100%TO Scaled feedforward equation:

Initial conditions:

y=

Ratio m0 := ⋅ 100 %CO Rmax

y=

fc fcmax

⋅ 100 %TO

Rmax 100%CO

Rmax⋅ fmax fcmax⋅ 100 %CO m0 = 32.8 %CO

⋅ r⋅

r=

fmax 100%TO

Ratio Rmax

⋅ 100 %CO

⋅x

⋅ r⋅ x f fmax fc

= 66.8 %TO

fcmax

= 50 %TO

To maintain the same loop gain, the feedback controller gain must be adjusted: Rmax⋅ fmax Rmax⋅ f f Kc = Kcnew ⋅ ⋅ ⋅ 100 %TO = ⋅ Kcnew fcmax⋅ 100 %CO fmax fcmax %CO Kc := −5 %TO

fcmax Kcnew := ⋅K Rmax⋅ f c

The Simulink diagram for the ratio control scheme is:

%CO Kcnew = −3.3 %TO

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 equalpercentage valve and the ratio is adjusting the coolant flow set point. This is also teh reason they move in opposite directions.

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Smith & Corripio, 3rd edition Chapter 13. Simulation of Process Control Systems 13-33. Cascade control of reactors in series of Problem 13-23 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 From the results of an open-loop step test on the slave controler output: %TO K2 := 0.58 %CO

τ 2 := 6min

Synthesis tuning from Table 7-4.1:

%CO := %

%TO := %

t0 := 0min

τ I2 := τ 2

Kc2 = tunable

Set at

%CO Kc2 := 5 %TO

With these slave settings, an open-loop step test on the master controller output: %TO K1 := 3.3 %CO

τ .1 := 5min

Use synthesis tuning for the master also:

t0 := 0min τ I1 := 5min

The Simulink diagram for the cascade control scheme is:

τ D1 := 0min

Set

%CO Kc1 := 5 %TO

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.

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13-34. Multivariable control of caustic blending tank of Problem 12-2. Design conditions:

klb := 1000lb mass% := %

klb

wP := 40 hr

SP AC

SP

3

FC

wc

xc

FT

3

Caustic

SP

x

2

x c := 50mass% Problem parameters: M := 10klb

FT

AT

FC Water

SP FC

1

wP

FT

ww

At initial steady state: wP⋅ x 0 klb wc := wc = 24 xc hr klb ww := wP − wc ww = 16 hr

Assume perfectly mixed tank with constant mass: Caustic balance: Total balance:

d⋅ x ( t) dt

=

1 M

(wc(t) ⋅ xc( t) − wP( t)⋅ x(t) )

wP( t) = wc( t) + ww( t)

The Simulink diagram to simulate tank is:

x 0 := 30mass%

x ( 0 ) = x0

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 constant • Two 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. ww klb Initial conditions: FC-1: 16 mw0 := mw0 = 26.667 %CO −1 hr 60klb⋅ hr FC-2:

FC-3:

AC-3:

24

40

klb hr

mc0 :=

30mass%

60klb⋅ hr

wP

klb hr

wc

60klb ⋅ hr

−1

x0 50mass%

−1

mc0 = 40 %CO

= 66.667 %TO

= 60 %TO

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) With these tuning parameters, the responses to a

(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.

(c) Linear decoupler design as in Example 12-3.1 To keep the flow constant, the caustic flow must be adjusted as follows:

(

)

∆wP = Kvw + D12⋅ Kvc ∆mAC = 0 −Kvw %CO D12 = = −1 Kvc %CO

(

Kvc = Kvw =

−1

100%CO

%CO D12 := −1 %CO

)

∆x = Kvc⋅ Kxc + D21⋅ Kvw⋅ Kxw ∆mFC −Kvc⋅ Kxc ww D21 = = Kvw⋅ Kxw wc

60⋅ klb ⋅ hr

ww Kxc = wP

ww D21 := wc

−wc Kxw = wP %CO D21 = 0.667 %CO

Initial outputs: To start at the proper decoupler outputs, biases must be added to the decouplers: mc0 = mc0 + D12⋅ mw0 + Bc

Bc := −D12⋅ mw0

Bc = 26.667 %CO

mw0 = mw0 + D21⋅ mc0 + Bw

Bw := −D21⋅ mc0

Bw = −26.667 %CO

The Simulink diagram for the linear decoupler is:

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

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13-35. Multivariable control of evaporator of Figure 12-3.4 Design conditions:

wV

Cond.

wF xF

wP x

T

wS

lb wF := 50000 hr

x F := 12mass%

x 0 := 70mass%

E := 0.95

M 0 := 747lb

h L0 := 50%TO

Process parameters: M min := 394lb

Ts Condensate

Developent of the model equations (assume perfect mixing): Total mass balance:

Sugar balance:

d⋅ M( t)

= wF( t) − wv ( t) − wP( t)

dt

d⋅ ( M( t) ⋅ x( t) ) dt

M( 0) = M0

= wF( t) ⋅ x F( t) − wP( t) ⋅ x ( t)

2 eqns. 2 unk.(M,x)

Expand and substitute total balance:

M( t) ⋅ Simplify:

d⋅ x ( t)

d⋅ x ( t) dt

dt =

)

1 w ( t) ⋅ xF( t) − x( t) + wv ( t) ⋅ x( t) M( t)  F

Economy relates steam and vapor flows: At the intial steady state: wF⋅ x F wP := wv := wF − wP x0 Maximum hold up:

(

+ x ( t) ⋅ wF( t) − wv ( t) − wP( t) = wF( t) ⋅ x F( t) − wP( t) ⋅ x ( t)

M max := M min +

(

)

x( 0) = x0

wv ( t) = E⋅ ws( t) wv ws := E M 0 − M min h L0

lb wP = 8571 hr

lb ws = 43609 hr M max = 1100 lb

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 constant • f401Vlv1: three control valves sizd for 100% overcapacity, one on the feed, one on the product, and one on the steam • f403PI: 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 =

Product =

Steam =

50000

lb

100%CO hr wP⋅ 2 100%CO

⋅ 2 = 1000

= 171.4

ws⋅ 2 100%CO

lb hr⋅ %CO

lb hr⋅ %CO

= 872.2

lb hr⋅ %CO

Initial valve positions are 50%. Initial transmitter outputs: x

wf The Simulink diagram for the evaporator is:

x 0 − 40mass% ( 90 − 40)mass% wF 70000lb⋅ hr

−1

⋅ 100 %TO = 60 %TO

⋅ 100 %TO = 71.4 %TO

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.

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13-36. Control of distillation column of Example 13-5.1 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. Flow control loop gains and initial conditions sized for 200% of design flows:

Reflux flow:

4802 KLrFC :=

lbmole

KWsFC :=

⋅2

100%TO 97873

Steam flow:

hr

lbmole := 453.59mole

lb hr

mD0 := 50%CO

lbmole KLrFC = 96.04 hr⋅ %TO

mB0 := 50%CO

lb KWsFC = 1957 hr⋅ %TO

⋅2

100%TO

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: Distillate composition:

%CO Kc := 5 %TO

τ I := 60min

Bottoms composition:

%CO Kc := −3 %TO

τ I := 60min

The Simulink disgram for the control system is:

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.

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13-37. Control of gasoline blending tank of Example 12-2.5 Assuming constant volume. perfectly mixed, and constant densities, the model equations are: Total balance: Octane balance:

kbl := 42000gal f ( t) = f1 ( t) + f2 ( t) + f3 ( t) d⋅ x ( t) 1 = f ( t) x 1 ( t) + f2 ( t) ⋅ x 2 ( t) + f3 ( t) ⋅ x 3 ( t) − f ( t) ⋅ x ( t) dt V

(

d⋅ y ( t) dt

=

1 V

)

(

⋅ f1 ( t) ⋅ y 1 ( t) + f2 ( t) ⋅ y 2 ( t) + f3 ( t) ⋅ y 3 ( t) − f ( t) ⋅ y ( t)

) y( 0) = y0

RVP balance: Design conditions:

f := 60

kbl day

3 eqns. 3 unk(f,x,y) x 0 := 87

At the initial steady state:

y 0 := 7

 97  x :=  80   92 

 1 1 1   f   7.5   x x x   kbl lsolve 0 1 2  ,  f ⋅ x 0  =  28.125  y y y  f ⋅ y   24.375  day  0 1 2   0  Process parameter:

x( 0) = x0

V := 0.450kbl

The Simulink diagram for the blender is

5 y :=  11  3 Alkylate Straight run Reformate

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 min • f407Trmr: 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 constant • f403PI: three PI controllers for the product flow, octane and RVP. Flow control loops gains and initial conditions:

7.5 Alkylate:

k T1 :=

kbl day

k T2 :=

k T3 :=

kbl day

kbl day

k T2 = 0.422

7.5

kbl day⋅ %TO

⋅ 1.5

100%TO

24.375 Reformate:

k T1 = 0.113

100%TO 28.125

Straight Run:

⋅ 1.5

kbl

k T3 = 0.366

day

k T1 28.125

day⋅ %TO

⋅ 1.5

100%TO

kbl

kbl day⋅ %TO

= 66.7 %TO

kbl day

k T2

24.375

= 66.7 %TO

kbl day

k T3

= 66.7 %TO

The controllers are tuned as follows: Product flow controller:

%CO Kc := 0.9 %TO

τ I := 0.1min

RVP controller:

%CO Kc := −20 %TO

τ I := 5min

Octane controller:

%CO Kc := 30 %TO

τ I := 3min

(a) With the correct pairing, the responses to a 10 kbl/day step increase in product flow set point are at 1 min:

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.

(c) Decoupler of Example 12-3.3 The decoupler developed in Example 12-3.3 gives the gains of the decoupler on the unscaled variables: f1 f3 f2

set

= m1 − 0.71⋅ m2 + 0.27⋅ m3

set

= −0.75m1 + m2 + 0.87m3

set

= −0.25m1 − 0.29m2 + m3

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: f1

set

= k T1 ⋅ u 1

f3

set

= k T3 ⋅ u 3

f2

set

= k T2 ⋅ u 2

m1 = k T1 ⋅ v 1

m2 = k T3 ⋅ v 2

m3 = k T2 ⋅ v 3

Substitute and solve for the scaled variables u and v: k T3 k T2 u 1 = v 1 − 0.71⋅ ⋅ v 2 + 0.27⋅ v k T1 k T1 3 u 3 = −0.75

k T1

k T2 v 1 + v 2 + 0.87 v k T3 k T3 3

−0.71⋅

−0.75

k T3 k T1 k T1 k T3

= −2.308

= −0.231

%TO %CO %CO %TO

0.27⋅

0.87⋅

k T2 k T1 k T2 k T3

= 1.013

= 1.004

%CO %TO %CO %TO

u 2 = −0.25

k T1 k T2

v 1 − 0.29⋅

k T3 k T2

v2 + v3

−0.25⋅

k T1 k T2

= −0.067

%CO %TO

−0.29⋅

k T3 k T2

= −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.

