Process Dynamics and Control, Ch. 11 Solution Manual

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11.1

11.2

 1   Gc ( s ) = K c 1 +  τI s  The closed-loop transfer function for set-point changes is given by Eq. 11 1   , 36 with Kc replaced by K c 1 +  τI s  H ′( s ) ′ (s) H sp H ′( s ) ′ (s) H sp

 1  1  K c K v K p K m 1 + τ I s  (τs + 1)  =  1  1  1 + K c K v K p K m 1 +  τ I s  (τs + 1) (τ I s + 1) = 2 τ 3 s + 2ζ 3 τ 3 s + 1

where ζ3,τ3 are defined in Eqs. 11-62, 11-63 , Kp = R = 1.0 min/ft2 , and τ = RA = 3.0 min Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp

11-1

 ft 3 / min  min  psi  1.0 2 1.7 K OL = K c K v K p K m = (4) 0.2  = 1.36 psi  ft  ft  

τ3 = 2

τ τ I (3 min)(3 min) = = 6.62 min 2 K OL 1.36

 1 + K OL 2 ζ3 τ3 =   K OL

 2.36 τ I = × 3 = 5.21 min 1.36 

H ′( s ) 3s + 1 1 = = ′ ( s ) (3.0s + 1) + (2.21s + 1) 2.21s + 1 H sp For H sp′ ( s ) =

(3 − 2) 1 = s s

h′(t ) = 1 − e −t / 2..21

t = −2.21ln[1 − h′(t )] h(t ) = 2.5 ft

h ′(t ) = 0.5 ft

t = 1.53 min

h(t ) = 3.0 ft

h ′(t ) = 1.0 ft

t →∞

Therefore, h(t = 1.53 min) = 2.5ft h(t → ∞) = 3.0 ft

11-2

11.3 Gc ( s ) = K c = 5 ma/ma Assume τm = 0, τv = 0, and K1 = 1, in Fig 11.7. a)

Offset = Tsp′ (∞) − T ′(∞) = 5 1 F − 4.14 1 F = 0.86 1 F

b)

 K  K m K c K IP K v  2  T ′( s )  τs + 1  = Tsp′ ( s )  K  1 + K m K c K IP K v  2   τs + 1 

Using the standard current range of 4-20 ma, Km =

20 ma − 4 ma = 0.32 ma/ 1 F 50 1 F

K v = 1.2 , K IP = 0.75 psi/ma , τ =5 min , Tsp′ ( s ) =

T ′( s ) =

7.20 K 2 s (5s + 1 + 1.440 K 2 )

T ′(∞) = lim sT ′( s ) = s →0

T ′(∞) = 4.14 1 F c)

7.20 K 2 (1 + 1.440 K 2 ) K 2 = 3.34 1 F / psi

From Fig. 11-7, since Ti′ = 0 Pt′(∞) K v K 2 = T ′(∞) , and

Pt′(∞) = 1.03 psi

Pt′K v K 2 + Ti K 1 = T , Pt = 3.74 psi Pt (∞) = Pt − Pt′(∞) = 4.77 psi

11-3

5 s

11.4

a)

b)

Gm ( s) = K m e − θm s

assuming τm = 0

(20 − 4)ma − 2 s  ma  − 2 s e e =  2.67 lb sol lb sol/ft 3   (9 − 3) 3 ft  1   Gc ( s ) = K c 1 +  τI s 

Gm ( s) =

G IP ( s ) = K IP = 0.3 psi/ma Gv ( s ) = K v =

(10 − 20) USGPM USGPM = −1.67 (12 − 6) psi psi

Overall material balance for the tank, USgallons  dh  = q1 + q 2 − C v h  7.481 A ft 3   dt

(1)

Component balance for the solute, 7.481 A

d (hC 3 ) = q1c1 + q 2 c 2 − (C v h )c3 dt

Linearizing (1) and (2) gives

11-4

(2)

 C dh ′ = q 2′ −  v dt 2 h

7.481 A

 h′  

(3)

 C dc′   dh′ 7.481 A c3 + h 3  = c2 q′2 + q2c′2 − c3  v dt   dt 2 h

(

Subtracting (3) times c3 from the above equation gives 7.481 Ah

(

)

dc3′ = (c 2 − c3 ) q ′2 + q 2 c ′2 − C v h c3′ dt

Taking Laplace transform and rearranging gives K1 K2 Q2′ ( s ) + C 2′ ( s ) τs + 1 τs + 1

