Process Dynamics and Control Solutions

Revised: 1-3-04

Chapter 15

15.1 For Ra=d/u Kp =

∂Ra d =− 2 ∂u u

which can vary more than Kp in Eq. 15-2, because the new Kp depends on both d and u.

15.2 By definition, the ratio station sets (um – um0) = KR (dm – dm0) Thus

u − u m0 K 2u 2 K 2  u  = = KR = m   d m − d m0 K1d 2 K1  d 

2

(1)

For constant gain KR, the values of u and d in Eq. 1 are taken to be at the desired steady state so that u/d=Rd, the desired ratio. Moreover, the transmitter gains are K1 =

(20 − 4)mA Sd

2

K2 =

,

(20 − 4)mA Su

2

Substituting for K1, K2 and u/d into (1) gives:

KR =

Su

2

Sd

2

Rd

2

 S  =  Rd d  Su  

2

Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp

15-1

15.3 (a)

The block diagram is the same as in Fig. 15.11 where Y ≡ H2, Ym ≡ H2m, Ysp ≡ H2sp, D ≡ Q1, Dm ≡ Q1m, and U ≡ Q3.

b)

(A steady-state mass balance on both tanks gives 0 = q1 – q3

or

Q1 = Q3 (in deviation variables)

(1)

From the block diagram, at steady state:

Q3 = Kv Kf Kt Q1 From (1) and (2), Kf =

1 K v Kt

(2)

c)

(No, because Eq. 1 above does not involve q2.

(b)

From the block diagram, exact feedforward compensation for Q1 would result when

15.4

Q1 + Q2 = 0 Substituting Gf = −

Q2 = KV Gf Kt Q1, 1 K v Kt

15-2

(c)

Same as part (b), because the feedforward loop does not have any dynamic elements.

(d)

For exact feedforward compensation Q4 + Q3 = 0 From the block diagram,

(1) Q2 = KV Gf Kt Q4

(2)

Using steady-state analysis, a mass balance on tank 1 for no variation in q1 gives Q2 − Q3 = 0

(3)

Substituting for Q3 from (3) and (2) into (1) gives Q4 + KV Gf Kt Q4 = 0 Gf = −

or

1 K v Kt

For dynamic analysis, find Gp1 from a mass balance on tank 1,

A1

dh1 = q1 + q2 − C1 h1 dt

15-3

Linearizing (4), noting that q1′ = 0, and taking Laplace transforms: A1

dh′ C = q2′ − 1 h1′ dt 2 h 1

or

(2 h1 / C1) H1′ ( s) = Q2′ ( s) (2 A h / C ) s + 1 1 1 1 q3 = C1 h1

Since

q3′ =

C1 2 h1

or

h1′

(5)

Q3′ ( s ) C = 1 H1′ ( s) 2 h 1

(6)

From (5) and (6),

Q3′ ( s) 1 = = GP1 Q2′ ( s) (2 A h / C ) s + 1 1 1 1 Substituting for Q3 from (7) and (2) into (1) gives

Q4 + (

or

1

) K v G f Kt Q4 = 0 (2 A1 h1 / C1 ) s + 1

Gf = −

1 [(2 A1 h1 / C1 ) s + 1] K v Kt

15.5

(a)

For a steady-state analysis: Gp=1,

Gd=2,

From Eq.15-21, Gf = (b)

− Gd −2 = = −2 Gv Gt G p (1)(1)(1)

Using Eq. 15-21,

15-4

Gv = Gm = Gt =1

(7)

−2 − Gd −2 ( s + 1)(5s + 1) = Gf = = 5s + 1 G v Gt G p  1  (1)(1)   s +1 (c)