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13-38. Three pump ad tank start-up problem

dp h(t) P1

P2

.fo(t)

L

P3

Development of the model equations:

Mass balance on tank:

3  π ⋅ D2  ft ⋅ h( t) = ρ ⋅ ⋅ ( f ( t) + f2 ( t) + f3 ( t) − fo ( t) ) 7.48⋅ gal 1 dt  4 

d

⋅ρ⋅

1 eqn. 2 unks.(h,f 0 ) Momentum balance on outlet pipe:

 ρ ⋅ π ⋅ d 2⋅ L⋅ v ( t)  = π d 2 p + ρ ⋅ g ⋅ h( t) − p − ∆p ( t)  ( a f ) dt  4 p  4 p a d

Outlet flow in gpm:

π

2 eqns. 4 unks.(v,∆pf)

7.48gal

2

fo ( t) = d p v ( t) ⋅ 4

ft

Pressure drop correlation: ∆p f ( t) = ρ ⋅ g ⋅  2.22⋅ 10 

3 eqns. 4 unks.

3

−3

fo ( t)  

−6 2

fo ( t) + 0.5184⋅ 10

4 eqns. 4 unks. Combine and simplify:

d⋅ h ( t) dt d ⋅ fo ( t) dt

Problem parameters:

4

=

ft

3

(

f ( t) + f2 ( t) + f3 ( t) − fo ( t) 2 7.48gal 1

πD

2

=

π ⋅ d p 7.48gal g

D := 3.41ft

4

ft

3

⋅  h ( t) − 2.22⋅ 10

−3

L 

d p := 1.19ft

L := 3170ft

)

h ( 0 ) = 0ft

−6

fo ( t) − 0.5184⋅ 10

ρ := 8.33

fo ( t)

fo ( 0 ) = 0gpm lb gal

2



g = 115.827 × 10

ft

3

h max := 9ft

2

min

Each pump flow changes from 0 to f when the flow is turned on (step function): 4

Numerical values of the coefficients:

ft

3

2 7.48gal

= 0.015

π⋅D a1 = 304

gal 2

ft⋅ min

− 3 ft⋅ min

a1 ⋅ 2.22⋅ 10

a1 ⋅ 0.5184⋅ 10

gal ft⋅ 

−6

= 0.675

min 

ft

gal min

2

π ⋅ d p 7.48gal g a1 := ⋅ 4 3 L ft

gal

1 min

2

 gal 

f := 750

= 157.6 × 10

−6 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.

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13-39. Ecological interaction of host-parasite populations Developemnt of the model equations: Rat population:

d ⋅ Rats( t) dt

= RRG( t) − RRD( t) = 5 ⋅ Rats( t) − 0.05⋅ Rats( t) ⋅ Fleas( t) Rats( 0 ) = 100

d ⋅ Fleas( t) dt

= RFG ( t) − RFD ( t) = 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.

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3 13-40. Environmental impact of chlorine barge accident degC := K kJ := 10 joule

kmole := 1000mol Problem parameters:

wc

Po

2

D := 4in

A := 212m

M 0 := 136000kg Ta := 25degC

P

kJ

U := 1000

T

Mc

Rg := 8314

Ta Properties of chlorine:

2

hr⋅ m degC joule kmole⋅ K

Po := 101300Pa

M w := 71.5

kg

λ := 288

kmole

kJ

cpL := 0.946

kg

kJ kg⋅ degC

Development of the model equations: Chlorine mass balance:

d ⋅ M c( t ) dt

Entahlpy balance:

= −wc( t)

M c( t) ⋅ cpL

d ⋅ T( t ) dt

M c( 0 ) = M 0 1 eqn. 2 unks. (Mc, wc)

(

)

= U⋅ A⋅ Ta − T( t) − λ ⋅ wc( t)

T( 0 ) = Ta

2 eqns. 3 unks. (T) 2

Flow through nozzle:

Subcritical flow:

wc( t) =

πD 4

(the smaller of the two) Critical flow:

⋅

2

wc( t) =

πD 4

⋅

3600s hr 3600s hr

⋅

⋅

(

)

2 ⋅ M w⋅ P( t) ⋅ P( t) − Po    Rg ⋅ ( T( t) + 273.16K) 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

15.961−

1978.32 T( t) + 273.16K − 27.01 101300Pa

⋅

P( T( t) ) = e

760mmHg

4 eqns. 4 nks. All the numbers given are in consistent units except for the diameter of the nozzle: D := 4in⋅ The Simulink diagram for the barge is:

0.3048m 12in

D = 0.102 m

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.

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13-41. Control of semi-batch reactor

TT

TT

Catalyst Hot water Butanol Anhydride

Cold water

Development of the model equations: Assume • Perfect mixing of ractor and jacket contents • The reactor is initially charged with the 2-btanol, essentially pure

Total mass balance:

2-butanol balance:

dMr( t) dt d dt

= FPA ( t) + FCat( t)

M r( 0 ) = M r0 1 eqn. 1 unk. (Mr)

(Mr( t)⋅ xr1( t)) = FCat( t)⋅ xfCat1 ⋅ −2⋅ rB(t) ⋅ Mw1

x r1( 0 ) = x r10

2 eqns. 3 unks. (xr1,rB) Propionic anhydride:

d dt

(Mr( t)⋅ xr2( t)) = FPA(t) ⋅ xPA − rB( t)⋅ Mw2

x r2( 0 ) = 0 3 eqs. 4 unks. (xr2)

Catalyst 1:

d dt

(Mr( t)⋅ xr4( t)) = FCat( t)⋅ xfCat4 − rCat1(t) ⋅ Mw4

x r4( 0 ) = 0

4 eqns. 6 unks. (xr4,rCat1) d

Catalyst 2:

dt

(Mr( t)⋅ xr5( t)) = rCat1( t)⋅ Mw4

x r5( t) = 0 5 eqns. 7 unks. (xr5)

d

Water:

dt

(Mr( t)⋅ xr3( t)) = FCat( t)⋅ xfCat3 + FPA( t)⋅ (1 − xPA) + rB( t)⋅ Mw3

x r3( 0 ) = 1 − x r10

6 eqns. 8 unks. (xr3) Enthalpy on reactor: d dt

(

)

⋅ M r( t) ⋅ cpavg( t) ⋅ Tr( t) =

 ( FPA( t) ⋅ cp2 + FCat( t) ⋅ cpcat) ⋅ ( Tf − T( t) ) + Uo⋅ A⋅ ( Tr( t) − Tj( t) )

− ∆Hr( t) ⋅ rB( t)

Tr( 0 ) = Tr0 7 eqns. 11 unks. (Tr,Tj,cpavg)

Enthalpy on jacket: M j⋅ cp3⋅

d ⋅ Tj ( t ) dt

(

)

(

)

(

= Uo ⋅ A⋅ Tr( t) − Tj( t) − Fcw( t) ⋅ cp3⋅ Tj( t) − Tcw + Fhw( t) ⋅ cp3⋅ Thw − Tj( t) Tj( 0 ) = Tj0

Volume of reactants:

Average density:

Vmr( t) =

ρ r( t) =

)

8 eqns. 11 unks.

M r( t) ρ r( t)

9 eqns. 13 unks. (Vmr,ρr)

 x r1( t) x r2( t) x r3( t) x rBP( t) xr4( t) + xr5( t)  + + + +  ρ2 ρ1 ρ3 ρ5 ρ4   10 eqns. 14 unks. (xrBP)

Butyl propionate:

x rBP( t) = 1 − x r1( t) − x r2( t) − x r3( t) − x r4( t) − x r5( t) 11 eqns. 14 unks.

Reaction rates:

(

)

rB( t) = Vmr( t) ⋅ k 1 ⋅ CA( t) + k 2 ⋅ CCat1 ( t) + k 3 ⋅ CCat2 ( t) CB( t) 12 eqns. 18 unks. (CA,CB, Ccat1, Ccat2) − H r( t)

rCat1 ( t) = Vmr( t) ⋅ k 4 ⋅ 10

(

CCat2 ( t) ⋅ CB( t)

)



13 eqns. 19 unks. (Hr) p4 

Hr( t) = − p 1 ⋅ CCat1 ( t) + p 2 ⋅ CCat2 ( t)  p 3 + T( t) + 273.16K  

14 eqns. 19 unks.

Concentrations:

CA( t) =

ρ r( t) ⋅ x r1( t)

CB( t ) =

M w1

CCat2 ( t) =

ρ r( t) ⋅ x r2( t) M w2

ρ r( t) ⋅ x r5( t)

CCat1 ( t) =

ρ r( t) ⋅ x r4( t) M w4

18 eqns. 19 unks.