C 3′ ( s ) =

where K1 =

K2 =

τ=

c 2 − c3

lb sol/ft 3 = 0.08 USGPM h

Cv

q2

= 0.6

Cv h

7.481A h = 15 min Cv

since A = πD 2 / 4 = 12.6 ft 2 , and q h =  3  Cv

2

  q + q2  =  1   Cv

2

  = 4 ft 

Therefore, G p ( s) =

0.08 15s + 1

Gd ( s) =

0 .6 15s + 1

11-5

)

 h′ − Cv h c3′ 

c)

The closed-loop responses for disturbance changes and for setpoint changes can be obtained using block diagram algebra for the block diagram in part (a). Therefore, these responses will change only if any of the transfer functions in the blocks of the diagram change. i.

c 2 changes. Then block transfer function G p (s ) changes due to K1. Hence Gc(s) does need to be changed, and retuning is required.

ii.

Km changes. Block transfer functions do change. Hence Gc(s) needs to be adjusted to compensate for changes in block transfer functions. The PI controller should be retuned.

iii.

Km remains unchanged. No block transfer function changes. The controller does not need to be retuned.

11.5

a) One example of a negative gain process that we have seen is the liquid level process with the outlet stream flow rate chosen as the manipulated variable

c LT

p

h

ω With an "air-to-open" valve, w increases if p increases. However, h decreases as w increases. Thus Kp 1 / 14 and K c > 1 For a 2nd order characteristic equation, these conditions are also sufficient. Therefore, K c > 1 for closed-loop stability.

11-12

11.11

a)

c)

Transfer Line: Volume of transfer line = π /4 (0.5 m)2(20m)= 3.93 m3 Nominal flow rate in the line = q A + q F = 7.5 m 3 / min Time delay in the line =

3.93 m 3 = 0.52 min 7.5 m 3 /min

GTL ( s ) = e −0.52 s Composition Transmitter: Gm ( s) = K m =

(20 − 4) ma ma = 0.08 3 (200 − 0) kg/m kg/m 3

Controller From the ideal controller in Eq. 8.14

[

]

 1  ~  E ( s ) + K c τ D s C sp′ ( s ) − C m′ ( s ) P ′( s ) = K c 1 +  τI s  ~ In the above equation, set C sp′ ( s ) = 0 in order to get the derivative on the process output only. Then,

11-13

 1   G PI ( s ) = K c 1 +  τI s  G D (s) = − K c τ D s with Kc >0 as the controller should be reverse-acting, since P(t) should increase when Cm(t) decreases. I/P transducer K IP =

(15 − 3) psig psig = 0.75 (20 − 4) ma ma

Control valve

Gv ( s ) =

Kv τv s + 1

5τ v = 1

,

Kv =

dq A dp v

τ v = 0.2 min = 0.03(1 / 12)(ln 20)(20)

pv = pv

q A = 0.5 = 0.17 + 0.03(20) 0.03(20)

pv −3 12

pv − 3 12

= 0.5 − 0.17 = 0.33

K v = (1 / 12)(ln 20)(0.33) = 0.082

Gv ( s ) =

pv −3 12

m 3 /min psig

0.082 0 .2 s + 1

Process Assume cA is constant for pure A. Material balance for A: V

dc = q A c A + q F c F − ( q A + q F )c dt

11-14

(1)

Linearizing and writing in deviation variable form V

dc ′ = c A q ′A + q F c ′F − (q A + q F )c ′ − c q ′A dt

Taking Laplace transform

[Vs + (q A + q F )]C ′(s) = (c A − c )Q ′A ( s) + q F C F′ ( s)

(2)

From Eq. 1 at steady state, dc / dt = 0 , c = (q A c A + q F c F ) /(q A + q F ) = 100 kg/m 3 Substituting numerical values in Eq. 2,

[5s + 7.5]C ′( s) = 700 Q ′A ( s) + 7 C F′ ( s) [0.67s + 1]C ′( s) = 93.3 Q′A ( s) + 0.93 C F′ (s) 93.3 0.67 s + 1 0.93 Gd (s ) = 0.67 s + 1 G p ( s) =

11.12

The stability limits are obtained from the characteristic Eq. 11-83. Hence if an instrumentation change affects this equation, then the stability limits will change and vice-versa. a)

The transmitter gain, Km, changes as the span changes. Thus Gm(s) changes and the characteristic equation is affected. Stability limits would be expected to change.

b)

The zero on the transmitter does not affect its gain Km. Hence Gm(s) remains unchanged and stability limits do not change.

c)

Changing the control valve trim changes Gv(s). This affects the characteristic equation and the stability limits would be expected to change as a result.

