Process Dynamics and Control Solutions

1234567898

7.1

In the absence of more accurate data, use a first-order transfer function as T '( s ) Ke −θs = Qi '( s ) τs + 1 o T (∞) − T (0) (124.7 − 120) F = = 0.118 ∆qi 540 − 500 gal/min θ = 3:09 am – 3:05 am = 4 min

K=

Assuming that the operator logs a 99% complete system response as “no change after 3:34 am”, 5 time constants elapse between 3:09 and 3:34 am. 5τ = 3:34 min − 3:09 min = 25 min τ = 25/5 min = 5 min Therefore, T '( s ) 0.188e−4 s = Qi '( s ) 5s + 1 To obtain a better estimate of the transfer function, the operator should log more data between the first change in T and the new steady state.

7.2 h(5.0) − h(0) (6.52 − 5.50) min = = 0.336 2 ∆qi 30.4 × 0.1 ft Output at 63.2% of the total change Process gain, K = a)

= 5.50 + 0.632(6.52-5.50) = 6.145 ft Interpolating between h = 6.07 ft

and

h = 6.18 ft

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

7-1

τ = 0.6 +

(0.8 − 0.6) (6.145 − 6.07) min = 0.74 min (6.18 − 6.07)

b)

dh h(0.2) − h(0) 5.75 − 5.50 ft ft ≈ = = 1.25 dt t = 0 0.2 − 0 0.2 min min Using Eq. 7-15, τ=

c)

KM 0.347 × (30.4 × 0.1) = = 0.84 min 1.25  dh     dt t =0 

 h(t i ) − h(0)  The slope of the linear fit between ti and z i ≡ ln 1 −  gives an  h ( ∞ ) − h ( 0)  approximation of (-1/τ) according to Eq. 7-13. Using h(∞) = h(5.0) = 6 .52, the values of zi are ti 0.0 0.2 0.4 0.6 0.8 1.0 1.2

zi 0.00 -0.28 -0.55 -0.82 -1.10 -1.37 -1.63

ti 1.4 1.6 1.8 2.0 3.0 4.0 5.0

zi -1.92 -2.14 -2.43 -2.68 -3.93 -4.62 -∞

Then the slope of the best-fit line, using Eq. 7-6 is

 1  13Stz − St S z slope =  −  = 2  τ  13Stt − ( St )

(1)

where the datum at ti = 5.0 has been ignored. Using definitions, St = 18.0 S z = −23.5

Stt = 40.4 Stz = −51.1

Substituting in (1),  1  −  = −1.213  τ

τ = 0.82 min

7-2

d) 6.8

6.6

6.4

6.2

6 Experimental data Model a) Model b) Model c)

5.8

5.6

5.4

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Figure S7.2. Comparison between models a), b) and c) for step response.

7.3

a)

T1′( s ) K1 = Q′( s ) τ1s + 1

T2′( s ) K2 = T1′( s ) τ2 s + 1

T2′( s ) K1 K 2 K1 K 2 e −τ2 s = ≈ Q′( s ) (τ1s + 1)(τ2 s + 1) (τ1s + 1)

(1)

where the approximation follows from Eq. 6-58 and the fact that τ1>τ2 as revealed by an inspection of the data. T1 (50) − T1 (0) 18.0 − 10.0 = = 2.667 ∆q 85 − 82 T (50) − T2 (0) 26.0 − 20.0 K2 = 2 = = 0.75 T1 (50) − T1 (0) 18.0 − 10.0

K1 =

Let z1, z2 be the natural log of the fraction incomplete response for T1,T2, respectively. Then,

7-3

 T (50) − T1 (t )  18 − T1 (t )  = ln  z1 (t ) = ln  1   8   T1 (50) − T1 (0)   T (50) − T2 (t )   26 − T2 (t )  z2 (t ) = ln  2 = ln    6   T2 (50) − T2 (0)  A graph of z1 and z2 versus t is shown below. The slope of z1 versus t line is –0.333 ; hence (1/-τ1)=-0.333 and τ1=3.0 From the best-fit line for z2 versus t, the projection intersects z2 = 0 at t≈1.15. Hence τ2 =1.15. T1 ' ( s ) 2.667 = Q ' ( s ) 3s + 1 T2 ' ( s ) 0.75 = T1 ' ( s ) 1.15s + 1

(2) (3)

0.0 -1.0

0

5

10

15

20

-2.0 z 1,z 2

-3.0 -4.0 -5.0 -6.0 -7.0 -8.0 time,t

Figure S7.3a. z1 and z2 versus t

By means of Simulink-MATLAB, the following simulations are obtained 28

26

24

22

20

T1 , T2

b)

18

16

T1 T2 T1 (experimental) T2 (experimental)

14

12

10

0

2

4

6

8

10

12

14

16

18

20

22

time

Figure S7.3b. Comparison of experimental data and models for step change

7-4

7.4 Y (s) = G( s) X ( s) =

2 1.5 × (5s + 1)(3s + 1)( s + 1) s

Taking the inverse Laplace transform y (t ) = -75/8*exp(-1/5*t)+27/4*exp(-1/3*t)-3/8*exp(-t)+3

a)

Fraction incomplete response  y (t )  z (t ) = ln 1 − 3   0.0 -1.0 0

10

20

30

40

50

-2.0

z(t)

-3.0 -4.0 -5.0 -6.0 -7.0 z(t) = -0.1791 t + 0.5734

-8.0 -9.0

time,t

Figure S7.4a. Fraction incomplete response; linear regression

From the graph, slope = -0.179 and intercept ≈ 3.2 Hence, -1/τ = -0.179 and τ = 5.6 θ = 3.2 G (s) = b)

2e −3.2 s 5.6 s + 1

In order to use Smith’s method, find t20 and t60 y(t20)= 0.2 × 3 =0.6 y(t60)= 0.6 × 3 =1.8 Using either Eq. 1 or the plot of this equation, t20 = 4.2 , t60 = 9.0 Using Fig. 7.7 for t20/ t60 = 0.47 ζ= 0.65 , t60/τ= 1.75, and τ = 5.14

7-5

(1)

G (s) ≈

2 26.4 s + 6.68s + 1 2

The models are compared in the following graph: 2.5

2

1.5

y(t)

Third-order model First order model Second order model 1

0.5

0

0

5

10

15

20

25

30

35

40

time,t

Figure S7.4b. Comparison of three models for step input

7.5 The integrator plus time delay model is K G(s) e −θs s In the time domain, y(t) = 0 t