M w4

Reactor specific heat:

(

)

cpavg( t) = x r1( t) cp1 + x r2( t) ⋅ cp2 + x r3( t) ⋅ cp3 + x r4( t) + x r5( t) ⋅ cp4 + x rBP( t) ⋅ .cp5 19 eqns. 19 unks. The reaction rate coefficients are Arrhenius functions of temperature: − Ei

k = A ⋅e i

Rg⋅ ( T( t) + 273.16K)

Problem parameters:  0.2022 

 0.3205  p :=   −21.38   1271   Rg = 8.314

kJ kmole⋅ K

M j := 415kg

 1.93⋅ 1011   14  3 1.01⋅ 10  m  A :=  14  kmole⋅ min  1.42⋅ 10   5.05⋅ 1011  

M r0 := 600kg

kJ ∆Hr := −80000 kmole

x PA := 100mass%

for i = 1..4

i

Tr0 := 20degC kJ Uo ⋅ A = 300 min⋅ degC

x r10 := 100mass%

Catalyst feed composition:

 80480  79160  kJ E=  69970  kmole  76620   Tj0 := 30degC

Tf := 20degC

Tcw := 15degC

Thw := 90degC

kg Fcw := 10000 hr

x fCat1 := 70mass%

x fCat3 := 10mass%

x fCat4 := 20mass%

Physical properties from Perry's, 7th ed., Tables 2-1 and 2-2: 2-butanol Propionic anhydride Water Sulfuric acid Butyl propionate

 74   130   kg M w :=  18   98  kmole   130 

 808    1012  kg ρ :=  1000   1829  m3   866 

 2.876   2.345   kJ cp :=  4.187   1.424  kg⋅ degC   2.345 

Hot water flow required for an inlet jacket temperature of 30 ºC:

(

)

Fcw⋅ cp3⋅ Tcw + Fhw⋅ cp3⋅ Thw = Fcw + Fhw cp3⋅ 30degC

30degC − Tcw Fhw := Fcw⋅ Thw − 30degC

kg Fhw = 2500 hr

The eight differential equations may be simplified as follows: d ⋅ x r1( t) 1 = F ( t) ⋅ xfCat1 − xr1( t) − FPA( t) ⋅ x r1( t) − 2 ⋅ rB( t) ⋅ Mw1 dt M r( t)  Cat d ⋅ x r2( t) 1 = ⋅ F ( t) ⋅ x fCat2 − x r2( t) + FPA( t) ⋅ x PA − x r2( t) − rB( t) ⋅ M w2  dt M r( t)  Cat d ⋅ x r3( t) 1 = ⋅ F ( t) ⋅ x fCat3 − x r3( t) + FPA( t) ⋅ 1 − x PA − x r3( t) + rB( t) ⋅ M w3  dt M r( t)  Cat d ⋅ x r4( t) 1 = F ( t) ⋅ xfCat4 − xr4( t) − FPA( t) ⋅ x r4( t) − rCat1 ( t) ⋅ Mw4 dt M r( t)  Cat d ⋅ x r5( t) 1 = ( t) + FPA ( t) ⋅ x r5( t) + rCat1 ( t) ⋅ M w4 − F  dt M r( t)  Cat1 d ⋅ Tr( t) =  FCat1 ( t) ⋅ cpcat + FPA( t) ⋅ cp2 ⋅ Tf − Tr( t) − Uo ⋅ A⋅ Tr( t) − Tj( t) − ∆Hr⋅ rB( t)   dt

(

)

(

)

(

(

)

(

(

)

(

(

)

)

)

)(

)

(

)

1 M ( t) ⋅ cpavg( t) d ⋅ Tj ( t ) dt d ⋅ Mj( t) dt

=

 Uo ⋅ A   c ⋅ ( Tr( t) − Tj( t) ) − Fcw( t) ⋅ ( Tj( t) − Tcw) + Fhw( t) ⋅ ( Thw − Tj( t) ) Mj  p3  1

= FCat1 ( t) + FPA ( t)

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purpose only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Smith & Corripio, 3rd edition Chapter 13. Simulation of Process Control Systems (continued) 13-18. Temperature control of stirred tank cooler of Problems 4-7 and 6-19 degC := K

The model equations, from the solution to Problem 4-7, are: d ⋅ T( t ) dt d ⋅ Tc ( t ) dt Design conditions:

Problem parameters:

=

=

f ( t) V

( Ti ( t ) − T( t ) ) −

fc( t) Vc

Ti := 70degC 3

V := 5m

U⋅ A V⋅ ρ ⋅ cp

(Tci − Tc( t)) +

( T( t ) − Tc ( t ) )

U⋅ A Vc⋅ ρ c⋅ cpc

At the initial steady state:

Tc( 0 ) = Tc0

3

f := 0.1

m

min

U := 200

kJ kg⋅ degC

T0 := 45degC

Tci := 25degC

kJ

2

A := 4m

2

ρ c := 1000

kg

(

(T0 − Tc0) ρ c⋅ cpc ( Tc0 − Tci) U⋅ A

The Simulink diagram for the cooler is:

3

m

f ⋅ ρ ⋅ cp Tc0 := T0 − Ti − T0 U⋅ A fc0 :=

T( 0 ) = T0

( T( t ) − Tc ( t ) )

min⋅ m degC cp := 3.8

kJ := 1000joule

)

cpc := 4.2

3

Vc := 1.1m

ρ := 800

kg 3

m

kJ kg⋅ degC

Tc0 = 35.5 degC 3

m fc0 = 0.172 min

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 %CO := % %TO := % From the solution to Problem 6-19: %CO Kcu := −75.4 Tu := 7.59min %TO From Table 7-1.1:

Kcu Kc := 1.7

τ I :=

Tu 2

Tu

τ D :=

8

%CO Kc = −44 τ I = 3.8 min %TO τ D = 0.95 min

Initial conditions:

Transmitter output:

45 − 20 70 − 20

= 50 %TO

fc0

Controller output:

3

0.008

m

min⋅ %CO

= 21.5 %CO

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.

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gal

gpm :=

13-19. Direct contact heater of Problem 4-5

degF := R

min

The model equations developed in the solution to Problem 4-5 are: d ⋅ T3 ( t ) dt

=

1

(

)

⋅ f ( t) ⋅ T1 ( t) − f3 ( t) T3 ( t) + V 1

hs V⋅ ρ ⋅ cp

⋅ w2 ( t)

T3 ( 0 ) = T30

w2 ( t) = 1.954 10⋅ vp( t) = k v ⋅ vp( t) w2 ( t)

f3 ( t) = f1 ( t) +

ρ

d ⋅ T3 ( t )

Substitute to eliminate f.3(t):

dt

Design conditions:

f1 := 25gpm

Problem parameters:

ρ := 7 k v :=

=

f1 ( t) V

( T1 ( t ) − T3 ( t ) ) +

T1 := 60degF

lb

cp := 0.8

gal

V⋅ ρ ⋅ cp

( h s − c p ⋅ T3 ( t ) )

T30 := 80degF

BTU lb⋅ degF

1.954 ⋅ 10 lb

w2 ( t)

V := 5gal k v = 0.0618

100%CO min

lb min⋅ %CO

Steam enthalpy, saturated at 1 atm., referenced to 32 F, is: 1150.4 BTU/lb h s := 1150.4

At the initial steady state:

w2 :=

BTU lb

+ cp ⋅ 32degF

(

f1 ⋅ ρ ⋅ cp ⋅ T30 − T1 h s − cp ⋅ T30

w2 vp0 := kv

)

h s = 1176

BTU lb

lb w2 = 2.518 min

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 τ v := 4s The Simulink block diagram is:

τ v = 0.067 min

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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.

13-20. Gas flow control valve of Example 5-2.2 The flow through a gas control valve can be calculated from Eq. 5-2.3: p1 3 fs = 836 ⋅ Cv ⋅ Cf ⋅ y − 0.148 ⋅ y G⋅ T

(

1.63

y=

Cf

⋅

∆p

G=

p1

Problem parameters from Example 5-2.2: Cv := 440

gpm

p 1 := 34.7psia

psi

Functions for valve flow:

)

Mw 29

psia := psi G := 0.621

scfh := Cf := 0.8

ft

3

hr

T := ( 259 + 460 )R

1.63 ∆p y p 1 , ∆p := ⋅ Cf p1

(

)

p1 R 3 scfh fs p 1 , ∆p := 836 ⋅ ⋅ Cv ⋅ Cf ⋅ ⋅  y p 1 , ∆p − 0.148 y p 1 , ∆p    gpm psi G⋅ T

(

)

(

)

(

)

5

6 .10

5

4 .10

(

)

fs p1 , ∆p⋅ psi scfh

5

2 .10

0

0

5

10

15

20

25

∆p Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

This is the plot also obtained with the following Simulink diagram:

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.

13-21. Temperature control of reactor of Section -2.3 and Problem 6-12 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-close • f405PIDs: 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 Valve maximum flow:

2 ⋅ 0.8771

ft

3

min

= 1.754

678.9 − 640

Transmitter initial condition:

700 − 640

ft

3

Initial position:

min

−

ln( 0.5) ln( 50)

= 17.72 %CO

= 64.8 %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: %CO Kcu := 10 %TO

Tu := 6.7min

From Table 7-1.1, the quarter decay rato tuning parameters are: Kcu Kc := 1.7

τ I :=

Tu 2

τ D :=

Tu 8

%CO Kc = 5.9 %TO

The Simulink block diagram for the temperature control loop is:

τ I = 3.4 min

τ D = 0.84 min

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.

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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. 18.3036−

p2 ( t) =

14.7psia 760mmHg

e

3816.44K

( Ts( t) +460R) K 1.8R

− 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.

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13-23. Concentration control of isothermal reactors in series of Problem 6-17 The model equation for each reactor is developed in the soution to Problem 6-17: d⋅ c ( t) j f ( t) For j = 1,2,3 = c ( 0 ) = cj0  c ( t) − c j( t) − k⋅ c j( t) j dt V  j−1 and

c ( t ) = ci ( t ) 0

Design conditions:

f := 100gpm

c := 4 0

f ⋅c At the initial steady state:

c := 1

0

f + k⋅ V

lb gal

−1

k := 0.1min f ⋅c

c := 2

V := 1000gal f ⋅c

1

c :=

f + k⋅ V

3

2

f + k⋅ V

 4  2  lb c=  1  gal  0.5  

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. From the solution to Problem 6-17:

Process gain:

Initial conditions:

gpm Kv := 2.46 %CO Kv ⋅ Ko ⋅

( 100 − 0 )%TO ( 1 − 0 )lb⋅ gal

f m0 := Kv

The Simulink diagram for the control loop is:

lb Ko := 0.0075 gal⋅ gpm = 1.845

−1

m0 = 40.65 %CO

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

τ I = τ = 9min

τ D = 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.