11-15

11.13

a)

Ga ( s) =

Kc K (τs + 1)( s + 1)

b)

Gb ( s ) =

K c K (τ I s + 1) τ I s (τs + 1)( s + 1)

For a)

D( s ) + N ( s ) = (τs + 1)( s + 1) + K c K = τs 2 + (τ + 1) s + 1 + K c K p Stability requirements: 1 + Kc K p > 0

∞ > K c K p > −1

or

For b) D( s ) + N ( s ) = τ I (τs + 1)( s + 1) + K c K (τ I s + 1)

= τ I τs 3 + τ I (τ + 1) s 2 + τ I (1 + K c K p ) s + K c K p Necessary condition: K c K p > 0 Sufficient conditions (Routh array):

τI τ

τ I (1 + K c K p )

τ I (τ + 1)

Kc K p

τ I (τ + 1)(1 + K c K p ) − τ I τK c K p 2

τ I (τ + 1) Kc K p

Additional condition is: τ I (τ + 1)(1 + K c K p ) − τ( K c K p ) > 0

(since τ I and τ are both positive)

11-16

τ I (τ + 1) + τ I (τ + 1) K c K p − τK c K p > 0

[τ I (τ + 1) − τ]K c K p > −τ I (τ + 1) Note that RHS is negative for all positive τ I and τ (∴ RHS is always negative)

Case 1: If

τ I (τ + 1) − τ > 0

τ   i.e., τ I > τ + 1  

 − τ I (τ + 1)  KcKp > 0 >    τ I (τ + 1) − τ  In other words, this condition is less restrictive than KcKp >0 and doesn't apply. then

Case 2:

If

then

τ I (τ + 1) − τ < 0

τ   i.e., τ I < τ + 1  

 − τ I (τ + 1)  KcKp <    τ I (τ + 1) − τ 

In other words, there would be an upper limit on KcKp so the controller gain is bounded on both sides 0 c)

< KcKp <

− τ I (τ + 1) τ I (τ + 1) − τ

Note that, in either case, the addition of the integral mode decreases the range of stable values of Kc.

11-17

11.14

From the block diagram, the characteristic equation is obtained as   4    (0.5) s + 3   2  1     1 + Kc  =0   4    s − 1  s + 10   1 + (0.5) s + 3    

that is,  2  2  1  1 + Kc    =0  s + 5   s − 1  s + 10 

Simplifying, s 3 + 14 s 2 + 35s + (4 K c − 50) = 0

The Routh Array is 1

35

14

4Kc-50

490 − (4 K c − 50) 14 4Kc – 50

For the system to be stable, 490 − (4 K c − 50) >0 14

and

or

4 K c − 50 > 0 or Kc > 12.5

Therefore

12.5 < Kc < 135

11-18

Kc < 135

11.15

a)

Kc K Kc K K K /(1 + K c K ) Y ( s) = 1 − τs = = c K K 1 − τs + K c K τ Ysp ( s ) − s +1 1+ c 1 + Kc K 1 − τs For stability

−

τ >0 1+ KcK

Since τ is positive, the denominator must be negative, i.e., 1+ Kc K < 0 K c K < −1 K c < −1 / K Kc K K CL = 1+ Kc K

Note that

b)

If K c K < −1 and 1 + K c K is negative, then CL gain is positive. ∴ it has the proper sign.

c)

K = 10 and τ = 20 and we want

or

−

τ = 10 1+ KcK

− 20 = 10 + (10)(10) K c − 30 = 100 K c K c = −0.3

Offset: K CL =

(−0.3)(10) −3 = = 1.5 1 + (−0.3)(10) − 2

∴ Offset = +1 − 1.5 = − 50% (Note this result implies overshoot)

11-19

d)

KcK (1 − τs )(τ m s + 1) Kc K Y ( s) = = Kc K Ysp ( s ) (1 − τs )(τ m s + 1) + K c K 1+ (1 − τs )(τ m s + 1)

=

=

Kc K − ττ m s + (τ m − τ) s + 1 + K c K 2

K c K /(1 + K c K ) ττ m τ −τ − s2 + m s +1 1 + Kc K 1 + Kc K

(standard form)

For stability, (1)

−

ττ m >0 1+ Kc K

(2) 1+ Kc K < 0 K c K < −1 1 Kc < − K

From (1)

Since

From (2)

Since 1 + K c K < 0

For

K = 10 ,

Y ( s) = Ysp ( s )

τm − τ >0 1+ Kc K

τm − τ < 0 − τ < −τ m τ > τm τ = 20 , Kc = –0.3 , τm = 5

1 .5 1 .5 = 2 (20)(5) 2 (5 − 20) 50 s + 2.5s + 1 − s + s +1 1− 3 (1 − 3)

Underdamped but stable.

