Process Dynamics and Control Solutions

1234567898

8.1

a)

For step response, M s  τD s +1   τ s +1  Ya′ ( s ) = K c  D  U ′( s ) = K c M   s (ατ D s + 1)   ατ D s + 1

input is u ′(t ) = M

Ya′ (s ) =

U ′( s ) =

,

K c Mτ D KcM + ατ D s + 1 s (ατ D s + 1)

Taking inverse Laplace transform y ′a (t ) =

K c M −t /( ατ D ) e + K c M (1 − e −t /( ατ D ) ) α

As α →0 ∞

e − t /( ατ D ) dt + K c M α t =0

y a′ (t ) = K c Mδ(t ) ∫

y a′ (t ) = K c Mδ(t )τ D

+ KcM

Ideal response, KM  τ s + 1 Yi′( s ) = Gi ( s )U ′( s ) = K c M  D = KcMτD + c  s  s 

yi′ (t ) = K c Mτ D δ(t ) + K c M Hence y a′ (t ) → y i′ (t ) as α → 0 For ramp response, input is u ′(t ) = Mt

,

U ′( s ) =

M s2

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

8-1

 τDs +1   τ s +1  Ya′ ( s ) = K c  D  U ′( s ) = K c M  2  ατ D s + 1  s (ατ D s + 1)  Ya′ (s ) =

K c Mτ D K M + 2 c s (ατ D s + 1) s (ατ D s + 1)  − ατ D 1 1 ατ D  (ατ D ) 2  K M = K c Mτ D  − + + +   c  s 2 ατ D s + 1  s ατ D s + 1  s

Taking inverse Laplace transform

[

]

[

y a′ (t ) = K c Mτ D 1 − e − t /( ατ D ) + K c M t + ατ D (e − t /( ατ D ) − 1)

]

As α →0 y a′ (t ) = K c Mτ D + K c Mt Ideal response, K c Mτ D K c M  τ s + 1 Yi′(s ) = K c M  D 2  = + 2 s s  s 

yi′ (t ) = K c Mτ D

+ K c Mt

Hence y a′ (t ) → y i′ (t ) as α → 0 b)

It may be difficult to obtain an accurate estimate of the derivative for use in the ideal transfer function.

c)

Yes. The ideal transfer function amplifies the noise in the measurement by taking its derivative. The approximate transfer function reduces this amplification by filtering the measurement.

8.2

a)

K1 K + K 2 τ1 s + K 2 P ′( s ) = + K2 = 1 E ( s ) τ1 s + 1 τ1 s + 1

8-2

 K 2 τ1   K + K s + 1 2  = ( K1 + K 2 )  1 τ1 s + 1      

b)

Kc = K1 + K2

K2 = Kc − K1

→

τ1 = ατ D τD =

K 2 τ1 K ατ = 2 D K1 + K 2 K1 + K 2

or

1=

K 2α K1 + K 2

K1 + K 2 = K 2 α K 1 = K 2 α − K 2 = K 2 (α − 1) Substituting, K 1 = ( K c − K 1 )(α − 1) = (α − 1) K c − (α − 1) K 1 Then,  α −1 K1 =  K c  α 

c)

If Kc = 3

, τD = 2

K1 =

,

α = 0.1

− 0 .9 × 3 = −27 0 .1

K 2 = 3 − (−27) = 30 τ1 = 0.1 × 2 = 0.2 Hence K1 + K2 = -27 + 30 = 3

K 2 τ1 30 × 0.2 = =2 K1 + K 2 3  2s + 1  Gc ( s ) = 3   0 .2 s + 1 

8-3

then,

8.3

a)

From Eq. 8-14, the parallel form of the PID controller is :   1 Gi ( s ) = K c′ 1 + + τ′D s   τ′I s  From Eq. 8-15, for α →0, the series form of the PID controller is:

 1  Ga ( s ) = K c 1 + [τ D s + 1]  τI s   τ  1 = K c 1 + D + + τ D s  τI τI s     τ D  τDs 1   = K c 1 + 1+ +   τI   τ   τ   1 + D τ I s 1 + D τI τI    

       

Comparing Ga(s) with Gi(s)

 τ  K c′ = K c 1 + D  τI    τ  τ′I = τ I 1 + D  τI   τ′D =

b)

τD τ 1+ D τI

 τ Since 1 + D  τI K c ≤ K c′ ,

  ≥ 1 for all τD, τI, therefore  τ I ≤ τ′I and τ D ≥ τ′D

c)

For Kc = 4, τI=10 min , τD =2 min

d)

K c′ = 4.8 , τ′I = 12 min , τ′D = 1.67 min Considering only first-order effects, a non-zero α will dampen all responses, making them slower.

8-4

8.4

Note that parts a), d), and e) require material from Chapter 9 to work. a)

System I (air-to-open valve) : Kv is positive. System II (air-to-close valve) : Kv is negative.

b)

System I : Flowrate too high → need to close valve → decrease controller output → reverse acting System II: Flow rate too high → need to close valve → increase controller output → direct acting.

c)

System I : Kc is positive System II : Kc is negative

d) System I : System II :

Kc

Kv

+ −

+ −

Kp Km + +

+ +

Kc and Kv must have same signs e)

Any negative gain must have a counterpart that "cancels" its effect. Thus, the rule: # of negative gains to have negative feedback = 0 , 2 or 4. # of negative gains to have positive feedback = 1 or 3.

8.5

a)

From Eqs. 8-1 and 8-2,

[

p(t ) = p + K c y sp (t ) − y m (t )

]

(1)

The liquid-level transmitter characteristic is ym(t) = KT h(t)

(2)

where h is the liquid level KT > 0 is the gain of the direct acting transmitter.

8-5

The control-valve characteristic is q(t) = Kvp(t)

(3)

where q is the manipulated flow rate Kv is the gain of the control valve. From Eqs. 1, 2, and 3

[

]

[

q(t ) − q = K v p (t ) − p = K V K c y sp (t ) − K T h(t ) KV K c =

]

q (t ) − q y sp − K T h(t )

For inflow manipulation configuration, q(t)> q when ysp(t)>KTh(t). Hence KvKc > 0 then for "air-to-open" valve (Kv>0), Kc>0 : and for "air-to-close" valve (Kv