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13-24. Temperature and level control of oil heater of Problem 6-24 The model equations from the solution to Problem 6-24 are: d⋅ h ( t) dt

ft

=

d ⋅ T3 ( t ) dt

3

4

7.48gal

=

f1 ( t)

(

f ( t) − f3 ( t) 2 1

πD

)

h( 0) = h0

U⋅ A ⋅ T1 ( t ) − T3 ( t ) + Ts( t) − T3 ( t) V( t) V( t) ⋅ ρ ⋅ cp

(

)

(

)

T3 ( 0 ) = T30

2

V( t) = 7.48⋅

gal π ⋅ D ⋅ h ( t) 3 4 ft

d ⋅ Ts( t)

1

dt

=

 λ ⋅ ws( t) − U⋅ A⋅ ( Ts( t) − T3( t) )

Ts( 0 ) = Ts0

CM 

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:

(

)

∆p v1 = p 1 + 14.7psi − p 2 − 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:

p 1 := 45psig

p 2 := 40psia

BTU

Problem parameters: U := 136

f1 := 100gpm

D := 3ft

2

T1 := 70degF

A := 974ft⋅ π ⋅ 0.5in

T30 := 200degF A = 127.5 ft

2

hr⋅ ft degF ρ := 53

lb ft

3

cp := 0.45 At the initial steady state:

CM := 974ft⋅ 0.178 BTU

p s := 115psig

lb⋅ degF

f3 := f1 ws :=

Ts0 := T30 +

(

)

U⋅ A⋅ Ts0 − T30

The Simulink diagram for the heater is:

λ

ρ ⋅ cp ⋅ f1 U⋅ A

lb ft

⋅ 0.12

BTU lb⋅ degF

λ := 872.9

(T30 − T1)

CM = 20.8

BTU degF

BTU lb Ts0 = 343.4 degF lb ws = 47.48 min

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. Intial conditions, from the results of Problem 6-24:: ws 46 wsmax := ⋅ 1.5⋅ ws ms0 := 41.8 wsmax

lb wsmax = 78.4 ms0 = 60.58 %CO min

32.2 m10 := 1.5⋅ 46

m10 = 46.67 %CO

Ultimate gain and period of temperature controller: Quarter decay tuning parameters from Table 7-1.1:

50% for both transmitters

%CO Kcu := 77.1 %TO Kcu Kc := 1.7 %CO Kc = 45 %TO

Tu := 1.12min

τ I :=

Tu 2

τ D :=

τ I = 0.56 min

Tu 8

τ D = 0.14 min

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.

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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:

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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: %CO Kc := −0.42 %TO

Cascade: −0.56

%CO %TO

τ I := 3.5min

1.0min

τ D := 0.86min

0.26min

13-26. Feedback vs. cascade control of Problem 9-3 Simple Feedback Loop. From the solution to Problem 9-3, the ultimate gain and period are: %CO Kcu := 2.38 %TO

Tu := 27.8min

From Table 7-1.1, the quaerte decay tuning parameters for a PI controller are: Kcu Kc := 2.2

τ I :=

Tu 1.2

Cascade Control. From the solution to Problem 9-3, with %CO Kcu := 13.5 %TO

Tu := 12.5min

Kcu Kc := 2.2

τ I :=

Tu 1.2

%CO Kc = 1.1 %TO

τ I = 23 min

%CO Kc2 := 2.2 %TO

%CO Kc = 6.1 %TO

τ I = 10 min

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: 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.

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13-27. Jacketed reactor of Problem 9-4. Simple feedback vs. cascade control 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: Simple Feedback Loop: %TO Kp := 0.285 %CO

τ := 13.5min

t0 := 3.5min

The tuning parameters for minimum IAE response, from Table 7-2.2:

 t0  Kc :=  Kp  τ  1.435

− 0.921

%CO Kc = 17 %TO

 t0  τ I :=  0.878  τ  τ

0.749

τ I = 5.6 min

 t0  τ D := 0.482 ⋅ τ ⋅  τ

1.137

τ D = 1.4 min

Cascade Control (with the secondary proportional controller gain set at 2%CO/%TO): %TO Kp := 0.385 %CO

 t0  Kc := ⋅ Kp  τ  1.435

%CO Kc = 19 %TO

t0 := 2.2min

τ := 12.8min − 0.921

 t0  τ I := ⋅ 0.878  τ  τ

0.749

τ I = 3.9 min

 t0  τ D := 0.482 ⋅ τ ⋅  τ

1.137

τ D = 0.83 min

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The responses to a 5ºC step increase in set point are:

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.

13-28. Feedforward temperature control of heater of Example 13-4.2 If the variation in the inlet temperature diturbance is neglected, the feedforward controller calculation becomes: set

w ⋅ ( t) =

ρ ⋅ cp ⋅ T



⋅ ( t) − Ti 

set

f ( t) = Ratio⋅ f ( t)

λ

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: ρ := 68

lb ft

3

cp := 0.8

The ratio for set point of

BTU

λ := 966

lb⋅ degF

BTU lb

Ti := 100degF

lb fmax := 75 min

Tset := 150degF Ratio :=

(

ρ ⋅ cp ⋅ Tset − Ti λ

) ⋅ 100%TO fmax

Ratio = 3.75

min⋅ %TO ft

3

The Simulink diagram for the ratio control scheme is obtained by a simple modification of the diagram of Fig. 13-4.11:

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 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.

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13-29. Ratio control of blending tank of Problems 3-18 and 13-11 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. Design conditions: 3 3 f2 m m f1 := 1.6 f2 := 2.4 Ratio := Ratio = 1.5 min min f1 3

Transmitter ranges:

m f1max := 4 min

3

m f2max := 5.42 Rmax := 3 min

(at 100%CO)

The signal from the flow transmitter must then be scaled by the factor: Rmax⋅ f1max 100%CO⋅ f2max The controller gain must be adjusted by the factor the output is multiplied by: %CO 100%CO f2max Kc := −147 ⋅ ⋅ %TO Rmax f1 The Simulink diagram for the ratio control scheme is:

%CO Kc = −166 %TO

= 0.0221

1 %CO

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.

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13-30. Cascade temperature control of reactor of Section 4-2.3 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. An open-loop step test in the slave controller results in the following parameters: Slave loop:

%TO K2 := 1.95 %CO

Synthesis tuning from Table 7-4.1: Master loop:

%TO Kcu := 10.5 %CO

τ := 3min τ I := τ

t0 := 0min Set Kc to 2%CO/%TO to obtain fast response without much oscillation.

Tu := 3.75min

Quarter decay ratio tuning from Table 7-1.1:

Kcu Kc := 1.7 τ I :=

Tu 2

The Simulink diagram for the cascade control loop is:

%CO Kc = 6 %TO τ D :=

Tu 8

τ I = 1.9 min τ D = 0.47 min

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.

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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: −GD FFC = 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: KDT 100%TO %TO HD = KDT := KDT = 50 3 −1 3 −1 0.1⋅ s + 1 2ft ⋅ min ft ⋅ min From step tests on the disturbance, the folowing parameters are obtained: − 5.5s

e %TO GD = 61.5 ⋅ 3 − 1 9s + 1 ft ⋅ min

−61.5%TO

FFC =

−1

3

− 3.25s

KDT⋅ ft min

%TO e GM = −0.98 ⋅ %CO 6.75s + 1

⋅

⋅ ( −0.98)

6.75s + 1 − ( 5.5−3.25)s e 9s + 1

−61.5%TO 3

Feedforward controller gain:

−1

KDT⋅ ft min FFC = 1.225

= 1.255

⋅ ( −0.98)

%CO %TO

%CO 6.75s + 1 − 2.25s+ 1 ⋅ ⋅e %TO 9.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

delay = 4.5 min

To start with a correct initial set point of the coolant flow controller, a bias is needed: Initial reactant transmitter output: Bias required for the same set point:

1.3364

ft

3

min

⋅ KDT = 66.82 %TO

m( 0 ) = m( 0 ) + 1.255 ⋅ 66.82 %TO + Bias = m( 0 ) Bias := −1.255 ⋅ 66.82 %TO

Bias = −83.86 %TO

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: 1.2 Kc := −0.98

 3.25    6.75 

−1

τ I := 2 ⋅ 3.25min

τ D :=

3.25min 2

τ I = 6.5 min The Simulink diagram for the linear feedforward controller is:

%CO Kc = −2.5 %TO τ D = 1.6 min

The responses to a 0.2 ft3/min step decrease in reactants flow at 1 min are

The feedforward responses are in magenta and the simple feedback responses are in yellow. The feedforward controller required some adjustments: Delay := 3.5min %CO Kc := −5 %TO τ I := 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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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: 1. Control objective: T(t) = Tset(t) (reactor temperature) 2. Manipulated variable: fc(t) (coolant flow to the jacket) 3. Disturbances: f(t) (reactants flow) 4. Steady state balances: 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) ⋅ ρ c⋅ cpc⋅ Tc − Tci Combine to eliminate rA(t): f ( t ) ⋅  ρ ⋅ c p ⋅  Ti − T





(

)(

( t)  + −∆Hr ⋅ cAi − cA 

set

Solve for the manipulated variable:

) = fc(t) ⋅ ρ c⋅ cpc⋅ (Tc − Tci)

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 changes 7. 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 := 55

lb ft

cpc := 1

3

cp := 0.88

BTU

Ratio at design conditions:

Ti := 635R

lb⋅ degF

cAi := 0.5975

BTU lb⋅ degF

lbmole ft

Ratio :=

Check:

BTU ∆Hr := −12000 lbmole T := 678.9R

cA := 0.2068

3

(

Tci := 540R

ft

)

3

min

ft

3

Tc := 602.7R

3

) ( )( ρ c⋅ cpc⋅ ( Tc − Tci) ft

lb

lbmole

ρ ⋅ cp ⋅ Ti − T + −∆Hr ⋅ cAi − cA

f := 1.3364

ρ c := 62.4

Ratio = 0.655 3

fc := Ratio⋅ f

ft fc = 0.876 min

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.

3

3

ft fmax := 2 min

Rmax := 2 The scaled variables are:

x=

ft fcmax := 1.75 min

f fmax

⋅ 100 %TO

y=

fcmax

Substitute into the feedforward equation:

100%TO Scaled feedforward equation:

Initial conditions:

y=

Ratio m0 := ⋅ 100 %CO Rmax

y=

fc fcmax

⋅ 100 %TO

Rmax 100%CO

Rmax⋅ fmax fcmax⋅ 100 %CO m0 = 32.8 %CO

⋅ r⋅

r=

fmax 100%TO

Ratio Rmax

⋅ 100 %CO

⋅x

⋅ r⋅ x f fmax fc

= 66.8 %TO

fcmax

= 50 %TO

To maintain the same loop gain, the feedback controller gain must be adjusted: Rmax⋅ fmax Rmax⋅ f f Kc = Kcnew ⋅ ⋅ ⋅ 100 %TO = ⋅ Kcnew fcmax⋅ 100 %CO fmax fcmax %CO Kc := −5 %TO

fcmax Kcnew := ⋅K Rmax⋅ f c

The Simulink diagram for the ratio control scheme is:

%CO Kcnew = −3.3 %TO

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 equalpercentage valve and the ratio is adjusting the coolant flow set point. This is also teh reason they move in opposite directions.