Using Eq. 12-19,

where

1 ~ ~ ~ G = Gv G p G m = = G+ G− s +1 1 G + = 1, G − = s +1

For τc=2, and r=1, Eq. 12-21 gives f=

1 2s + 1

From Eq. 12-20  1  s +1 Gc* = G − −1 f = ( s + 1)  =  2s + 1  2s + 1

From Eq. 12-16 s +1 s +1 = 2s + 1 = Gc = ∗~ 2s 1 − Gc G 1 − 1 2s + 1 Gc

(d)

∗

For feedforward control only, Gc=0. For a unit step change in disturbance, D(s) = 1/s. Substituting into Eq. 15-20 gives Y(s) = (Gd+GtGfGvGp)

1 s

For the controller of part (a)  2  1  1 + (1)(−2)(1) Y(s) =    s + 1  s  ( s + 1)(5s + 1)

15-5

Y(s) = or

5 / 2 − 25 / 2 2.5 − 2.5 − 10 = − = + ( s + 1)(5s + 1) s + 1 5s + 1 s + 1 s + 1 / 5

y(t) = 2.5 (e-t – e- t/5)

For the controller of part (b)  2  − 2   1  1 + (1) Y(s) =  (1)  = 0  5s + 1   s + 1   s  ( s + 1)(5s + 1)

or

y(t) = 0

The plots are shown in Fig. S15.5a below.

Figure S15.5a. Closed-loop response using feedforward control only.

(e)

Using Eq. 15-20: For the controller of parts (a) and (c), 2   1   ( s + 1)(5s + 1) + (1)(−2)(1)  s + 1   1   Y(s) =   s +1  1   s 1+   (1)   (1)    2s   s + 1 

15-6

5 25 / 3 − 40 / 3 − 20s = + + ( s + 1)(5s + 1)(2s + 1) s + 1 5s + 1 2 s + 1 5 20 / 3 5/3 = − + s + 1 s + 1/ 2 s + 1/ 5

or

Y(s) =

or

y(t) = 5e-t –

20 -t/2 5 - t/5 e + e 3 3

and for controllers of parts (b) and (c)  2  − 2   1   ( s + 1)(5s + 1) + (1) 5s + 1 (1) s + 1   1     =0 Y(s) =    s +1  1  s 1+  (1) (1)    2s   s + 1    or

y(t) = 0

The plots of the closed-loop responses are shown in Fig. S15.5b.

Figure S15.5b. Closed-loop response for feedforward-feedback control.

15.6

(a)

The steady-state energy balance for both tanks takes the form 0 = w1 C T1 + w2 C T2 − w C T4 + Q 15-7

where Q is the power input of the heater C is the specific heat of the fluid. Solving for Q and replacing unmeasured temperatures and flow rates by their nominal values, Q = C ( w1T1 + w 2 T 2 − wT 4 )

(1)

Neglecting heater and transmitter dynamics, Q = Kh p

(2)

T1m = T1m0 + KT(T1-T10)

(3)

wm = wm0 + Kw(w-w0)

(4)

Substituting into (1) for Q, T1, and w from (2), (3), and (4), gives

p=

(b)

C 1 1 [ w1 (T10 + (T1m − T1m0 )) + w2 T2 − T4 ( w0 + ( wm − wm0 ))] Kh KT Kw

Dynamic compensation is desirable because the process transfer function Gp= T4(s)/P(s) is different from each of the disturbance transfer functions, Gd1= T4(s)/T1(s), and Gd2= T4(s)/w(s); this is more so for Gd1 which has a higher order.

15.7

(a)

(b)

A steady-state material balance for both tanks gives,

15-8

0 = q1 + q2 + q4 − q5 Because q ′2 = q ′4 = 0, the above equation gives 0 = q1′ – q5′ or 0 = Q1 – Q5

(1)

