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<0 y(t)= K (t-θ) t≥0 Thus a straight line tangent to the point of inflection will approximate the step response. Two parameters must be found: K and θ (See Fig. S7.5 a) 1.- The process gain K is found by calculating the slope of the straight line. 1 K= = 0.074 13.5 2.- The time delay is evaluated from the intersection of the straight line and the time axis (where y = 0). θ = 1.5 7-6
Therefore the model is G(s) =
0.074 −1.5 s e s
y(t)
Slope = KM
θ
Figure S7.5a. Integrator plus time delay model; parameter evaluation
From Fig. E7.5, we can read these values (approximate): Time 0 2 4 5 7 8 9 11 14 16.5 30
Data 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Model -0.111 0.037 0.185 0.259 0.407 0.481 0.555 0.703 0.925 1.184 2.109
Table.- Output values from Fig. E7.5 and predicted values by model
A graphical comparison is shown in Fig. S7.5 b 1 0.9 0.8
Output
0.7 0.6 0.5 0.4
Experimental data Integrator plus time delay model
0.3 0.2 0.1 0 0
5
10
15
20
25
30
Time
Figure S7.5b. Comparison between experimental data and integrator plus time delay model.
7-7
7.6
a)
b)
Drawing a tangent at the inflection point which is roughly at t ≈5, the intersection with y(t)=0 line is at t ≈1 and with the y(t)=1 line at t ≈14. Hence θ =1 , τ = 14−1=13 e−s G1 ( s ) ≈ 13s + 1 Smith’s method From the graph, t20 = 3.9 , t60 = 9.6 ; using Fig 7.7 for t20/ t60 = 0.41 ζ = 1.0 , G (s) ≈
t60/τ= 2.0 ,
hence τ = 4.8 and τ1 = τ2 = τ = 4.8
1 (4.8s + 1) 2
Nonlinear regression From Figure E7.5, we can read these values (approximated): Time 0.0 2.0 4.0 5.0 7.0 8.0 9.0 11.0 14.0 17.5 30.0
Output 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Table.- Output values from Figure E7.5
In accounting for Eq. 5-48, the time constants were selected to minimize the sum of the squares of the errors between data and model predictions. Use Excel Solver for this Optimization problem: τ1 =6.76 and G (s) ≈
τ2 = 6.95
1 (6.95s + 1)(6.76s + 1))
The models are compared in the following graph:
7-8
1
0.9
0.8
0.7
Output
0.6
0.5
0.4
0.3
0.2 Non linear regression model First-order plus time delay model Second order model (Smith's method)
0.1
0
0
5
10
15
20
25 time
30
35
40
45
50
Figure S7.6. Comparison of three models for unit step input
7.7
a)
From the graph, time delay θ = 4.0 min Using Smith’s method, from the graph, t20 + θ ≈ 5.6 , t60 + θ ≈ 9.1 t20 = 1.6 , t60 = 5.1 , t20 / t60 = 1.6 / 5.1 = 0.314 From Fig.7.7 , ζ = 1.63 , t60 / τ = 3.10 , τ = 1.645 Using Eqs. 5-45, 5-46, τ1 = 4.81 , τ2 = 0.56
b)
Overall transfer function 10e −4 s G (s) = , τ1 > τ2 (τ1s + 1)(τ2 s + 1) Assuming plug-flow in the pipe with constant-velocity,
7-9
G pipe ( s ) = e
−θ p s
, θp =
3 1 × = 0.1min 0.5 60
Assuming that the thermocouple has unit gain and no time delay GTC ( s ) =
1 (τ2 s + 1)
since τ2 << τ1
Then 10e −3 s GHE ( s ) = , so that (τ1s + 1)
10e −3 s −0.1s 1 G ( s ) = GHE ( s )G pipe ( s )GTC ( s ) = (e ) τ1s + 1 τ2 s + 1
7.8
a)
To find the form of the process response, we can see that Y (s) =
K K M K M U ( s) = = s (τs + 1) s (τs + 1) s (τs + 1) s 2
Hence the response of this system is similar to a first-order system with a ramp input: the ramp input yields a ramp output that will ultimately cause some process component to saturate. b)
By applying partial fraction expansion technique, the domain response for this system is A B C + 2+ hence y(t) = -KMτ + KMt − KMτe-t/τ s s τs + 1 In order to evaluate the parameters K and τ, important properties of the above expression are noted:
Y(s) =
1.- For large values of time (t>>τ) , 2.- For t = 0, y′(0) = −KMτ
y(t) ≈ y′(t ) = KM (t-τ)
These equations imply that after an initial transient period, the ramp input yields a ramp output with slope equal to KM. That way, the gain K is
7-10
obtained. Moreover, the time constant τ is obtained from the intercept in Fig. S7.8
y(t) Slope = KM
−ΚΜτ
Figure S7.8. Time domain response and parameter evaluation