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Smith & Corripio, 3rd edition Chapter 13. Simulation of Process Control Systems 13-33. Cascade control of reactors in series of Problem 13-23 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 From the results of an open-loop step test on the slave controler output: %TO K2 := 0.58 %CO

τ 2 := 6min

Synthesis tuning from Table 7-4.1:

%CO := %

%TO := %

t0 := 0min

τ I2 := τ 2

Kc2 = tunable

Set at

%CO Kc2 := 5 %TO

With these slave settings, an open-loop step test on the master controller output: %TO K1 := 3.3 %CO

τ .1 := 5min

Use synthesis tuning for the master also:

t0 := 0min τ I1 := 5min

The Simulink diagram for the cascade control scheme is:

τ D1 := 0min

Set

%CO Kc1 := 5 %TO

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.

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13-34. Multivariable control of caustic blending tank of Problem 12-2. Design conditions:

klb := 1000lb mass% := %

klb

wP := 40 hr

SP AC

SP

3

FC

wc

xc

FT

3

Caustic

SP

x

2

x c := 50mass% Problem parameters: M := 10klb

FT

AT

FC Water

SP FC

1

wP

FT

ww

At initial steady state: wP⋅ x 0 klb wc := wc = 24 xc hr klb ww := wP − wc ww = 16 hr

Assume perfectly mixed tank with constant mass: Caustic balance: Total balance:

d⋅ x ( t) dt

=

1 M

(wc(t) ⋅ xc( t) − wP( t)⋅ x(t) )

wP( t) = wc( t) + ww( t)

The Simulink diagram to simulate tank is:

x 0 := 30mass%

x ( 0 ) = x0

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 constant • Two 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. ww klb Initial conditions: FC-1: 16 mw0 := mw0 = 26.667 %CO −1 hr 60klb⋅ hr FC-2:

FC-3:

AC-3:

24

40

klb hr

mc0 :=

30mass%

60klb⋅ hr

wP

klb hr

wc

60klb ⋅ hr

−1

x0 50mass%

−1

mc0 = 40 %CO

= 66.667 %TO

= 60 %TO

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) With these tuning parameters, the responses to a

(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.

(c) Linear decoupler design as in Example 12-3.1 To keep the flow constant, the caustic flow must be adjusted as follows:

(

)

∆wP = Kvw + D12⋅ Kvc ∆mAC = 0 −Kvw %CO D12 = = −1 Kvc %CO

(

Kvc = Kvw =

−1

100%CO

%CO D12 := −1 %CO

)

∆x = Kvc⋅ Kxc + D21⋅ Kvw⋅ Kxw ∆mFC −Kvc⋅ Kxc ww D21 = = Kvw⋅ Kxw wc

60⋅ klb ⋅ hr

ww Kxc = wP

ww D21 := wc

−wc Kxw = wP %CO D21 = 0.667 %CO

Initial outputs: To start at the proper decoupler outputs, biases must be added to the decouplers: mc0 = mc0 + D12⋅ mw0 + Bc

Bc := −D12⋅ mw0

Bc = 26.667 %CO

mw0 = mw0 + D21⋅ mc0 + Bw

Bw := −D21⋅ mc0

Bw = −26.667 %CO

The Simulink diagram for the linear decoupler is:

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

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13-35. Multivariable control of evaporator of Figure 12-3.4 Design conditions:

wV

Cond.

wF xF

wP x

T

wS

lb wF := 50000 hr

x F := 12mass%

x 0 := 70mass%

E := 0.95

M 0 := 747lb

h L0 := 50%TO

Process parameters: M min := 394lb

Ts Condensate

Developent of the model equations (assume perfect mixing): Total mass balance:

Sugar balance:

d⋅ M( t)

= wF( t) − wv ( t) − wP( t)

dt

d⋅ ( M( t) ⋅ x( t) ) dt

M( 0) = M0

= wF( t) ⋅ x F( t) − wP( t) ⋅ x ( t)

2 eqns. 2 unk.(M,x)

Expand and substitute total balance:

M( t) ⋅ Simplify:

d⋅ x ( t)

d⋅ x ( t) dt

dt =

)

1 w ( t) ⋅ xF( t) − x( t) + wv ( t) ⋅ x( t) M( t)  F

Economy relates steam and vapor flows: At the intial steady state: wF⋅ x F wP := wv := wF − wP x0 Maximum hold up:

(

+ x ( t) ⋅ wF( t) − wv ( t) − wP( t) = wF( t) ⋅ x F( t) − wP( t) ⋅ x ( t)

M max := M min +

(

)

x( 0) = x0

wv ( t) = E⋅ ws( t) wv ws := E M 0 − M min h L0

lb wP = 8571 hr

lb ws = 43609 hr M max = 1100 lb

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 constant • f401Vlv1: three control valves sizd for 100% overcapacity, one on the feed, one on the product, and one on the steam • f403PI: 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 =

Product =

Steam =

50000

lb

100%CO hr wP⋅ 2 100%CO

⋅ 2 = 1000

= 171.4

ws⋅ 2 100%CO

lb hr⋅ %CO

lb hr⋅ %CO

= 872.2

lb hr⋅ %CO

Initial valve positions are 50%. Initial transmitter outputs: x

wf The Simulink diagram for the evaporator is:

x 0 − 40mass% ( 90 − 40)mass% wF 70000lb⋅ hr

−1

⋅ 100 %TO = 60 %TO

⋅ 100 %TO = 71.4 %TO

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.

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13-36. Control of distillation column of Example 13-5.1 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. Flow control loop gains and initial conditions sized for 200% of design flows:

Reflux flow:

4802 KLrFC :=

lbmole

KWsFC :=

⋅2

100%TO 97873

Steam flow:

hr

lbmole := 453.59mole

lb hr

mD0 := 50%CO

lbmole KLrFC = 96.04 hr⋅ %TO

mB0 := 50%CO

lb KWsFC = 1957 hr⋅ %TO

⋅2

100%TO

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: Distillate composition:

%CO Kc := 5 %TO

τ I := 60min

Bottoms composition:

%CO Kc := −3 %TO

τ I := 60min

The Simulink disgram for the control system is:

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.

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13-37. Control of gasoline blending tank of Example 12-2.5 Assuming constant volume. perfectly mixed, and constant densities, the model equations are: Total balance: Octane balance:

kbl := 42000gal f ( t) = f1 ( t) + f2 ( t) + f3 ( t) d⋅ x ( t) 1 = f ( t) x 1 ( t) + f2 ( t) ⋅ x 2 ( t) + f3 ( t) ⋅ x 3 ( t) − f ( t) ⋅ x ( t) dt V

(

d⋅ y ( t) dt

=

1 V

)

(

⋅ f1 ( t) ⋅ y 1 ( t) + f2 ( t) ⋅ y 2 ( t) + f3 ( t) ⋅ y 3 ( t) − f ( t) ⋅ y ( t)

) y( 0) = y0

RVP balance: Design conditions:

f := 60

kbl day

3 eqns. 3 unk(f,x,y) x 0 := 87

At the initial steady state:

y 0 := 7

 97  x :=  80   92 

 1 1 1   f   7.5   x x x   kbl lsolve 0 1 2  ,  f ⋅ x 0  =  28.125  y y y  f ⋅ y   24.375  day  0 1 2   0  Process parameter:

x( 0) = x0

V := 0.450kbl

The Simulink diagram for the blender is

5 y :=  11  3 Alkylate Straight run Reformate

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 min • f407Trmr: 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 constant • f403PI: three PI controllers for the product flow, octane and RVP. Flow control loops gains and initial conditions:

7.5 Alkylate:

k T1 :=

kbl day

k T2 :=

k T3 :=

kbl day

kbl day

k T2 = 0.422

7.5

kbl day⋅ %TO

⋅ 1.5

100%TO

24.375 Reformate:

k T1 = 0.113

100%TO 28.125

Straight Run:

⋅ 1.5

kbl

k T3 = 0.366

day

k T1 28.125

day⋅ %TO

⋅ 1.5

100%TO

kbl

kbl day⋅ %TO

= 66.7 %TO

kbl day

k T2

24.375

= 66.7 %TO

kbl day

k T3

= 66.7 %TO

The controllers are tuned as follows: Product flow controller:

%CO Kc := 0.9 %TO

τ I := 0.1min

RVP controller:

%CO Kc := −20 %TO

τ I := 5min

Octane controller:

%CO Kc := 30 %TO

τ I := 3min

(a) With the correct pairing, the responses to a 10 kbl/day step increase in product flow set point are at 1 min:

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.

(c) Decoupler of Example 12-3.3 The decoupler developed in Example 12-3.3 gives the gains of the decoupler on the unscaled variables: f1 f3 f2

set

= m1 − 0.71⋅ m2 + 0.27⋅ m3

set

= −0.75m1 + m2 + 0.87m3

set

= −0.25m1 − 0.29m2 + m3

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: f1

set

= k T1 ⋅ u 1

f3

set

= k T3 ⋅ u 3

f2

set

= k T2 ⋅ u 2

m1 = k T1 ⋅ v 1

m2 = k T3 ⋅ v 2

m3 = k T2 ⋅ v 3

Substitute and solve for the scaled variables u and v: k T3 k T2 u 1 = v 1 − 0.71⋅ ⋅ v 2 + 0.27⋅ v k T1 k T1 3 u 3 = −0.75

k T1

k T2 v 1 + v 2 + 0.87 v k T3 k T3 3

−0.71⋅

−0.75

k T3 k T1 k T1 k T3

= −2.308

= −0.231

%TO %CO %CO %TO

0.27⋅

0.87⋅

k T2 k T1 k T2 k T3

= 1.013

= 1.004

%CO %TO %CO %TO

u 2 = −0.25

k T1 k T2

v 1 − 0.29⋅

k T3 k T2

v2 + v3

−0.25⋅

k T1 k T2

= −0.067

%CO %TO

−0.29⋅

k T3 k T2

= −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.