From the block diagram, Q5 = Kv Gf Kt Q1 Substituting for Q5 into (1) gives 0 = Q1 − Kv Gf Kt Q1 (c)

or

Gf =

1 Kv Kt

To find Gd and Gp, the mass balance on tank 1 is A1

dh1 = q1 + q 2 − C1 h1 dt

where A1 is the cross-sectional area of tank 1. Linearizing and setting q 2′ = 0 leads to A1

dh1′ C = q1′ − 1 h1′ dt 2 h1

Taking the Laplace transform, H1′ ( s) R1 = Q1′ ( s) A1R1s + 1

Linearizing q3 = C1 h1

q3′ =

1 h1′ R1

or

where R1 ≡

C1

(2)

gives

Q3′ ( s) 1 = H1′ ( s) R1

Mass balance on tank 2 is

A2

2 h1

dh2 = q3 + q 4 − q 5 dt

15-9

(3)

Using deviation variables, setting q ′4 = 0, and taking Laplace transform A2 sH 2′ ( s ) = Q3′ ( s ) − Q5′ ( s )

H 2′ ( s) 1 = Q3′ ( s ) A2 s

(4)

and

H 2′ ( s ) 1 =− = G p ( s) Q5′ ( s) A2 s Gd ( s) =

H 2′ ( s) H 2′ ( s) Q3′ ( s) H1′ ( s ) 1 = = Q1′ ( s ) Q3′ ( s ) H1′ ( s ) Q1′ ( s ) A2 s ( A1R 1 s + 1)

upon substitution from (2), (3), and (4). Using Eq. 15-21,

1 − Gd A2 s ( A1 R1 s + 1) Gf = = Gt Gv G p K t K v (−1 / A2 s ) −

=+

1 1 K v K t A1 R1 s + 1

15.8

For the process model in Eq. 15-22 and the feedforward controller in Eq. 15-29, the correct values of τ1 and τ2 are given by Eq. 15-42 and (15-43). Therefore, τ1 − τ2 = τp − τL

(1)

for a unit step change in d, and no feedback controller, set D(s)=1/s, and Gc= 0 in Eq. 15-20 to obtain

[

Y(s) = Gd + Gt G f Gv G p

]1s

Setting Gt = Gv = 1, and using Eqs. 15-22 and 15-29,

15-10

 K  K p  1 d + (1)  − K d / K P (τ1s + 1)  (1) Y (s) =      τ2s + 1  τ d s + 1    τ p s + 1   s

1 (τ1 − τ p )τ p 1  τ 1 τ (τ − τ ) 1 = Kd  − d − − 2 1 2 −  τ p − τ 2 τ p s + 1   s τ d s + 1 s (τ 2 − τ p ) τ 2 s + 1  τ1 − τ p −t / τ p  (τ − τ ) y(t) = K d  −e −t / τ − 1 2 e −t / τ 2 − e  τ2 − τ p τ p − τ2  

or

∞

∫0

e(t )dt = ∫

=

∞ 0

 τ (τ − τ ) τ p (τ1 − τ p )  y (t )dt = − K d  τ d + 2 1 2 +  τ2 − τ p τ p − τ 2  

−Kd  τ d τ 2 − τ d τ p + τ 2 τ1 − τ 2 2 − τ p τ1 + τ p 2 + (τ p τ 2 − τ p τ 2 )    τ2 − τ p

= − K d  (τ1 − τ 2 ) − (τ p − τ d )  = 0 when (1) holds.

15.9

(a)

For steady-state conditions Gp=1,

Gd=2,

Gv = Gm = Gt =1

Using Eq. 15-21, Gf = (b)

− Gd −2 = = −2 Gv Gt G p (1)(1)(1)

Using Eq. 15-21,

15-11

− 2e − s − Gd −2 ( s + 1)(5s + 1) = Gf = = G v Gt G p  1  − s 5s + 1 (1)(1) e  s + 1 (c)

Using Eq. 12-19, e −s ~ ~ ~ G = Gv G p G m = = G+ G− s +1 1 ~ ~ , where G+ = e − s G− = s +1 For τc=2, and r = 1, Eq. 12-21gives f=

1 2s + 1

From Eq. 12-20 1 s +1 1 Gc * = ~ f = ( s + 1) = 2s + 1 2s + 1 G−

From Eq. 12-16 s +1 2s + 1 = s + 1 Gc = ~= 2s 1 − Gc * G 1 − 1 2s + 1 Gc *

(d)

For feedforward control only, Gc=0. For a unit step disturbance, D(s) = 1/s. Substituting into Eq. 15-20 gives Y(s) = (Gd+GtGfGvGp)

1 s

For the controller of part (a)

  e − s  1 2e − s  + (1)(−2)(1) Y(s) =   s + 1  s  ( s + 1)(5s + 1)

15-12

= or

− 10e − s ( s + 1)(5s + 1)

y(t) = 2.5 (e-(t-1) – e-( t-1)/5)S(t-1)

For the controller of part (b) −s  2e − s  − 2   e  1  = 0 + (1) Y(s) =  (1)  5s + 1   s + 1   s  ( s + 1)(5s + 1)

or

y(t) = 0

The plots are shown in Fig. S15.9a below.

Figure S15.9a. Closed-loop response using feedforward control only.

(e)

Using Eq. 15-20: For the controllers of parts (a) and (c),   e −s   2e − s  + (1)(−2)(1)  s + 1   1 ( s + 1)(5s + 1)   Y(s) = −s  s  s + 1  e  (1) 1+    (1)  2s   s + 1   

15-13

and for the controllers of parts (b) and (c),  2  − 2   1   ( s + 1)(5s + 1) + (1) 5s + 1 (1) s + 1   1     =0 Y(s) =    s + 1  1  s 1+  (1) (1)    2s   s + 1    or

y(t) = 0

The plots of the closed-loop responses are shown in Fig. S15.9b.

Figure S15.9b. Closed-loop response for the feedforward-feedback control.

15.10

(a)

For steady-state conditions Gp=Kp,

Gd=Kd,

Gv = Gm = Gt =1

Using Eq. 15-21,

15-14

Gf =

− Gd − 0.5 = = −0.25 Gv Gt G p (1)(1)(2)

(b)

Using Eq. 15-21,

(c)

− 0.5e −30 s − Gd (95s + 1) −10 s 60 s + 1 = = −0.25 e Gf = − 20 s (60s + 1) G v Gt G p  2e   (1)(1)  95s + 1  Using Table 12.1, a PI controller is obtained from equation G, Kc =

τ 1 1 95 = = 0.95 K p τ c + θ 2 (30 + 20)

τ I = τ = 95 (d)

As shown in Fig.S15.10a, the dynamic controller provides significant improvement.

(e) 0.08 Controller of part (a) Controller of part (b)

0.06

0.04 y(t) 0.02

0

-0.02

-0.04 0

100

200

300

400

500

time

Figure S15.10a. Closed-loop response using feedforward control only.

15-15

0.06 Controllers of part (a) and (c) Controllers of part (b) and (c)

0.04

0.02 y(t)

0

-0.02

-0.04

-0.06 0

100

200

300

400

500

time

Figure S15.10b. Closed-loop response for feedforward-feedback control.

f)

As shown in Fig. S15.10b, the feedforward configuration with the dynamic controller provides the best control.

15.11

Energy Balance: ρVC

dT = wC (Ti − T ) − U (1 + q c ) A(T − Tc ) − U L AL (T − Ta ) (1) dt

Expanding the right hand side, dT = wC (Ti − T ) − UA(T − Tc ) dt − UAqc T + UAqc Tc − U L AL (T − Ta )

ρVC

(2)

Linearizing,

q cT ≈ q cT + q cT ′ + T qc′ Substituting (3) into (2), subtracting the steady-state equation, and introducing deviation variables,

15-16

(3)

dT ′ = wC (Ti′ − T ′) − UAT ′ − UAT q c′ − UAq c T ′ dt + UATc qc′ − U L ALT ′

ρVC

(4)

Taking the Laplace transform and assuming steady-state at t = 0 gives,

ρVCsT ′( s) = wCTi′( s) + UA(Tc − T − )Qc′ ( s ) − ( wC + UA + UAq c + U L AL )T ′( s)

(5)

Rearranging, T ′( s ) = GL ( s )Ti′( s ) + G p ( s )Qc′ ( s )

(6)

where:

Gd ( s) = G p ( s) =

KL τs + 1 Kp

τs + 1 wC Kd = K UA(Tc − T ) Kp = K ρVC τ= K K = wC + UA + UAq c + U L AL

(7)

The ideal FF controller design equation is given by,

GF =

−Gd Gt GvG p

(17-27)

But, Gt = K t e − θs and Gv=Kv

(8)

Substituting (7) and (8) gives,

GF =

− wCe + θs K t K vUA(Tc − T )

(9)

In order to have a physically realizable controller, ignore the e+θs term,

15-17

− wC K t K vUA(Tc − T )

GF =

(10)

15.12

a)

A component balance in A gives:

V

dc A = qc Ai − qc A − Vkc A dt

(1)

At steady state, 0 = q c Ai − q c A − Vkc A

(2)

Solve for q ,

q=

kVC A C Ai − C A

(3)

For an ideal FF controller, replace C Ai by C Ai , q by q1 and C A by C Asp : q=

b)

kVC Asp C Ai − C Asp

Linearize (1):

V

dc A = q ciA+ qc′Ai + c Ai q′ − q c A − qc′A − c Aq′ − Vkc A dt

Subtract (2),

V

dc′A = qc′iA+ c Ai q′ − qc′A − c Aq′ − Vkc′A dt

Take the Laplace transform,

sVc′A ( s ) = qc′iA( s ) + c Ai Q′( s ) − qc′A ( s) − c AQ′( s) − Vkc′A ( s) Rearrange, 15-18

C ′A ( s) =

c −c q C ′Ai ( s ) + Ai A Q′( s) sV + q + Vk sV + q + Vk

(6)

or C ′A ( s ) = Gd ( s )C ′Ai ( s ) + G p ( s )Q′( s )

(7)

The ideal FF controller design equation is,

GF ( s) = −

Gd ( s) Gv ( s )G p ( s )Gt ( s )

(8)

Substitute from (6) and (7) with Gv(s)=Kv and Gt(s)=Kt :

GF ( s) = − .

q K v (c Ai − c A ) Kt

(9)

Note: G F ( s) = P ′( s) / C ′Ai m ( s ) where P is the controller output and cAim is the measured value of cAi.

15.13

(a)

Steady-state balances: 0 = q5 + q1 − q3

(1)

0 = q3 + q 2 − q 4

(2) 0

0 = x5 q5 + x1 q1 − x3 q3

(3)

0 = x3 q 3 + x 2 q 2 − x 4 q 4

(4)

Solve (4) for x3 q3 and substitute into (3), 0 = x5 q5 + x 2 q 2 − x 4 q 4 Rearrange,

15-19

(5)

q2 =

x 4 q 4 − x5 q 5 x2

(6)

In order to derive the feedforward control law, let x4 → x4 sp ,

x2 → x2 (t ),

x5 → x5 (t ),

and

q 2 → q 2 (t )

Thus,

q 2 (t ) =

x 4 sp q 4 − x5 (t )q5 (t ) x2

(7)

Substitute numerical values:

q 2 (t ) =

(3400) x 4 sp − x5 (t )q5 (t ) 0.990

(8)

or q 2 (t ) = 3434 x 4 sp − 1.01x5 (t )q 5 (t )

(9)

Note: If transmitter and control valve gains are available, then an expression relating the feedforward controller output signal, p(t), to the measurements , x5m(t) and q5m(t), can be developed. (b)

Dynamic compensation: It will be required because of the extra dynamic lag preceding the tank on the left hand side. The stream 5 disturbance affects x3 while q3 does not.

15-20