11-20

11.16

 1   Gc ( s ) = K c 1 +  τI s  Gv ( s ) =

Kv − 1 .3 = (10 / 60) s + 1 0.167 s + 1

G p ( s) = −

1 1 =− As 22.4 s

since A = 3 ft 2 = 22.4

gal ft

Gm ( s) = K m = 4 Characteristic equation is

 1   − 1.3  − 1    1 + K c 1 +   ( 4) = 0  τ I s   0.167 s + 1  22.4 s  (3.73τ I ) s 3 + (22.4τ I ) s 2 + (5.2 K c τ I ) s + (5.2 K c ) = 0

The Routh Array is

3.73τ I

5.2 K c τ I

22.4τ I

5.2 K c

5.2 K c τ I − 0.867 K c 5.2 K c For stable system, τI > 0 ,

5.2 K c τ I − 0.867 K c > 0

That is, Kc > 0 τ I > 0.167 min

11-21

Kc > 0

11.17

 τ s + 1  5  GOL ( s ) = K c  I 2  τ I s  (10s + 1)

 N ( s)  =  D( s)

D( s ) + N ( s ) = τ I s (100 s 2 + 20 s + 1) + 5 K c (τ I s + 1) = 0 = 100τ I s 3 + 20τ I s 2 + (1 + 5 K c )τ I s + 5 K c = 0

a)

Analyze characteristic equation for necessary and sufficient conditions Necessary conditions: 5K c > 0

→

(1 + 5 K c )τ I > 0 →

Kc > 0

τI > 0

and

Kc > −

Sufficient conditions obtained from Routh array

100τ I

(1 + 5 K c )τ I

20τ I

5K c

20τ I (1 + 5 K c ) − 500τ I K c 20τ I 2

5K c Then,

20τ I (1 + 5K c ) − 500τ I K c > 0 2

τ I (1 + 5 K c ) > 25 K c b)

or

τI >

25K c 1 + 5K c

Sufficient condition is appropriate. Plot is shown below.

11-22

1 5

7

6

Stability region

5

τI 4

3

2

1

0

0

1

2

3

4

5

6

7

Kc

c)

Find τ I as K c → ∞

 25K c   25  lim   = Kclim  =5 → ∞ 1 + 5K c  1 / K c + 5 

Kc → ∞

∴ τ I > 5 guarantees stability for any value of Kc. Appelpolscher is wrong yet again.

11.18 Gc ( s ) = K c KV GV ( s ) = τV s + 1 Kv =

dws dp

= p =12

5τ v = 20 sec G p ( s) =

0.6 2 12 − 4

= 0.106

lbm/sec ma

τ v = 4 sec −s

2.5e 10s + 1

Gm ( s) = K m =

(20 − 4) ma ma = 0 .4 1 1 (160 − 120) F F

Characteristic equation is

11-23

−s  0.106  2.5e  (0.4) = 0 1 + ( K c )   4 s + 1  10s + 1 

a)

(1)

Substituting s=jω in (1) and using Euler's identity e-jω=cosω – j sin ω gives -40ω2 +14jω + 1 + 0.106 Kc (cosω – jsinω)=0 Thus

and

-40ω2 + 1 + 0.106 Kc cosω = 0

(2)

14ω - 0.106Kc sinω =0

(3)

From (2) and (3), tan ω =

14ω 40ω 2 − 1

(4)

Solving (4), ω = 0.579 by trial and error. Substituting for ω in (3) gives Kc = 139.7 = Kcu Frequency of oscillation is 0.579 rad/sec b)

Substituting the Pade approximation e −s ≈

1 − 0 .5 s 1 + 0 .5 s

into (1) gives 20 s 3 + 47 s 2 + (14.5 − 0.053K c ) s + (1 + 0.106 K c ) = 0

The Routh Array is 20

14.5 –0.053 Kc

47

1+ 0.106 Kc

14.07 – 0.098 Kc 1 + 0.106 Kc

11-24

For stability,

and

14.07 − 0.098 K c > 0

or Kc < 143.4

1 + 0.106 K c > 0

or Kc > -9.4

Therefore, the maximum gain, Kcu = 143.4, is a satisfactory approximation of the true value of 139.7 in (a) above.

