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