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13-38. Three pump ad tank start-up problem

dp h(t) P1

P2

.fo(t)

L

P3

Development of the model equations:

Mass balance on tank:

3  π ⋅ D2  ft ⋅ h( t) = ρ ⋅ ⋅ ( f ( t) + f2 ( t) + f3 ( t) − fo ( t) ) 7.48⋅ gal 1 dt  4 

d

⋅ρ⋅

1 eqn. 2 unks.(h,f 0 ) Momentum balance on outlet pipe:

 ρ ⋅ π ⋅ d 2⋅ L⋅ v ( t)  = π d 2 p + ρ ⋅ g ⋅ h( t) − p − ∆p ( t)  ( a f ) dt  4 p  4 p a d

Outlet flow in gpm:

π

2 eqns. 4 unks.(v,∆pf)

7.48gal

2

fo ( t) = d p v ( t) ⋅ 4

ft

Pressure drop correlation: ∆p f ( t) = ρ ⋅ g ⋅  2.22⋅ 10 

3 eqns. 4 unks.

3

−3

fo ( t)  

−6 2

fo ( t) + 0.5184⋅ 10

4 eqns. 4 unks. Combine and simplify:

d⋅ h ( t) dt d ⋅ fo ( t) dt

Problem parameters:

4

=

ft

3

(

f ( t) + f2 ( t) + f3 ( t) − fo ( t) 2 7.48gal 1

πD

2

=

π ⋅ d p 7.48gal g

D := 3.41ft

4

ft

3

⋅  h ( t) − 2.22⋅ 10

−3

L 

d p := 1.19ft

L := 3170ft

)

h ( 0 ) = 0ft

−6

fo ( t) − 0.5184⋅ 10

ρ := 8.33

fo ( t)

fo ( 0 ) = 0gpm lb gal

2



g = 115.827 × 10

ft

3

h max := 9ft

2

min

Each pump flow changes from 0 to f when the flow is turned on (step function): 4

Numerical values of the coefficients:

ft

3

2 7.48gal

= 0.015

π⋅D a1 = 304

gal 2

ft⋅ min

− 3 ft⋅ min

a1 ⋅ 2.22⋅ 10

a1 ⋅ 0.5184⋅ 10

gal ft⋅ 

−6

= 0.675

min 

ft

gal min

2

π ⋅ d p 7.48gal g a1 := ⋅ 4 3 L ft

gal

1 min

2

 gal 

f := 750

= 157.6 × 10

−6 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.

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13-39. Ecological interaction of host-parasite populations Developemnt of the model equations: Rat population:

d ⋅ Rats( t) dt

= RRG( t) − RRD( t) = 5 ⋅ Rats( t) − 0.05⋅ Rats( t) ⋅ Fleas( t) Rats( 0 ) = 100

d ⋅ Fleas( t) dt

= RFG ( t) − RFD ( t) = 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.

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3 13-40. Environmental impact of chlorine barge accident degC := K kJ := 10 joule

kmole := 1000mol Problem parameters:

wc

Po

2

D := 4in

A := 212m

M 0 := 136000kg Ta := 25degC

P

kJ

U := 1000

T

Mc

Rg := 8314

Ta Properties of chlorine:

2

hr⋅ m degC joule kmole⋅ K

Po := 101300Pa

M w := 71.5

kg

λ := 288

kmole

kJ

cpL := 0.946

kg

kJ kg⋅ degC

Development of the model equations: Chlorine mass balance:

d ⋅ M c( t ) dt

Entahlpy balance:

= −wc( t)

M c( t) ⋅ cpL

d ⋅ T( t ) dt

M c( 0 ) = M 0 1 eqn. 2 unks. (Mc, wc)

(

)

= U⋅ A⋅ Ta − T( t) − λ ⋅ wc( t)

T( 0 ) = Ta

2 eqns. 3 unks. (T) 2

Flow through nozzle:

Subcritical flow:

wc( t) =

πD 4

(the smaller of the two) Critical flow:

⋅

2

wc( t) =

πD 4

⋅

3600s hr 3600s hr

⋅

⋅

(

)

2 ⋅ M w⋅ P( t) ⋅ P( t) − Po    Rg ⋅ ( T( t) + 273.16K) 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

15.961−

1978.32 T( t) + 273.16K − 27.01 101300Pa

⋅

P( T( t) ) = e

760mmHg

4 eqns. 4 nks. All the numbers given are in consistent units except for the diameter of the nozzle: D := 4in⋅ The Simulink diagram for the barge is:

0.3048m 12in

D = 0.102 m

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.

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13-41. Control of semi-batch reactor

TT

TT

Catalyst Hot water Butanol Anhydride

Cold water

Development of the model equations: Assume • Perfect mixing of ractor and jacket contents • The reactor is initially charged with the 2-btanol, essentially pure

Total mass balance:

2-butanol balance:

dMr( t) dt d dt

= FPA ( t) + FCat( t)

M r( 0 ) = M r0 1 eqn. 1 unk. (Mr)

(Mr( t)⋅ xr1( t)) = FCat( t)⋅ xfCat1 ⋅ −2⋅ rB(t) ⋅ Mw1

x r1( 0 ) = x r10

2 eqns. 3 unks. (xr1,rB) Propionic anhydride:

d dt

(Mr( t)⋅ xr2( t)) = FPA(t) ⋅ xPA − rB( t)⋅ Mw2

x r2( 0 ) = 0 3 eqs. 4 unks. (xr2)

Catalyst 1:

d dt

(Mr( t)⋅ xr4( t)) = FCat( t)⋅ xfCat4 − rCat1(t) ⋅ Mw4

x r4( 0 ) = 0

4 eqns. 6 unks. (xr4,rCat1) d

Catalyst 2:

dt

(Mr( t)⋅ xr5( t)) = rCat1( t)⋅ Mw4

x r5( t) = 0 5 eqns. 7 unks. (xr5)

d

Water:

dt

(Mr( t)⋅ xr3( t)) = FCat( t)⋅ xfCat3 + FPA( t)⋅ (1 − xPA) + rB( t)⋅ Mw3

x r3( 0 ) = 1 − x r10

6 eqns. 8 unks. (xr3) Enthalpy on reactor: d dt

(

)

⋅ M r( t) ⋅ cpavg( t) ⋅ Tr( t) =

 ( FPA( t) ⋅ cp2 + FCat( t) ⋅ cpcat) ⋅ ( Tf − T( t) ) + Uo⋅ A⋅ ( Tr( t) − Tj( t) )

− ∆Hr( t) ⋅ rB( t)

Tr( 0 ) = Tr0 7 eqns. 11 unks. (Tr,Tj,cpavg)

Enthalpy on jacket: M j⋅ cp3⋅

d ⋅ Tj ( t ) dt

(

)

(

)

(

= Uo ⋅ A⋅ Tr( t) − Tj( t) − Fcw( t) ⋅ cp3⋅ Tj( t) − Tcw + Fhw( t) ⋅ cp3⋅ Thw − Tj( t) Tj( 0 ) = Tj0

Volume of reactants:

Average density:

Vmr( t) =

ρ r( t) =

)

8 eqns. 11 unks.

M r( t) ρ r( t)

9 eqns. 13 unks. (Vmr,ρr)

 x r1( t) x r2( t) x r3( t) x rBP( t) xr4( t) + xr5( t)  + + + +  ρ2 ρ1 ρ3 ρ5 ρ4   10 eqns. 14 unks. (xrBP)

Butyl propionate:

x rBP( t) = 1 − x r1( t) − x r2( t) − x r3( t) − x r4( t) − x r5( t) 11 eqns. 14 unks.

Reaction rates:

(

)

rB( t) = Vmr( t) ⋅ k 1 ⋅ CA( t) + k 2 ⋅ CCat1 ( t) + k 3 ⋅ CCat2 ( t) CB( t) 12 eqns. 18 unks. (CA,CB, Ccat1, Ccat2) − H r( t)

rCat1 ( t) = Vmr( t) ⋅ k 4 ⋅ 10

(

CCat2 ( t) ⋅ CB( t)

)



13 eqns. 19 unks. (Hr) p4 

Hr( t) = − p 1 ⋅ CCat1 ( t) + p 2 ⋅ CCat2 ( t)  p 3 + T( t) + 273.16K  

14 eqns. 19 unks.

Concentrations:

CA( t) =

ρ r( t) ⋅ x r1( t)

CB( t ) =

M w1

CCat2 ( t) =

ρ r( t) ⋅ x r2( t) M w2

ρ r( t) ⋅ x r5( t)

CCat1 ( t) =

ρ r( t) ⋅ x r4( t) M w4

18 eqns. 19 unks.

M w4

Reactor specific heat:

(

)

cpavg( t) = x r1( t) cp1 + x r2( t) ⋅ cp2 + x r3( t) ⋅ cp3 + x r4( t) + x r5( t) ⋅ cp4 + x rBP( t) ⋅ .cp5 19 eqns. 19 unks. The reaction rate coefficients are Arrhenius functions of temperature: − Ei

k = A ⋅e i

Rg⋅ ( T( t) + 273.16K)

Problem parameters:  0.2022 

 0.3205  p :=   −21.38   1271   Rg = 8.314

kJ kmole⋅ K

M j := 415kg

 1.93⋅ 1011   14  3 1.01⋅ 10  m  A :=  14  kmole⋅ min  1.42⋅ 10   5.05⋅ 1011  

M r0 := 600kg

kJ ∆Hr := −80000 kmole

x PA := 100mass%

for i = 1..4

i

Tr0 := 20degC kJ Uo ⋅ A = 300 min⋅ degC

x r10 := 100mass%

Catalyst feed composition:

 80480  79160  kJ E=  69970  kmole  76620   Tj0 := 30degC

Tf := 20degC

Tcw := 15degC

Thw := 90degC

kg Fcw := 10000 hr

x fCat1 := 70mass%

x fCat3 := 10mass%

x fCat4 := 20mass%

Physical properties from Perry's, 7th ed., Tables 2-1 and 2-2: 2-butanol Propionic anhydride Water Sulfuric acid Butyl propionate

 74   130   kg M w :=  18   98  kmole   130 

 808    1012  kg ρ :=  1000   1829  m3   866 

 2.876   2.345   kJ cp :=  4.187   1.424  kg⋅ degC   2.345 

Hot water flow required for an inlet jacket temperature of 30 ºC:

