1234567898
7.1In the absence of more accurate data, use a first-order transfer function as '( ) '( ) 1 s i T s Ke Q s s −θ = τ + o ( ) (0) (124.7 120) F 0.118 540 500 gal/min i T T K q ∞ − − = = = ∆ − θ = 3:09 am – 3:05 am = 4 min
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, 4 '( ) 0.188 '( ) 5 1 s i T s e Q s s − = +
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
Process gain, (5.0) (0) (6.52 5.50) 0.336 min2
30.4 0.1 ft i h h K q − − = = = ∆ ×
a) Output at 63.2% of the total change = 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,
(0.8 0.6) 0.6 (6.145 6.07) min 0.74 min (6.18 6.07) − τ = + − = − b) 0 (0.2) (0) 5.75 5.50 ft ft 1.25 0.2 0 0.2 min min t dh h h dt = − − ≈ = = − Using Eq. 7-15, 0 t KM dh dt = τ = = min 84 . 0 25 . 1 ) 1 . 0 4 . 30 ( 347 . 0 = × ×
c) The slope of the linear fit between ti and
− ∞ − − ≡ ) 0 ( ) ( ) 0 ( ) ( 1 ln h h h t h zi i gives an
approximation of (-1/τ) according to Eq. 7-13. Using h(∞) = h(5.0) = 6 .52, the values of zi are
ti zi ti zi 0.0 0.00 1.4 -1.92 0.2 -0.28 1.6 -2.14 0.4 -0.55 1.8 -2.43 0.6 -0.82 2.0 -2.68 0.8 -1.10 3.0 -3.93 1.0 -1.37 4.0 -4.62 1.2 -1.63 5.0 -∞
Then the slope of the best-fit line, using Eq. 7-6 is
2 13 1 13 ( ) tz t z tt t S S S slope S S − = − = τ − (1)
where the datum at ti = 5.0 has been ignored.
Using definitions, 0 . 18 = t S Stt =40.4 5 . 23 − = z S Stz =−51.1 Substituting in (1), 1 1.213 − = − τ τ =0.82 min
d) 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.4 5.6 5.8 6 6.2 6.4 6.6 6.8 Experimental data Model a) Model b) Model c)
Figure S7.2. Comparison between models a), b) and c) for step response.
7.3 a) 1 1 1 ( ) ( ) 1 T s K Q s s ′ = ′ τ + 2 2 1 2 ( ) ( ) 1 T s K T s s ′ = ′ τ + 2 2 1 2 1 2 1 2 1 ( ) ( ) ( 1)( 1) ( 1) s T s K K K K e Q s s s s −τ ′ = ≈ ′ τ + τ + τ + (1)
where the approximation follows from Eq. 6-58 and the fact that τ1>τ2 as
revealed by an inspection of the data.
667 . 2 82 85 0 . 10 0 . 18 ) 0 ( ) 50 ( 1 1 1 = − − = ∆ − = q T T K 75 . 0 0 . 10 0 . 18 0 . 20 0 . 26 ) 0 ( ) 50 ( ) 0 ( ) 50 ( 1 1 2 2 2 = − − = − − = T T T T K
Let z1, z2 be the natural log of the fraction incomplete response for T1,T2,
− = − − = 8 ) ( 18 ln ) 0 ( ) 50 ( ) ( ) 50 ( ln ) ( 1 1 1 1 1 1 t T T T t T T t z − = − − = 6 ) ( 26 ln ) 0 ( ) 50 ( ) ( ) 50 ( ln ) ( 2 2 2 2 2 2 t T T T t T T t z
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. 1 3 667 . 2 ) ( ' ) ( ' 1 + = s s Q s T (2) 1 15 . 1 75 . 0 ) ( ' ) ( ' 1 2 + = s s T s T (3)
Figure S7.3a. z1 and z2 versus t
b) By means of Simulink-MATLAB, the following simulations are obtained
0 2 4 6 8 10 12 14 16 18 20 22 10 12 14 16 18 20 22 24 26 28 time T1 , T2 T 1 T 2 T1 (experimental) T 2 (experimental)
Figure S7.3b. Comparison of experimental data and models for step change -8.0 -7.0 -6.0 -5.0 -4.0 -3.0 -2.0 -1.0 0.0 0 5 10 15 20 time,t z1 ,z2
7.4 s s s s s X s G s Y 1.5 ) 1 )( 1 3 )( 1 5 ( 2 ) ( ) ( ) ( × + + + = =
Taking the inverse Laplace transform
( ) -75/8*exp(-1/5*t)+27/4*exp(-1/3*t)-3/8*exp(-t)+3
y t = (1)
a) Fraction incomplete response
− = 3 ) ( 1 ln ) (t y t z
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 3.2 2 ( ) 5.6 1 s e G s s − = +
b) 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 z(t) = -0.1791 t + 0.5734 -9.0 -8.0 -7.0 -6.0 -5.0 -4.0 -3.0 -2.0 -1.0 0.0 0 10 20 30 40 50 time,t z (t )
1 68 . 6 4 . 26 2 ) ( 2 + + ≈ s s s G
The models are compared in the following graph:
0 5 10 15 20 25 30 35 40 0 0.5 1 1.5 2 2.5 time,t y (t ) Third-order model First order model Second order model
Figure S7.4b. Comparison of three models for step input
7.5
The integrator plus time delay model is
G(s) K e s s
−θ
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. K = 0.074 5 . 13 1 =
2.- The time delay is evaluated from the intersection of the straight line and the time axis (where y = 0).
