Revised: 1-3-04
Chapter 15
15.1
For R For Ra=d /u /u K p
=
∂ Ra d =− 2 ∂u u
which can vary more than K than K p in Eq. 15-2, because the new K new K p depends on both d and and u.
15.2
By definition, the ratio station sets (um – um0) = K = K R (d m – d m0 m0)
− u m0 K 2 u 2 K 2 u 2 Thus K R = = = d m − d m 0 K 1 d 2 K 1 d um
(1)
For constant gain K gain K R, 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 K 1
=
(20 − 4)mA S d
2
,
K 2
=
(20 − 4)mA S u
2
Substituting for K for K 1, K 2 and u/d into into (1) gives:
K R
=
S u
2
S d
2
R
2 d
S = Rd d S u
2
Solution Manual for Process Dynamics and Control, 2 nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. F. Edgar and Duncan A. Mellichamp Mellichamp
15-1
15.3
(a)
The block diagram is the same as in Fig. 15.11 where Y ≡ H 2, Y m ≡ H 2m, Y sp ≡ H 2 sp, D ≡ Q1, Dm ≡ Q1m, and U ≡ Q3.
b)
(A steady-state mass balance on both tanks gives 0 = q1 – q3
Q1 = Q3 (in deviation variables)
or
(1)
From the block diagram, at steady state: Q3 = K = K v K f K t Q1 From (1) and (2), K f =
1 K v K t
(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 = K = K V G f K t Q1, 1 K v K t
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
(1)
From the block diagram,
Q2 = K V G f K t 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 + K V G f K t Q4 = 0 G f = −
or
1 K v K t
For dynamic analysis, find G p1 from a mass balance on tank 1,
dh1 1
dt
= q1 + q 2 −C 1
h1
15-3
Linearizing (4), noting that q1′ = 0, and taking Laplace transforms: A1
or
dh′
= q2′ −
dt
H1′ (s )
2 h1
h1′
(2 h1 / C 1)
=
Q2′ ( s)
C 1
(2 1 h1 / C1 ) s + 1
(5)
q3 = C1 h1
Since
q3′ =
C 1 2 h1
Q3′ ( s )
or
h1′
H1′ ( s )
=
C 1
(6)
2 h1
From (5) and (6),
Q3′ ( s)
Q2′ ( s)
=
1 (2 A1 h1 / C1) s + 1
= G P
(7)
1
Substituting for Q3 from (7) and (2) into (1) gives
Q4 + (
or
G f
1 (2 A1 h1 / C1 ) s + 1
=−
1
K v K t
) K vG f K t Q4
=0
[(2 A1 h1 / C1 ) s + 1]
15.5
(a)
For a steady-state analysis: G p=1,
Gd =2,
Gv = Gm = Gt =1
From Eq.15-21, G f =
(b)
− Gd Gv Gt G p
=
−2 (1)(1)(1)
Using Eq. 15-21,
15-4
= −2
−2 − Gd
G f =
(c)
Gv Gt G p
=
( s + 1)(5 s + 1)
1 (1)(1) + s 1
=
−2 5 s + 1
Using Eq. 12-19,
~ G where
1
= Gv G p Gm =
G +
= 1, G − =
s + 1 1
~ ~
= G + G−
+1
For τc=2, and r =1, Eq. 12-21 gives f =
1 2 s + 1
From Eq. 12-20 Gc*
1 s + 1 = G − −1 f = ( s + 1) = 2s + 1 + 2 s 1
From Eq. 12-16
Gc
(d)
=
Gc
s + 1
∗
~ 1 − Gc G ∗
s + 1 = 2 s + 1 = 1 2 s 1− 2 s + 1
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 +Gt G f GvG p)
1 s
For the controller of part (a) Y ( s) =
2 1 1 + − ( 1 )( 2 )( 1 ) ( s + 1)(5 s + 1) s + 1 s
15-5
− 10
Y ( s) =
or
( s + 1)(5 s + 1) -t
=
5/ 2 s + 1
+
− 25 / 2 2.5 − 2.5 = − 5 s + 1 s + 1 s + 1 / 5
- t /5
y(t ) = 2.5 (e – e
)
For the controller of part (b) Y ( s) =
or
2 − 2 1 1 ( s + 1)(5 s + 1) + (1) 5 s + 1 (1) s + 1 s = 0
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) 2 s s + 1
15-6
or
Y ( s) =
5 25 / 3 − 40 / 3 − 20 s = + + ( s + 1)(5 s + 1)(2 s + 1) s + 1 5 s + 1 2 s + 1
5 20 / 3 5/3 − + = s + 1 s + 1 / 2 s + 1 / 5 or
5 - t /5 -t 20 -t /2 y(t ) = 5e – e + e 3 3
and for controllers of parts (b) and (c)
Y (s) =
or
2 − 2 (1) 1 ( 1 ) + ( s + 1)(5 s + 1) 5 s + 1 s + 1 1 =0 s 1 1 + (1) (1) s 1+ 2 s s + 1
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 T 1 + w2 C T 2 − w C T 4 + 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 ( w1T 1 + w 2 T 2
− wT 4 )
(1)
Neglecting heater and transmitter dynamics, Q = K h p
(2)
0
0
T 1m = T 1m + K T(T 1-T 1 ) 0
(3)
0
wm = wm + K w(w-w )
(4)
Substituting into (1) for Q, T 1, and w from (2), (3), and (4), gives
p =
(b)
C K h
0
[ w1 (T1
+
1 KT
0
(T1m − T1m )) + w2 T2 − T4 (w
+
1 K w
(a)
A steady-state material balance for both tanks gives,
15-8
0
(wm − wm ))]
Dynamic compensation is desirable because the process transfer function G p= T 4( s)/ P ( s) is different from each of the disturbance transfer functions, Gd1= T 4( s)/T 1( s), and Gd 2= T 4( s)/w( s); this is more so for Gd 1 which has a higher order.
15.7
(b)
0
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 = K v G f K t Q1 Substituting for Q5 into (1) gives 0 = Q1 − K v G f K t Q 1
(c)
or
G f =
1 K v K t
To find Gd and G p, the mass balance on tank 1 is A1
dh1
= q1 + q2 − C 1
dt
h1
where A1 is the cross-sectional area of tank 1. Linearizing and setting q ′2 = 0 leads to A1
dh1′
C 1
= q1′ −
dt
2 h1
h1′
Taking the Laplace transform, H1′ ( s) Q1′ ( s)
=
R1 A1R1s + 1
where R1
≡
2 h1 C 1
(2)
Linearizing q3 = C 1 h1 gives q3′
=
1 R1
h1′
or
Q3′ ( s)
H1′ ( s)
Mass balance on tank 2 is A2
dh2 dt
= q3 + q 4 − q5
15-9
=
1 R1
(3)
Using deviation variables, setting q ′4 = 0, and taking Laplace transform 2 sH 2′
H 2′ ( s )
( s ) = Q3′ ( s ) − Q5′ ( s )
=
Q3′ ( s )
1 A2 s
(4)
and H 2′ ( s) Q5′ ( s)
=−
Gd ( s) =
1 A2 s
H 2′ ( s) Q1′ ( s )
= G p ( s) =
H 2′ ( s) Q3′ ( s) H1′ ( s) Q3′ ( s) H1′ ( s) Q1′( s)
=
1 A2 s( A1 R1 s +1)
upon substitution from (2), (3), and (4). Using Eq. 15-21,
G f
=
− Gd Gt Gv G p
=+
1
− =
1 A2 s( A1 R1 s + 1)
K t K v ( −1 / A2 s ) 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
=
τ p − τ L
(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 1 Y ( s) = [Gd + Gt G f Gv G p ] s Setting Gt = Gv = 1, and using Eqs. 15-22 and 15-29,
15-10
K − Kd / K P ( τ1s + 1) K p 1 d Y ( s) = + (1) (1) τ s + 1 s τ s +1 τ d s + 1 2 p
1 (τ − τ )τ τ 1 τ (τ − τ ) 1 = K d − d − − 2 1 2 − 1 p p τ s τ d s + 1 s (τ 2 − τ p ) τ2 s + 1 p − τ2 or
τ −τ (τ − τ ) = K d −e−t / − 1 2 e−t / − 1 p e−t / τ −τ τ 2 p p − τ2
∞
∫0
e(t ) dt
τ
τ
y(t )
=∫
=
∞ 0
y(t ) dt = − K d
2
− K d 2 2 τ τ τ τ τ τ −τ τ τ +τ − + − + ( τ p τ2 − τ p τ2 ) d 2 d p 2 1 2 p 1 p τ 2 − τ p
= 0 when (1) holds.
15.9
For steady-state conditions G p=1,
Gd =2,
Gv = Gm = Gt =1
Using Eq. 15-21, G f =
(b)
p
τ (τ − τ ) τ (τ − τ ) p 2 1 2 + p 1 + τ d τ −τ τ 2 p p − τ2
= − K d ( τ1 − τ 2 ) − ( τ p − τ d )
(a)
τ
τ s +1 p 1
− Gd Gv Gt G p
=
−2 (1)(1)(1)
Using Eq. 15-21,
15-11
= −2
G f =
(c)
− Gd Gv Gt G p
− 2e − s −2 ( s + 1)(5 s + 1) = = 1 e − s 5 s + 1 (1)(1) s + 1
Using Eq. 12-19, ~ G
= Gv G p Gm = ~ G+
where
= e − s
e − s
~ ~
s + 1
= G+ G− ~ G−
,
=
1 s + 1
For τc=2, and r = 1, Eq. 12-21gives f =
1 2 s + 1
From Eq. 12-20 s + 1 1 1 Gc * = ~ f = ( s + 1) = 2 s + 1 2 s + 1 G− From Eq. 12-16 s + 1 Gc
(d)
=
Gc *
~ 1 − Gc * G
s + 1 = 2 s + 1 = 1 2 s 1− 2 s + 1
For feedforward control only, Gc=0. For a unit step disturbance, D( s) = 1/ s. Substituting into Eq. 15-20 gives 1 Y ( s) = (Gd +Gt G f GvG p) s For the controller of part (a)
Y ( s) =
e − s 1 2e − s + (1)(−2)(1) s + 1 s ( s + 1)(5 s + 1)
15-12
− 10e − s ( s + 1)(5 s + 1)
=
or
-(t -1)
y(t ) = 2.5 (e
-( t -1)/5
– e
)S(t -1)
For the controller of part (b)
Y (s) =
or
2e − s − 2 e − s 1 = 0 + (1) (1) ( s 1 )( 5 s 1 ) 5 s 1 s 1 + + + + s
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),
Y ( s) =
e − s 2e − s + (1)(−2)(1) ( s 1 )( 5 s 1 ) s 1 + + + 1 s s + 1 e − s 1+ (1) (1) 2 s s + 1
15-13
and for the controllers of parts (b) and (c),
Y (s) =
or
2 − 2 1 ( s + 1)(5 s + 1) + (1) 5 s + 1 (1) s + 1 1 =0 s + 1 1 s + 1 ( 1 ) ( 1 ) 2 s s + 1
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 G p=K p ,
Gd =K d ,
Gv = Gm = Gt =1
Using Eq. 15-21,
15-14
G f =
(b)
Gv Gt G p
=
− 0.5 (1)(1)(2)
= −0.25
Using Eq. 15-21,
G f =
(c)
− Gd
− Gd Gv Gt G p
− 0.5e −30 s (95 s + 1) −10 s 60 s + 1 = = − 0 . 25 e − 20 s ( 60 s 1 ) + 2e (1)(1) + 95 s 1
Using Table 12.1, a PI controller is obtained from equation G,
τ 1 95 = = 0.95 K p τ c + θ 2 (30 + 20) τ I = τ = 95
K c
(d)
=
1
As shown in Fig.S15.10a, the dynamic controller provides significant improvement.
(e) 0.08 Controller of p art (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 p art (a) and (c) Controllers of p art (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 dt
= wC (T i − T ) − U (1 + qc ) A(T − T c ) − U L A L (T − T a ) (1)
Expanding the right hand side,
ρVC
dT
= wC (T i − T ) − UA(T − T c ) dt − UAqc T + UAqc T c − U L A L (T − T a )
(2)
Linearizing, q c T ≈ q c T + q c T ′ + T q c′ Substituting (3) into (2), subtracting the steady-state equation, and introducing deviation variables,
15-16
(3)
d T ′
= wC (T i′ − T ′) − UAT ′ − UAT qc′ dt + UAT c qc′ − U L A LT ′
ρVC
− UAqc T ′ (4)
Taking the Laplace transform and assuming steady-state at t = 0 gives, ρ VCsT
′( s) = wCTi′( s) + UA(Tc − T − ) Qc′ ( s)
− ( wC + UA + UAq c + U L A L )T ′( s)
(5)
Rearranging, T ′( s) = G L ( s)Ti′( s) + G p ( s)Qc′ ( s)
(6)
where: Gd ( s ) = G p ( s) = K d = K p
τ=
=
K L s + 1 K p
τ
τ s + 1
wC
K UA(T c
ρVC
(7)
− T )
K
K K = wC + UA + UAq c
+ U L A L
The ideal FF controller design equation is given by, G F =
−Gd Gt Gv G p
(17-27)
But, Gt = K t e − θ s and Gv=K v
(8)
Substituting (7) and (8) gives, G F
− wCe + θ s = K t K vUA(T c − T )
(9)
+ s In order to have a physically realizable controller, ignore the e θ term,
15-17
=
G F
− wC K t K vUA(T c
− T )
(10)
15.12
a)
A component balance in A gives: V
dc A dt
= qc Ai − qcA − VkcA
(1)
At steady state,
0=qc i
− q cA − VkcA
(2)
Solve for q ,
q
=
kV C 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 dt
= q ci + qc′Ai + cAi q′ − q cA − qc′A − cA q′ − VkcA
Subtract (2), V
′ dc A dt
= qc′iA+ cAi q′ − qc′A − cA q′ − Vkc′A
Take the Laplace transform,
′ ( s) = qc′iA( s) + cAi Q′( s) − qc′A ( s) − cA Q′( s) − VkcA′ ( s) sVc A Rearrange,
15-18
− cA Q′( s) sV + q + Vk c Ai
(6)
′ ( s) = Gd ( s)C A′i ( s) + G p ( s)Q′( s) C A
(7)
′ ( s) = C A
q sV
+ q + Vk
C ′Ai ( s ) +
or
The ideal FF controller design equation is, G F ( s) = −
Gd ( s ) Gv ( s )G p ( s)Gt (s )
(8)
Substitute from (6) and (7) with Gv( s)=K v and Gt ( s)=K t : G F ( s) = − K v (cAi .
q
− cA ) K t
(9)
′ m ( s ) where P is the controller output and c Aim Note: G F ( s) = P ′( s) / C Ai is the measured value of c Ai.
15.13
(a)
Steady-state balances: 0 = q5
+ q1 − q3
(1)
0 = q3
+ q2 − q4
(2)
0
0 = x5 q5 + x1 q1 − x3 q3
(3)
0 = x3 q 3 + x2 q 2 − x4 q 4
(4)
Solve (4) for x3 q 3 and substitute into (3), 0 = x5 q 5 + x 2 q 2 − x 4 q 4 Rearrange,
15-19
(5)
q2
=
x4 q 4 − x5 q5 x2
(6)
In order to derive the feedforward control law, let x4
→ x4 sp ,
x2
→ x2 (t ),
5
→ x5 (t ),
and
q2
→ q 2 (t )
Thus, q 2 (t ) =
x4 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 x4 sp − 1.01 x5 (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