(

)

Fcw⋅ cp3⋅ Tcw + Fhw⋅ cp3⋅ Thw = Fcw + Fhw cp3⋅ 30degC

30degC − Tcw Fhw := Fcw⋅ Thw − 30degC

kg Fhw = 2500 hr

The eight differential equations may be simplified as follows: d ⋅ x r1( t) 1 = F ( t) ⋅ xfCat1 − xr1( t) − FPA( t) ⋅ x r1( t) − 2 ⋅ rB( t) ⋅ Mw1 dt M r( t)  Cat d ⋅ x r2( t) 1 = ⋅ F ( t) ⋅ x fCat2 − x r2( t) + FPA( t) ⋅ x PA − x r2( t) − rB( t) ⋅ M w2  dt M r( t)  Cat d ⋅ x r3( t) 1 = ⋅ F ( t) ⋅ x fCat3 − x r3( t) + FPA( t) ⋅ 1 − x PA − x r3( t) + rB( t) ⋅ M w3  dt M r( t)  Cat d ⋅ x r4( t) 1 = F ( t) ⋅ xfCat4 − xr4( t) − FPA( t) ⋅ x r4( t) − rCat1 ( t) ⋅ Mw4 dt M r( t)  Cat d ⋅ x r5( t) 1 = ( t) + FPA ( t) ⋅ x r5( t) + rCat1 ( t) ⋅ M w4 − F  dt M r( t)  Cat1 d ⋅ Tr( t) =  FCat1 ( t) ⋅ cpcat + FPA( t) ⋅ cp2 ⋅ Tf − Tr( t) − Uo ⋅ A⋅ Tr( t) − Tj( t) − ∆Hr⋅ rB( t)   dt

(

)

(

)

(

(

)

(

(

)

(

(

)

)

)

)(

)

(

)

1 M ( t) ⋅ cpavg( t) d ⋅ Tj ( t ) dt d ⋅ Mj( t) dt

=

 Uo ⋅ A   c ⋅ ( Tr( t) − Tj( t) ) − Fcw( t) ⋅ ( Tj( t) − Tcw) + Fhw( t) ⋅ ( Thw − Tj( t) ) Mj  p3  1

= FCat1 ( t) + FPA ( t)

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purpose 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.

Chapter 13. Simulation of Process Control Systems Solutions to Problems 13-1 to 13-17

13-1. Simulation of Bird Mobile of Problem 2-9. The differential equation representing the motion of the bird mobile is, from Problem 2-9: 2

M⋅

d y ( t) 2

= −M ⋅ g − k ⋅ y ( t) + f ( t)

dt

Solve for the highest derivative:

2

d y ( t) 2

= −g +

dt

Integration of this second derivative results in: From the solution of problem 2-9:

f ( t) M

d⋅ y ( t)

M := 50gm

dt

−

k M

y( t)

and a second integartion gives y ( t)

g = 980.7

cm 2

s For this problem:

y ( 0 ) = 100cm

f ( t) = 0dynes

k :=

M⋅ g 27cm

k = 1816

gm 2

s

The period of oscillation is, as in the solution to problem 2-9:

Period := 2 ⋅ π ⋅

M k

Period = 1.043 s The number of complete cycles in 10 seconds is: 10s Period

= 9.592

The simulation plot shows the same result.

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13-2. Simulation of a Pendulum Application of Newton's Second Law of Motion: −M ⋅ g ⋅ tan( θ ) = M ⋅

2

d x ( t)

2

2

dt

L

For small angles θ, from the geometry: tan( θ ) = sin( θ ) =

x ( t) L

Mg sin2

Substitute and simplify to obtain: 2

d x ( t)

g

 L

2

dt g = 9.807

= −

m 2

x ( t)

M := 0.5kg

x(t) L := 0.8m

s

g r1. = i⋅ L

g r2 = −i⋅ L

The frequency of oscillation is:

x ( 0 ) = 0.1m

 s2 + g  X( s) = 0  L 

The solution of the differential equation:

Roots:

Mg

g L

 g t  L 

x ( t) = x ( 0 ) cos

= 3.501 Hz

Period :=

2⋅ π g L

(Table 2-1.1)

Period = 1.795 s

The number of oscillations in 10 s is: 10s Period

= 5.572

The simulation plot shows the same result.

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13-3. Simulation of Punctured Air Tank of Problem 2-23. In Problem 2-23 the mass balance on the tank produced the following equation: V Flow through the orifice:

dρ( t) dt

= wi( t) − wo ( t)

(

wo ( t) = Ao ⋅ 2 ρ ( t) p ( t) − p o

d⋅ p ( t) dt 3

V := 1.5m

Problem parameters:

=

Rg ⋅ T V⋅ M

Ao := 0.785cm

2

3

Rg := 8.314 ⋅

M

ρ ( t) =

Ideal gas law, assuming constant temperature:

Substitute and solve for dp(t)/dt:

)

Pa⋅ m

mole⋅ K

Rg ⋅ T

p( t)

(wi(t) − wo(t) ) M := 29

gm mole

T := ( 70 + 273.16)K

p o := 101300Pa

p ( 0 ) = ( 500000 + 101300)Pa

From the linearization of Problem 2-23, we know that the time constant is: V⋅ 2 ⋅ τ :=

M Rg⋅ T

⋅ 601300Pa⋅ 500000Pa

(

Ao ⋅ 2 ⋅ 601300Pa − p o

)

τ = 42.895 s

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: wi := Ao ⋅ 2 ⋅

M ⋅ 601300Pa Rg ⋅ T

(601300Pa − po)

kg wi = 0.194 s

As predicted by the linearized model, the pressure reaches steady state in about 200 s.

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13-4. Simulation of the turkey temperature response of Problem 2-24. From the solution to Problem 2-24, the differential equation obtained from an energy balace on the turkey is: M ⋅ cv ⋅

dT ( t) dt

= σ ⋅ ε ⋅ A Ts( t) − T( t) M := 12lb

The parameters, given in this problem are: cv := 0.95

4



BTU



A := 3.5ft

T( 0 ) = 535R

lb⋅ R

4 2

Ts := 800R −8

σ := 0.1718⋅ 10

ε := 0.6 BTU 2 4

hr⋅ ft R

By the linearization done in Problem 2-24, the time constant of the turkey is: τ :=

M ⋅ cv 4 ⋅ σ ⋅ ε ⋅ A⋅ ( 535R)

Integrate the differential equation :

τ = 5.16 hr

3

d ⋅ T( t ) dt

=

σ ⋅ε ⋅A 

4 4 Ts( t) − T( t)    M ⋅ cv

Run the simulation for 25 hrs (five time constants). Simulate the oven as a step function from an inital temperature of 535ºR to 800ºR.

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: M ⋅ cv τ := τ = 1.54 hr 3 4 ⋅ σ ⋅ ε ⋅ A⋅ ( 800R) 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.

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13-5. Non-isothermal Chemical Reactor of Section 4-2.3 lbmole := 453.59mole

Rearranging the model equations from Section 4-2.3: d ⋅ cA( t) f ( t) = c ( t) − cA( t) − rA( t) dt V Ai

(

)

cA( 0 ) = 0.2068

lbmole ft

−E

rA( t) = k o ⋅ e d ⋅ T( t )

=

dt d ⋅ Tc ( t ) dt

=

Design conditions: Parameters:

f ( t) V

2

⋅ cA ( t )

3

lb ft

∆Hr

( Ti ( t ) − T( t ) ) −

fc( t) Vc

cAi := 0.5975

Rg := 1.987

⋅ rA( t) −

ρ ⋅ Cp

(Tci( t) − Tc( t)) +

V := 13.46ft

ft fc := 0.8771 min ρ c := 62.4

Rg⋅ T( t)

V⋅ ρ ⋅ Cp

U⋅ A

lbmole ft

U⋅ A

Vc⋅ ρ c⋅ Cpc

( T( t ) − Tc ( t ) )

3

3

ft

8

ρ := 55

lbmole⋅ R

U := 75

ft

3

Tci := 540R

min

3

E := 27820

lbmole⋅ min

lb ft

Tc( 0 ) = 602.7R

f := 1.3364

k o := 8.33⋅ 10 BTU

T( 0 ) = 678.9R

( T( t ) − Tc ( t ) )

Ti := 633.5R

BTU ∆Hr := −12000 lbmole

3

3

Cp := 0.88

3

BTU 2

hr⋅ ft ⋅ R

BTU lb⋅ R

Vc := 1.56ft

3

A := 36ft

Cpc := 1

BTU lbmole

2

BTU lb⋅ R

Check that initial conditions are at steady state (derivatives = 0): −E

rA := k o ⋅ e f V

Rg⋅ 678.9R

⋅  cAi − 0.2068

f

 

Vc

 

ft

3

∆Hr ρ ⋅ Cp

(Tci − 602.7R) +

lbmole  ft

lbmole 

(T − 678.9R) − V i fc

⋅  0.2068



3

2

lbmole rA = 0.039 3 ft ⋅ min

 − 4 lbmole

− rA = −5.87 × 10

⋅ rA −

U⋅ A V⋅ ρ ⋅ Cp

U⋅ A Vc⋅ ρ c⋅ Cpc

3

ft ⋅ min

( 678.9 − 602.7 )R = −7.915 × 10

( 678.9 − 602.7 )R = −0.027

The following is the Simulink diagram for the reactor:

R min

−3 R

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. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

13-6. Mixing Process of Problem 3-1 The model equation, from the solution to Problem 3-1: d ⋅ cA( t) dt Problem parameters:

Total mass balance:

At the initial steady state:

=

f1 ( t) ⋅ cA1( t) + f2 ( t) ⋅ cA2( t) − f ( t) ⋅ cA( t) A⋅ h

cA1 := 0.01 f1 + f2 = f

mole cm

3

cA2 := 0.05

mole cm

3

cA( 0 ) = 0.025 f := 100

gal min

mole cm

Ah := 200gal

(assuming constant volume)

f1 ⋅ cA1 + f2 ⋅ cA2 − f ⋅ cA = 0

 c − 0.025mole⋅ cm− 3  A2  f1 := f ⋅ c −c

gal f1 = 62.5 min

f2 := f − f1

gal f2 = 37.5 min

A2

The folowing is the Simulink diagram for the mixer:

A1

3

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.

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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.

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13-8. Isothermal reactor of Problem 3-2 The model equations, from the solution to Problem 3-2, are: d ⋅ cA2( t) dt

=

f c ( t) − cA2( t) − kcA2( t) V A1

(

)

(

cA3( t) = cA2 t − to Design conditions:

Problem parameters:

f := 50

) ft

3

cA1 := 2

min

V := 150ft

Initial conditions at steady state:

cA2( 0 ) = 0.5

k :=

3

lbmole ft

3

Lp := 400ft

f ⋅  cA1 − 0.5lbmole⋅ ft



V⋅ 0.5lbmole⋅ ft

−3

The Simulink diagram is given by:

The responses to a 5 ft3/min step increase in process flow are:

Di := 5.5in − 3



−1

k = 1 min

lbmole ft

3

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: τ :=

V f + k⋅ V

τ = 0.75 min

2

to :=

π ⋅ Di ⋅ Lp 4f

to = 1.32 min

 c − 0.5 lbmole   A1 3 3 ft  ft  ∆cA2 := ⋅5 f + k⋅ V min ∆cA2 = 0.038

lbmole ft

3

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13-9. Flash drum of Problem 3-11 kmole := 1000mole

Rearranging the model equation developed in Problem 3-11: d⋅ x ( t)

=

dt

1 M

( F⋅ z( t) − V⋅ y ( t) − L⋅ x ( t) )

x ( 0 ) = 0.4

F=V+ L y( t) =

α ⋅ x( t)

1 + ( α − 1) x( t) kmole

Design conditions:

F := 10

Problem parameters:

M := 500kmole

At initial steady state:

y 0 :=

s α ⋅ 0.4

L := 5

kmole s

V := F − L

V= 5

kmole s

α := 2.5

1 + ( α − 1 ) 0.4

z0 :=

V⋅ y 0 + L⋅ 0.4 F

y 0 = 0.625 z0 = 0.513

The Simulink diagram for this problem is:

The responses to a 0.1 step increase in feed composition are:

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.

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13-10. Distillation tray of Problem 3-12 The model equations from the solution to Problem 3-12 are: d⋅ h ( t) dt

=

1

(

f ( t) − fo ( t) S i

fo ( t) = 0.415 ⋅ w⋅ h ( t) Problem parameters:

1.5

w := 3ft

Initial steady state conditions:

)

⋅ 2⋅ g

S := 11.2ft ft

2

1

3

fi := 30 min

fo := fi

fo   h 0 :=   0.415 ⋅ w⋅ 2⋅ g 

1.5

h 0 = 0.136 ft As the response is fast, convert time units to s by multiplying the derivative by 60 s/min. The Simulink diagram for the tray is:

The responses to a 20 ft3/min increase in inlet flow are:

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: −1

3

20ft ⋅ min 2

−1

= 0.06 ft

close!

331.4ft ⋅ min

The students can check the results for the change in 10 ft3/min.

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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: 3 %CO := % kg kg kg m 3 c1 := 80 c2 := 30 c0 := 50 f := 4 V := 40m %TO := % 3 3 3 min m m m f = f1 + f2

At the intial steady state:

f ⋅ c0 = f1 ⋅ c1 + f2 ⋅ c2

c0 − c1 f2 := f ⋅ c2 − c1

3

m f1 = 1.6 min

f1 := f − f2

3

m f2 = 2.4 min

From the results of Problem 6-11, the ultimate gain and period are: %CO Kcu := −250 %TO

Tu := 3.01min

From Table 7-1.1, for a series PID controller tuned for quarter decay ratio response: Kcu Kc := 1.7

τ I :=

Tu 2

τ D :=

Tu 8

%CO Kc = −147 %TO

τ I = 1.5 min

τ D = 0.38 min

3

Control valve (from F401Vlv1):

m Kv := 0.0542 min⋅ %CO Initial position:

Transmitter (from F407Trmr):

τ T := 3min

f2 Kv

τ v := 0.1min

(Solution of Problem 6-11)

= 44.28 %CO

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:

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.

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13-12. Non-interacting tanks in series of Fig. 4-1.1 The model equations developed in Section 4-1.1 are: d⋅ h 1( t) 1 = f ( t) − fo ( t) − f1 ( t) dt A1 i

(

d⋅ h 2( t) dt

=

1

(

f ( t) − f2 ( t) A2 1

h 10 := 2.5m

)

h 20 := 2.5m

f1 ( t) = Cv1⋅ h 1 ( t) 3

)

f2 ( t) = Cv2⋅ h 2 ( t) 3

Design conditions:

m fi := 5 min

m fo := 2 min

Problem parameters:

A1 := 9m

A2 := 9m

At initial steady state:

f1 := fi − fo

f2 := f1

2

2 3

Cv1 :=

f1

Cv2 :=

h 10

m f1 = 3 min 2.5

f2

Cv1 = 1.897

h 20

m

min 2.5

Cv2 = 1.897

m

The linearized gains and time constants are: K1 :=

2 ⋅ h 10 Cv1

Cv1 h 20 K2 := ⋅ Cv2 h 10

τ 1 :=

τ 2 :=

The Simulink diagram for this problem is:

2 ⋅ A1 ⋅ h 10 Cv1 2 ⋅ A2 ⋅ h 20 Cv2

K1 = 1.667

K2 = 1

min 2

m

τ 1 = 15 min

τ 2 = 15 min

min

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: 3

m K1 ⋅ 0.2 = 0.333 m min The responses for the second tank are second order with the same steady-state change in level, meaning that the gain K 2 is unity as predicted by the linear model.

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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⋅ h 1 ( t) − h 2 ( t) Design conditions are the same except for the initial condition in tank 1: At the initial steady state:

Cv1 :=

f1 h 10 − h 20

h 10 := 5m 2.5

Cv1 = 1.897

m

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. min K1 = 1.667 2 m The following is the Simulink diagram for the interacting tanks is series:

The responses to a 0.2 m3/min step increase in inlet flow are: 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 h 10 = 2.6 m, Cv1 :=

f1 2.6m − h 20 2.5

Cv1 = 9.487

m

min

The response of the second tank becomes first-order and the two tanks behave as a single tank.

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13-14. Non-interacting thermal tanks in series of Fig. 4-1.5 The model equations developed in Section 4-1.2 are: d ⋅ T2 ( t ) dt d ⋅ T4 ( t ) dt

=

= 3

fA V1

( T1 ( t ) − T2 ( t ) )

1

(

)

 f ⋅ T ( t) + fB⋅ T3( t) − fA + fB ⋅ T4 ( t)  V2  A 2 3

Design conditions:

m fA := 1 min

m 3 3 V1 := 5m V2 := 5m fB := 0 min

At initial steady state:

T2 := T1

T4 :=

fA⋅ T2 + fB⋅ T3 fA + fB

This problem is linear with a gains and time constants: fB K2 := fA + fB

τ 1 :=

V1 fA

τ 2 :=

The Simulink diagram for this problem is:

V2 fA + fB

T1 := 500K T3 := 500K

T2 = 500 K

fA K1 := fA + fB K2 = 0

T4 = 500 K K1 = 1

τ 1 = 5 min

τ 2 = 5 min

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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 ) dt d ⋅ T4 ( t ) dt

=

=

1

(

)

 f ⋅ T ( t) + fR⋅ T4( t) − fA + fR ⋅ T2 ( t)  V1  A 1 1

(

)(

f + fR ⋅ T1 ( t) − T4 ( t) V2 A

)

The design conditions and problem parameters are the same as in Problem 13-14, plus the recycle flow: 3

m fR := 1 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 Simulink diagram for this problem is:

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.

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13-16. Reactors with recycle of Problems 4-9 and 6-15 The model equatios are developed in the solution to Problem 4-9: d ⋅ cA1( t)

1 f ⋅ c ( t) + fR⋅ cA2( t) − f1 ⋅ cA1( t) − k 1 ⋅ cA1( t) V1 o Ao d ⋅ cA2( t) f1 = c ( t) − cA2( t) − k 2 ⋅ cA2( t) dt V2 A1 dt

=

( (

)

cA1( 0 ) = cA10

)

Design conditions from Problem 6-15:

cAo := 7

lbmole ft

3

cA2( 0 ) = cA20 −1

k 1 := 0.2min

3

ft fo := 10 min

V1 := 125ft

3

−1

k 2 := 0.2min

V2 := 125ft

3

3

For the base case let:

ft fR := 10 min

At the initial steady state:

f1 := fo + fR

3

cA20 =

ft f1 = 20 min

f1 ⋅ cA10 f1 + k 2 ⋅ V2 fo ⋅ cAo

cA10 :=

f1

cA10 = 1.726

f1 + k 1 ⋅ V1 − fR⋅ f1 + k 2 ⋅ V2 cA20 := The Simulink diagram for the reactors is:

f1 ⋅ cA10 f1 + k 2 ⋅ V2

cA20 = 0.767

lbmole ft

3

lbmole ft

3

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.

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13-17. Extraction process of Problem 4-6 The model equations developed in the solution to Problem 4-6 are: d ⋅ c1 ( t ) 2 ⋅ f1 ( t) = ci( t) − c1 ( t) − Ka⋅ c1 ( t) − me⋅ c2 ( t) dt V

(

d ⋅ c2 ( t ) dt

)

(

(

)

= Ka⋅ c1 ( t) − me⋅ c2 ( t) −

2 ⋅ f2 ( t) V

3

Design conditions:

Problem parameters:

)

ci := 0.4

me := 3.95

Ka := 3.646min

kmole m

c10 := ( 1 − Rec) ⋅ ci

f2 :=

Rec := 90%

3 −1

c20 :=

c2 ( 0 ) = c20

c2 ( t )

m f1 := 5 min

At the initial steady state:

(

c10 = 0.04

)

)

Ka⋅ V⋅ c10 − me⋅ c20 2 ⋅ c20

kmole 3

m

− 4 kmole

c20 = 1.2776 × 10

Ka⋅ me⋅ V

(

3

V := 25m

−2 ⋅ f1 ⋅ ci − c10 + Ka⋅ V⋅ c10

The Simulnk block diagram for the extractor is:

c1 ( 0 ) = c10

3

m f2 = 14088.6 min

3

m

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

−1

Ka := 209min

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