Therefore the model is G(s) = e s s 5 . 1 074 . 0 −
Figure S7.5a. Integrator plus time delay model; parameter evaluation
From Fig. E7.5, we can read these values (approximate): Time Data Model
0 0 -0.111 2 0.1 0.037 4 0.2 0.185 5 0.3 0.259 7 0.4 0.407 8 0.5 0.481 9 0.6 0.555 11 0.7 0.703 14 0.8 0.925 16.5 0.9 1.184 30 1 2.109
Table.- Output values from Fig. E7.5 and predicted values by model
A graphical comparison is shown in Fig. S7.5 b
Figure S7.5b. Comparison between experimental data and integrator plus time delay model. Slope = KM y(t) θ 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 20 25 30 Time O u tp u t Experimental data
7.6
a) 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 1 13 ) ( 1 + ≈ − s e s G s b) Smith’s method
From the graph, t20 = 3.9 , t60 = 9.6 ; using Fig 7.7 for t20/ t60 = 0.41
ζ= 1.0 , t60/τ= 2.0 , hence τ= 4.8 and τ1 = τ2 = τ= 4.8 2 ) 1 8 . 4 ( 1 ) ( + ≈ s s G Nonlinear regression
From Figure E7.5, we can read these values (approximated):
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 τ2 = 6.95 )) 1 76 . 6 )( 1 95 . 6 ( 1 ) ( + + ≈ s s s G
The models are compared in the following graph:
Time Output 0.0 0.0 2.0 0.1 4.0 0.2 5.0 0.3 7.0 0.4 8.0 0.5 9.0 0.6 11.0 0.7 14.0 0.8 17.5 0.9 30.0 1.0
0 5 10 15 20 25 30 35 40 45 50 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 time Ou tp u t
Non linear regression model First-order plus time delay model
Second order model (Smith's method)
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 6 . 1 20 = t , 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, τ =1 4.81 , τ =2 0.56
b) Overall transfer function
4 1 2 10 ( ) ( 1)( 1) s e G s s s − = τ + τ + , τ > τ1 2
( ) ps pipe G s =e−θ , 3 1 0.1min 0.5 60 p θ = × =
Assuming that the thermocouple has unit gain and no time delay
2 1 ( ) ( 1) TC G s s = τ + since τ << τ2 1 Then 3 1 10 ( ) ( 1) s HE e G s s − = τ + , so that 3 0.1 1 2 10 1 ( ) ( ) ( ) ( ) ( ) 1 1 s s HE pipe TC e G s G s G s G s e s s − − = = τ + τ + 7.8
a) To find the form of the process response, we can see that
2 ( ) ( ) ( 1) ( 1) ( 1) K K M K M Y s U s s s s s s s s = = = τ + τ + τ +
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 Y(s) = 2 1 A B C s +s +τ +s hence y(t) = -KMτ + KMt − KMτe -t/τ
In order to evaluate the parameters K and τ, important properties of the above expression are noted:
1.- For large values of time (t>>τ) , y(t) ≈ y t′( ) = KM (t-τ) 2.- For t = 0, y′(0)= −KMτ
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
obtained. Moreover, the time constant τ is obtained from the intercept in Fig. S7.8
Figure S7.8. Time domain response and parameter evaluation
7.9
For underdamped responses,
τ ζ − ζ − ζ + τ ζ − − = −ζ τ t t e KM t y t 2 2 2 / 1 sin 1 1 cos 1 ) ( (5-51)
a) At the response peaks,
2 2 / 2 1 1 cos sin 1 t dy KM e t t dt −ζ τ ζ − ζ ζ − ζ = + τ τ − ζ τ 2 2 2 / 1 1 1 sin cos 0 t e−ζ τ t t − ζ − ζ ζ − ζ − − + = τ τ τ τ Since KM ≠0 and e−ζt/τ ≠0 τ ζ − τ ζ − + ζ − τ ζ + τ ζ − τ ζ − τ ζ = t t 2 2 2 2 2 1 sin 1 1 1 cos 0 −ΚΜτ Slope = KM y(t)
π = τ ζ − =sin 1 t sinn 0 2 , t 2 1−ζ πτ =n
where n is the number of peak.
Time to the first peak,
2 1−ζ πτ = p t b) Graphical approach: Process gain, ( ) (0) 9890 9650 lb 80 hr 95 92 psig D D w w K Ps ∞ − − = = = ∆ − Overshoot = 0.333 9650 9890 9890 9970 = − − = b a 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 2 2 2 80 ( ) 2 1 0.29 0.36 1 K G s s s s s = = τ + ζτ + + + Analytical approach
The gain K doesn’t change: 80 lb hr
psig
K =
To obtain the ζ and τ values, Eqs. 5-52 and 5-53 are used:
Overshoot = 0.333 9650 9890 9890 9970 = − − = b a = exp(-ζπ/(1-ζ2)1/2) Resolving, ζ = 0.33
2 1.754 hence 0.527 hr 1 p t = πτ = τ = − ζ 2 2 2 80 ( ) 2 1 0.278 0.35 1 K G s s s s s = = τ + ζτ + + + c) 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 t
s) + 0.353e (-0.633 ts) sin(1.7973 ts)
Solve by trial and error……… ts≈ 6.9 hrs
7.10 a) '( ) 2 2 '( ) 2 1 T s K W s =τ s + ζτ + o ( ) (0) 156 140 C 0.2 80 Kg/min T T K w ∞ − − = = = ∆
From Eqs. 5-53 and 5-55,
Overshoot = 0.344 140 156 156 5 . 161 = − − = b a = exp(-ζπ (1-ζ2)1/2
By either solving the previous equation or from Figure 5.11, ζ= 0.322 (dimensionless)
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≈ 9.5 10min 5 . 3 33 − ≈
2.- From the plot of the output,
Period =
2
2 1
P= πτ
− ζ = 67 min and hence τ =10 min
Therefore the transfer function is
1 44 . 6 100 2 . 0 ) ( ' ) ( ' ) ( 2 + + = = s s s W s T s G
b) After an initial period of oscillation, the ramp input yields a ramp output with slope equal to KB. The MATLAB simulation is shown below:
0 10 20 30 40 50 60 70 80 90 100 140 142 144 146 148 150 152 154 156 158 160 time O u tp u t
We know the response will come from product of G(s) and Xramp = B/s2 Then ( ) 2 2 2 ( 2 1) KB Y s s s s = τ + ζτ +
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:
3 4 1 2 2 2 2 2 2 2 ( ) ( 2 1) 2 1 s KB Y s s s s s s s s α + α α α = = + + τ + ζτ + τ + ζτ +
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
2 2 2 2 3 2
1 ( 2 1) 2( 2 1) 3 4
KB= α τs s + ζτ + + α τs s + ζτ + + αs s + α s
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
/ 5 / 5
( ) t ( 1) [1 t ]6 ( 1)
y k =e−∆ y k− + −e−∆ u k−
b) Replacing τ by 5, and K by 6 in Eq. 7-32
( ) (1 ) ( 1) 6 ( 1)
5 5
t t
y k = −∆ y k− +∆ u k−
In the integrated results tabulated below, the values for ∆t = 0.1 are shown only at integer values of t, for comparison.
t y(k) (exact) y(k) (1t=1) y(k) (1t=0.1) 0 3 3 3 1 2.456 2.400 2.451 2 5.274 5.520 5.296 3 6.493 6.816 6.522 4 6.404 6.653 6.427 5 5.243 5.322 5.251 6 4.293 4.258 4.290 7 3.514 3.408 3.505 8 2.877 2.725 2.864 9 2.356 2.180 2.340 10 1.929 1.744 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 b , use the given first order model to minimize 1 10 2 1 1 1 ( ( ) ( 1) ( 1)) n J y k a y k b x k = ′ =
∑
− − − − 0 ) 1 ( ))( 1 ( ) 1 ( ) ( ( 2 1 1 10 1 1 = − − − − − ′ − = ′ ∂ ∂∑
= k y k x b k y a k y a J n 10 1 1 1 1 2( ( ) ( 1) ( 1))( ( 1)) 0 n J y k a y k b x k x k b = ∂ = − ′ − − − − − = ∂∑
Solving simultaneously for a1′ and b gives 1
10 10 1 1 1 1 10 2 1 ( ) ( 1) ( 1) ( 1) ( 1) n n n y k y k b y k x k a y k = = = − − − − ′ = −
∑
∑
∑
10 10 10 10 2 1 1 1 1 1 10 10 10 2 2 2 1 1 1 ( 1) ( ) ( 1) ( 1) ( 1) ( 1) ( ) ( 1) ( 1) ( 1) ( 1) n n n n n n n x k y k y k y k x k y k y k b x k y k y k x k = = = = = = = − − − − − − = − − − − − ∑
∑
∑
∑
∑
∑
∑
Using the given data, 212 . 35 ) ( ) 1 ( 10 1 = −
∑
= k y k x n , ( 1) ( ) 188.749 10 1 = −∑
= n k y k y 14 ) 1 ( 10 1 2 = −∑
= n k x , ( 1) 198.112 10 1 2 = −∑
= n k y 409 . 24 ) 1 ( ) 1 ( 10 1 = − −∑
= n k x k ySubstituting into expressions for a1′ and b1gives
1
a′= 0.8187 , b = 1.0876 1
Fitted model is y(k+1)=0.8187y(k)+1.0876x(k)
or y(k)=0.8187y(k−1)+1.0876x(k−1) (1) