Chapter 15
15.1 For Ra=d/u 2 u d u R K a p ∂ =− ∂ =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 2 1 2 2 1 2 2 0 0 = = − − = d u K K d K u K d d u u K m m m m R (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
2 1 mA ) 4 20 ( d S K = − , 2 (20 42)mA u S K = −
Substituting for K1, K2 and u/d into (1) gives: 2 2 2 2 = = u d d d d u R S S R R S S K
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp
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
v t
K K (2)
c) (No, because Eq. 1 above does not involve q2.
15.4
(b) From the block diagram, exact feedforward compensation for Q1 would
result when Q1 + Q2 = 0 Substituting Q2 = KV Gf Kt Q1, Gf = − 1 v t K K
(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 = 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
or Gf = − 1
v t
K K
For dynamic analysis, find Gp1 from a mass balance on tank 1,
1
1 1 2 1 1
dh
A q q C h
Linearizing (4), noting that q′1 = 0, and taking Laplace transforms: 1 1 2 1 1 2 C dh A q h dt h ′ ′ ′ = − or 1 1 1 2 1 1 1 (2 / ) ( ) ( ) (2 / ) 1 h C H s Q s A h C s ′ = ′ + (5) Since 3 1 1 1 3 1 1 2 q C h C q h h = ′ = ′ or 31 1 1 ( ) ( ) 2 Q s C H s h ′ = ′ (6) From (5) and (6), 1 3 2 1 1 1 ( ) 1 ( ) (2 / ) 1 P Q s G Q s A h C s ′ = = ′ + (7)
Substituting for Q3 from (7) and (2) into (1) gives
4 4 1 1 1 1 ( ) 0 (2 / ) 1 v f t Q K G K Q A h C s + = + or 1 1 1
1
[(2
/ )
1]
f v tG
A h C s
K K
= −
+
15.5(a) For a steady-state analysis:
Gp=1, Gd=2, Gv = Gm = Gt =1 From Eq.15-21, Gf = 2 ) 1 )( 1 )( 1 ( 2 =− − = − p t v d G G G G (b) Using Eq. 15-21,
Gf = 1 5 2 1 1 ) 1 )( 1 ( ) 1 5 )( 1 ( 2 + − = + + + − = − s s s s G G G G p t v d (c) Using Eq. 12-19, = + − + = = G G s G G G G v p m ~ ~ 1 1 ~ where 1, 1 1 G G s + = − = +
For τc=2, and r=1, Eq. 12-21 gives f = 1 2 1 + s From Eq. 12-20 * 1 ( 1) 1 1 2 1 2 1 − − + = = + = + + c s G G f s s s From Eq. 12-16 s s s s s G G G G c c c 2 1 1 2 1 1 1 2 1 ~ 1 + = + − + + = − = ∗ ∗
(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)
s
1
For the controller of part (a)
Y(s) = s s s s 1 1 1 ) 1 )( 2 )( 1 ( ) 1 5 )( 1 ( 2 + − + + +
Y(s) = 5 / 1 5 . 2 1 5 . 2 1 5 2 / 25 1 2 / 5 ) 1 5 )( 1 ( 10 + − − + = + − + + = + + − s s s s s s or y(t) = 2.5 (e-t – e- t/5) For the controller of part (b)
Y(s) = 1 0 1 1 ) 1 ( 1 5 2 ) 1 ( ) 1 5 )( 1 ( 2 = + + − + + + s s s s s 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),
Y(s) = 2 1 (1)( 2)(1) 1 ( 1)(5 1) 1 1 1 1 (1) (1) 2 1 s s s s s s s + − + + + + + +
or Y(s) = = + + + − ) 1 2 )( 1 5 )( 1 ( 20 s s s s 1 2 3 / 40 1 5 3 / 25 1 5 + − + + + + s s s = 5 / 1 3 / 5 2 / 1 3 / 20 1 5 + + + − + s s s or y(t) = 5e-t – 3 20 e-t/2 + 3 5 e- t/5 and for controllers of parts (b) and (c)
Y(s) = s s s s s s s s 1 ) 1 ( 1 1 ) 1 ( 2 1 1 1 1 ) 1 ( 1 5 2 ) 1 ( ) 1 5 )( 1 ( 2 + + + + + − + + + = 0 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
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+w2T2 −wT4) (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
0 0 0 0 1 1 1 1 1 2 2 4 1 [ ( ( m m )) ( ( m m ))] h T w C p w T T T w T T w w w K K K = + − + − + −
(b) 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)
0 = q1 + q2 + q4 − q5
Because q′2 = q′4 = 0, the above equation gives
0 = q′1 – q′ or 0 = Q5 1 – Q5 (1)
From the block diagram,
Q5 = Kv Gf Kt Q1
Substituting for Q5 into (1) gives
0 = Q1− Kv Gf Kt Q1 or Gf =
t vK K
1
(c) To find Gd and Gp, the mass balance on tank 1 is 1 1 2 1 1 1 q q C h dt dh A = + −
where A1 is the cross-sectional area of tank 1.