11.19

a)

G (s) =

4(1 − 5s ) (25s + 1)(4 s + 1)(2 s + 1)

Gc ( s ) = K c D( s ) + N ( s ) = (25s + 1)(4s + 1)(2 s + 1) + 4 K c (1 − 5s ) = 0 100 s 2 + 29 s + 1 2s + 1 200 s 3 + 58s 2 + 2 s 100 s 2 + 29 s + 1

200 s 3 + 158s 2 + 31s + 1 + 4 K c − 20 K c s = 0 200 s 3 + 158s 2 + (31 − 20 K c ) s + 1 + 4 K c = 0

Routh array: 200

31-20 Kc

158

1+4 Kc 4898 –3160Kc –200 –800Kc

158(31-20Kc)-200(1+4Kc) = 158

158

1+ 4 Kc ∴ 4698 –3960 Kc > 0

11-25

or

Kc < 1.2

b)

(25s + 1)(4 s + 1)(2 s + 1) + 4 K c = 0 Routh array: 200 s 3 + 158s 2 + 31s + (1 + 4 K c ) = 0

200

31

158

1 + 4 Kc

158 (31) − 200(1+4Kc) = 4898 –200 –800Kc 1+ 4 Kc ∴ 4698 –800 Kc > 0

c)

or

Kc < 5.87

Because Kc can be much higher without the RHP zero present, the process can be made to respond faster.

11.20

The characteristic equation is 1+ a)

0.5 K c e −3s =0 10 s + 1

(1)

Using the Pade approximation e −3 s ≈

1 − (3 / 2) s 1 + (3 / 2) s

in (1) gives 15s 2 + (11.5 − 0.75 K c ) s + (1 + 0.5 K c ) = 0

For stability,

and

11.5 − 0.75 K c > 0

or

K c < 15.33

1 + 0.5 K c > 0

or

K c > −2

11-26

Therefore − 2 < K c < 15.33 b)

Substituting s = jω in (1) and using Euler's identity. e −3 jω = cos(3ω) − j sin(3ω) gives 10 jω + 1 + 0.5 K c [cos(3ω) − j sin(3ω)] = 0 Then,

and

1 + 0.5K c cos(3ω) = 0

(2)

10ω − 0.5 K c sin(3ω) = 0

(3)

From (3), one solution is ω = 0, which gives Kc = -2 Thus, for stable operation Kc > -2 From (2) and (3) tan(3ω) = -10ω Eq. 4 has infinite number of solutions. The solution for the range π/2 < 3ω < 3π/2 is found by trial and error to be ω = 0.5805. Then from Eq. 2, Kc = 11.78 The other solutions for the range 3ω > 3π/2 occur at values of ω for which cos(3ω) is smaller than cos(3 × 5.805). Thus, for all other solutions of ω, Eq. 2 gives values of Kc that are larger than 11.78. Hence, stability is ensured when -2 < Kc < 11.78

11-27

11.21

a)

To approximate GOL(s) by a FOPTD model, the Skogestad approximation technique in Chapter 6 is used. Initially,

3K c e − (1.5+0.3+ 0.2 ) s 3K c e −2 s GOL ( s ) = = (60 s + 1)(5s + 1)(3s + 1)(2s + 1) (60s + 1)(5s + 1)(3s + 1)(2s + 1) Skogestad approximation method to obtain a 1st -order model: Time constant ≈ 60 + (5/2) Time delay ≈ 2 +(5/2) + 3 + 2 =9.5 Then GOL ( s ) ≈

b)

3K c e −9.5 s 62.5s + 1

The only way to apply the Routh method to a FOPTD transfer function is to approximate the delay term. e −9.5 s ≈

− 4.75s + 1 4.75s + 1

(1st order Pade-approximation)

Then GOL ( s ) ≈

3K c (−4.75s + 1) N ( s) ≈ D( s ) (62.5s + 1)(4.75s + 1)

The characteristic equation is: D( s ) + N ( s ) = (62.5s + 1)(4.75s + 1) + 3K c (−4.75s + 1) 297 s 2 + 67.3s + 1 − 14.3K c s + 3K c = 0 297 s 2 + (67.3 − 14.3K c ) s + (1 + 3K c ) = 0

11-28

Necessary conditions: 67.3 − 14.3K c > 0

1 + 3K c > 0

− 14.3K c > −67.3 K c < 4.71 Range of stability: c)

3 K c > −1 K c > −1 / 3 − 1 / 3 < K c < 4.71

Conditional stability occurs when K c = K cu = 4.71 With this value the characteristic equation is: 297 s 2 + (67.3 − 14.3 × 4.71) s + (1 + 3 × 4.71) = 0 297 s 2 + 15.13 = 0

s2 =

− 15.13 297

We can find ω by substituting jω → s ω = 0.226

at the maximum gain.

11-29