7.9
For underdamped responses, 1 − ζ2 y (t ) = KM 1 − e − ζt / τ cos τ
a)
1 − ζ2 ζ t + sin τ 1 − ζ2
t
At the response peaks, 1 − ζ2 dy ζ = KM e −ζ t / τ cos dt τ τ
1− ζ2 ζ t+ sin τ 1 − ζ2
1− ζ2 1− ζ2 −e −ζt / τ − sin τ τ
ζ 1 − ζ2 t + cos τ τ
t t = 0
Since KM ≠ 0 and e − ζt / τ ≠ 0 2 ζ ζ 1− ζ 0 = − cos τ τ τ
ζ2 1 − ζ2 t + + τ 1 − ζ2 τ
7-11
1 − ζ2 sin τ
t
(5-51)
1− ζ2 0 = sin t = sin nπ , τ where n is the number of peak. Time to the first peak,
b)
tp =
t= n
πτ 1 − ζ2
πτ 1 − ζ2
Graphical approach: Process gain,
K=
wD (∞) − wD (0) 9890 − 9650 lb = = 80 hr ∆Ps 95 − 92 psig
Overshoot =
a 9970 − 9890 = = 0.333 b 9890 − 9650
From Fig. 5.11,
ζ ≈ 0.33
tp can be calculated by interpolating Fig. 5.8 For ζ ≈ 0.33 , tp ≈ 3.25 τ Since tp is known to be 1.75 hr , τ = 0.54
G (s) =
K 80 = 2 τ s + 2ζτs + 1 0.29s + 0.36 s + 1 2 2
Analytical approach The gain K doesn’t change: K = 80
lb hr
psig To obtain the ζ and τ values, Eqs. 5-52 and 5-53 are used: Overshoot =
a 9970 − 9890 = = 0.333 = exp(-ζπ/(1-ζ2)1/2) b 9890 − 9650
Resolving, ζ = 0.33
7-12
tp =
πτ 1− ζ2
G (s) =
c)
= 1.754
hence
τ = 0.527 hr
K 80 = 2 τ s + 2ζτs + 1 0.278s + 0.35s + 1 2 2
Graphical approach From Fig. 5.8, ts/τ = 13 so ts = 2 hr (very crude estimation) Analytical approach From settling time definition, y = ± 5% KM
so
9395.5 < y < 10384.5
(KM ± 5% KM) = KM[ 1-e(-0.633)[cos(1.793ts)+0.353sin(1.793ts)]] 1 ± 0.05 = 1 – e(0.633 ts) cos(1.793 ts) + 0.353e (-0.633 ts) sin(1.7973 ts)
Solve by trial and error……………………
ts ≈ 6.9 hrs
7.10
a)
T '( s ) K = 2 2 W '( s ) τ s + 2ζτ + 1
K=
o T (∞) − T (0) 156 − 140 C = = 0.2 ∆w 80 Kg/min
From Eqs. 5-53 and 5-55, a 161.5 − 156 = = 0.344 = exp(-ζπ (1-ζ2)1/2 b 156 − 140 By either solving the previous equation or from Figure 5.11, ζ= 0.322 (dimensionless)
Overshoot =
7-13
There are two alternatives to find the time constant τ : 1.- From the time of the first peak, tp ≈ 33 min. One could find an expression for tp by differentiating Eq. 5-51 and solving for t at the first zero. However, a method that should work (within required engineering accuracy) is to interpolate a value of ζ=0.35 in Figure 5.8 and note that tp/τ ≈ 3 Hence τ ≈
33 ≈ 9.5 − 10 min 3 .5
2.- From the plot of the output, Period = P =
2πτ 1− ζ2
= 67 min and hence τ =10 min
Therefore the transfer function is G (s) =
After an initial period of oscillation, the ramp input yields a ramp output with slope equal to KB. The MATLAB simulation is shown below:
160
158
156
154
152 Output
b)
T ' (s) 0.2 = 2 W ' ( s ) 100 s + 6.44s + 1
150
148
146
144
142
140
0
10
20
30
40
50 time
60
70
80
Figure S7.10. Process output for a ramp input
7-14
90
100
We know the response will come from product of G(s) and Xramp = B/s2 KB Then Y ( s ) = 2 2 2 s (τ s + 2ζτs + 1) From the ramp response of a first-order system we know that the response will asymptotically approach a straight line with slope = KB. Need to find the intercept. By using partial fraction expansion: Y (s) =
α s + α4 KB α α = 1 + 22 + 2 2 3 s (τ s + 2ζτs + 1) s s τ s + 2ζτs + 1 2
2 2
Again by analogy to the first-order system, we need to find only α1 and α2. Multiply both sides by s2 and let s→ 0, α2 = KB (as expected) Can’t use Heaviside for α1, so equate coefficients KB = α1s (τ2 s 2 + 2ζτs + 1) + α 2 (τ2 s 2 + 2ζτs + 1) + α 3 s 3 + α 4 s 2
We can get an expression for α1 in terms of α2 by looking at terms containing s. s: 0 = α1+α22ζτ
→ α1 = -KB2ζτ
and we see that the intercept with the time axis is at t = 2ζτ. Finally, presuming that there must be some oscillatory behavior in the response, we sketch the probable response (See Fig. S7.10)
7.11
a)
Replacing τ by 5, and K by 6 in Eq. 7-34 y (k ) = e−∆t / 5 y (k − 1) + [1 − e−∆t / 5 ]6u (k − 1)
b)
Replacing τ by 5, and K by 6 in Eq. 7-32 y (k ) = (1 −
∆t ∆t ) y (k − 1) + 6u (k − 1) 5 5
In the integrated results tabulated below, the values for ∆t = 0.1 are shown only at integer values of t, for comparison.
7-15
t 0 1 2 3 4 5 6 7 8 9 10
y(k) (exact) 3 2.456 5.274 6.493 6.404 5.243 4.293 3.514 2.877 2.356 1.929
y(k) (1t=1) 3 2.400 5.520 6.816 6.653 5.322 4.258 3.408 2.725 2.180 1.744
y(k) (1t=0.1) 3 2.451 5.296 6.522 6.427 5.251 4.290 3.505 2.864 2.340 1.912
Table S7.11. Integrated results for the first order differential equation
Thus ∆t = 0.1 does improve the finite difference model bringing it closer to the exact model.
7.12 To find a1′ and b1 , use the given first order model to minimize 10
J = ∑ ( y (k ) −a1′ y (k − 1) − b1 x(k − 1)) 2 n =1
10 ∂J = ∑ 2( y (k ) −a1′ y (k − 1) − b1 x(k − 1))(− y (k − 1) = 0 ∂a1′ n =1 10 ∂J = ∑ 2( y (k ) −a1′ y (k − 1) − b1 x(k − 1))(− x(k − 1)) = 0 ∂b1 n =1 Solving simultaneously for a1′ and b1 gives
10
a1′ =
10
∑ y (k )y (k − 1) − b1 ∑ y (k − 1)x(k − 1) n =1
n =1
10
∑ y (k − 1)
2
n =1
10
b1 =
10
10
10
∑ x(k − 1) y(k )∑ y(k − 1)2 − ∑ y (k − 1)x(k − 1)∑ y(k − 1) y(k ) n =1
n =1
n =1
n =1
x(k − 1) ∑ y (k − 1) − ∑ y (k − 1)x(k − 1) ∑ n =1 n =1 n =1 10
10
10
2
2
7-16
2
Using the given data, 10
10
∑ x(k − 1) y(k ) = 35.212 ,
∑ y(k − 1) y(k ) = 188.749
n =1 10
n =1
10
∑ x(k − 1) 2 = 14
∑ y(k − 1)
,
n =1
2
= 198.112
n =1
10
∑ y (k − 1) x(k − 1) = 24.409 n =1
Substituting into expressions for a1′ and b1 gives a1′ = 0.8187
,
b1 = 1.0876
Fitted model is
y (k + 1) = 0.8187 y (k ) + 1.0876 x(k )
or
y (k ) = 0.8187 y (k − 1) + 1.0876 x(k − 1)
(1)
Let the first-order continuous transfer function be Y ( s) K = X ( s ) τs + 1 From Eq. 7-34, the discrete model should be y (k ) = e −∆t / τ y (k − 1) + [1 − e −∆t / τ ]Kx(k − 1) Comparing Eqs. 1 and 2, for ∆t=1, gives τ=5
and K = 6
Hence the continuous transfer function is 6/(5s+1)
7-17
(2)
8 actual data fitted model 7
6
y(t)
5
4
3
2
0
1
2
3
4
5
6
7
8
9
10
time,t
Figure S7.12. Response of the fitted model and the actual data
7.13
To fit a first-order discrete model
y (k ) = a1′ y (k − 1) + b1 x(k − 1) Using the expressions for a1′ and b1 from the solutions to Exercise 7.12, with the data in Table E7.12 gives
a1′ = 0.918
, b1 = 0.133
Using the graphical (tangent) method of Fig.7.5 . K = 1 , θ = 0.68 , and τ = 6.8
The response to unit step change for the first-order model given by e −0.68 s 6.8s + 1
is
y (t ) = 1 − e −( t −0.68) / 6.8
7-18
y(t)
1 0,9 0,8 0,7 0,6 0,5 0,4 0,3 0,2 0,1 0
actual data fitted model graphical method
0
2
4 time,t 6
8
10
Figure S7.13- Response of the fitted model, actual data and graphical method
7-19