Linearizing and setting q′2 = 0 leads to
1 1 1 1 1 1 2 dh C A q h dt h ′ ′ ′ = −
Taking the Laplace transform,
1 1 1 1 1 ( ) ( ) 1 H s R Q s A R s ′ = ′ + where 1 1 1 2 C h R ≡ (2) Linearizing q3 = C1 h gives 1 3 1 1 1 q h R ′ = ′ or 3 1 1 ( ) 1 ( ) Q s H s R ′ = ′ (3) Mass balance on tank 2 is
5 4 3 2 2 q q q dt dh A = + −
Using deviation variables, setting q′4 = 0, and taking Laplace transform
( )
( )
( )
2 2 3 5 A sH s′ =Q s′ −Q s′ 2 3 2 ( ) 1 ( ) H s Q s A s ′ = ′ (4) and 2 5 2 ( ) 1 ( ) ( ) p H s G s Q s A s ′ = − = ′ 2 2 3 1 1 3 1 1 2 1 1 ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) ( 1) d H s H s Q s H s G s Q s Q s H s Q s A s A R s ′ ′ ′ ′ = = = ′ ′ ′ ′ +upon substitution from (2), (3), and (4). Using Eq. 15-21, ) / 1 ( ) 1 ( 1 2 1 1 2 s A K K s R A s A G G G G G v t p v t d f − + − = − = 1 1 1 1 1 + + = s R A K Kv t 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) =
[
]
s G G G G Gd + t f v p 11 2 / (τ 1) 1 ( ) (1) (1) τ 1 τ 1 τ 1 p d d P d p K K K K s Y s s s s s − + = + + + + 1 2 1 2 2 2 2 (τ τ )τ τ τ (τ τ ) 1 1 1 1 τ 1 (τ τ ) τ 1 τ τ τ 1 p p d d d p p p K s s s s s − − = − − − − + − + − + or y(t) / τ 1 2 / τ2 1 / τ 2 2 τ τ (τ τ ) τ τ τ τ − − − − − = − − − − − p p t t t d p p K e e e 0 e t dt( ) 0 y t dt( ) ∞ ∞ =
∫
∫
2 1 2 1 2 2 τ (τ τ ) τ (τ τ ) τ τ τ τ τ p p d d p p K − − = − + + − − 2 2 1 22 1 2 2 2 2 τ τ τ τ τ τ τ τ τ τ (τ τ τ τ ) τ τ d d d p p p p p p K − = − + − − + + − − = −K d (τ1−τ ) (τ2 − 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 = 2 ) 1 )( 1 )( 1 ( 2 =− − = − p t v d G G G G (b) Using Eq. 15-21,
Gf = 1 5 2 1 1 ) 1 )( 1 ( ) 1 5 )( 1 ( 2 + − = + + + − = − − − s e s s s e G G G G s s p t v d (c) Using Eq. 12-19, − + − = + = = G G s e G G G G s m p v ~ ~ 1 ~ where 1 1 ~ , ~ + = = − − + s G e G s
For τc=2, and r = 1, Eq. 12-21gives
f = 1 2 1 + s From Eq. 12-20 1 2 1 1 2 1 ) 1 ( ~1 * + + = + + = = − s s s s f G Gc From Eq. 12-16 s s s s s G G G G c c c 2 1 1 2 1 1 1 2 1 ~ * 1 * = + + − + + = − =
(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)
s
1
For the controller of part (a)
Y(s) = s s e s s e s s 1 1 ) 1 )( 2 )( 1 ( ) 1 5 )( 1 ( 2 + − + + + − −
= ) 1 5 )( 1 ( 10 + + − − s s e s or y(t) = 2.5 (e-(t-1) – e-( t-1)/5)S(t-1) For the controller of part (b)
Y(s) = 1 0 1 ) 1 ( 1 5 2 ) 1 ( ) 1 5 )( 1 ( 2 = + + − + + + − − s s e s s s e s s 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),
Y(s) = s s e s s s e s s e s s s 1 ) 1 ( 1 ) 1 ( 2 1 1 1 ) 1 )( 2 )( 1 ( ) 1 5 )( 1 ( 2 + + + + − + + + − − −
and for the controllers of parts (b) and (c), Y(s) = s s s s s s s s 1 ) 1 ( 1 1 ) 1 ( 2 1 1 1 1 ) 1 ( 1 5 2 ) 1 ( ) 1 5 )( 1 ( 2 + + + + + − + + + = 0 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,
Gf = 0.25 ) 2 )( 1 )( 1 ( 5 . 0 =− − = − p t v d G G G G (b) Using Eq. 15-21, Gf = s s s p t v d e s s s e s e G G G G 10 20 30 ) 1 60 ( ) 1 95 ( 25 . 0 1 95 2 ) 1 )( 1 ( 1 60 5 . 0 − − − + + − = + + − = −
(c) Using Table 12.1, a PI controller is obtained from equation G, 95 . 0 ) 20 30 ( 95 2 1 1 = + = θ + τ τ = c p c K K τI = =τ 95
(d) As shown in Fig.S15.10a, the dynamic controller provides significant improvement. (e) 0 100 200 300 400 500 -0.04 -0.02 0 0.02 0.04 0.06 0.08 time y(t)
Controller of part (a) Controller of part (b)
0 100 200 300 400 500 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 time y(t)
Controllers of part (a) and (c) Controllers of part (b) and (c)
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: ) ( ) ( ) 1 ( ) (Ti T U qc AT Tc ULAL T Ta wC dt dT VC = − − + − − − ρ (1)
Expanding the right hand side,
) ( ) (Ti T UAT Tc wC dt dT VC = − − − ρ −UAqcT +UAqcTc −ULAL(T −Ta) (2) Linearizing, c c c cT q T q T Tq q ≈ + ′+ ′ (3)
Substituting (3) into (2), subtracting the steady-state equation, and introducing deviation variables,
T q UA q T UA T UA T T wC dt T d VC ′ = i′− ′ − ′− c′ − c ′ ρ ( ) +UATcqc′ −ULALT′ (4)
Taking the Laplace transform and assuming steady-state at t = 0 gives,
ρVCsT s′( )=wCT si′( )+UA T( c−T Q s−) c′( ) −(wC+UA+UAqc +ULAL)T′(s) (5) Rearranging, ( ) L( ) ( )i p( ) c( ) T s′ =G s T s′ +G s Q s′ (6) where: ( ) τ 1 L d K G s s = + 1 ) ( + τ = s K s Gp p d wC K K = (7) K T T UA K c p ) ( − = K VC ρ = τ L L c U A q UA UA wC K = + + +
The ideal FF controller design equation is given by, d F t v p G G G G G − = (17-27) But, s t t K e G = −θ and Gv=Kv (8)
Substituting (7) and (8) gives,
) (T T UA K K wCe G c v t s F − − = +θ (9)
) (T T UA K K wC G c v t F − − = (10) 15.12
a) A component balance in A gives: A Ai A A dc V qc qc Vkc dt = − − (1) At steady state, 0=q cAi−q cA−VkcA (2) Solve for ,q A Ai A C C C kV q − = (3)
For an ideal FF controller, replace C by Ai C , q by qAi 1 and C by A CAsp: Asp Ai Asp kVC q C C = − b) Linearize (1): A iA Ai Ai A A A A dc V q c qc c q q c qc c q Vkc dt = + ′ + ′− − ′ − ′− Subtract (2), A iA Ai A A A dc V qc c q qc c q Vkc dt ′ ′ ′ ′ ′ ′ = + − − −
Take the Laplace transform,
( ) ( ) ( ) ( ) ( ) ( )
A iA Ai A A A
sVc s′ =qc s′ +c Q s′ −qc s′ −c Q s′ −Vkc s′
( )
( ) Ai A ( ) A q Ai c c C s C s Q s sV q Vk sV q Vk − ′ = ′ + ′ + + + + (6) or ( ) ( ) ( ) ( ) ( ) A d iA p C s′ =G s C s′ +G s Q s′ (7)The ideal FF controller design equation is, ( ) ( ) ( ) ( ) ( ) d F v p t G s G s G s G s G s = − (8)
Substitute from (6) and (7) with Gv(s)=Kv and Gt(s)=Kt : ( ) ( ) F v Ai A t q G s K c c K = − − (9) .
Note: GF(s)=P′(s)/C′Aim(s) where P is the controller output and cAim is the measured value of cAi.
15.13
(a) Steady-state balances:
3 1 5 0=q +q −q (1) 4 2 3 0=q +q −q (2) 0 3 3 1 1 5 5 0=x q +x q −x q (3) 4 4 2 2 3 3 0=x q +x q −x q (4)
Solve (4) for x3q3 and substitute into (3), 4 4 2 2 5 5 0=x q +x q −x q (5) Rearrange,
2 5 5 4 4 2 x q x q x q = − (6)
In order to derive the feedforward control law, let 4 4sp, x →x x2 →x t2( ), x5→x t5( ), and q2 →q2(t) Thus, 2 5 5 4 4 2 ) ( ) ( ) ( x t q t x q x t q = sp − (7)
Substitute numerical values:
990 . 0 ) ( ) ( ) 3400 ( ) ( 4 5 5 2 t q t x x t q = sp − (8) or ) ( ) ( 01 . 1 3434 ) ( 4 5 5 2 t x x t q t q = sp − (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.
