This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1234567898 7.1 In the absence of more accurate data, use a firstorder 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(6.525.50) = 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. 71 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. 715,
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. 713.
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 bestfit line, using Eq. 76 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 72 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. 658 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, 73 T (50) − T1 (t ) 18 − T1 (t ) z1 (t ) = ln 1 = ln 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 bestfit 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 SimulinkMATLAB, 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 74 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 75 (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) Thirdorder 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 (tθ)
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
76 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. 77 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. 548, 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: 78 1 0.9 0.8 0.7 Output 0.6 0.5 0.4 0.3 0.2
Non linear regression model
Firstorder 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. 545, 546, τ1 = 4.81 , τ2 = 0.56 b) Overall transfer function
10e −4 s
G (s) =
, τ1 > τ2
(τ1s + 1)(τ2 s + 1)
Assuming plugflow in the pipe with constantvelocity, 79 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 firstorder 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 − KMτet/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 (tτ) 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 710 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 711 1 − ζ2 sin τ t (551) 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. 552 and 553 are used:
Overshoot = a 9970 − 9890
=
= 0.333 = exp(ζπ/(1ζ2)1/2)
b 9890 − 9650 Resolving, ζ = 0.33 712 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[ 1e(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. 553 and 555,
a 161.5 − 156
=
= 0.344 = exp(ζπ (1ζ2)1/2
b
156 − 140
By either solving the previous equation or from Figure 5.11, ζ= 0.322
(dimensionless) Overshoot = 713 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. 551 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 714 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 firstorder 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 firstorder 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. 734
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. 732
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. 715 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 2 716 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 firstorder continuous transfer function be
Y ( s)
K
=
X ( s ) τs + 1
From Eq. 734, 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) 717 (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 firstorder 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 firstorder model given by
e −0.68 s
6.8s + 1 is y (t ) = 1 − e −( t −0.68) / 6.8 718 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 719 ...
View
Full
Document
This note was uploaded on 04/10/2008 for the course CHE 242 taught by Professor Cummings during the Spring '08 term at Vanderbilt.
 Spring '08
 Cummings

Click to edit the document details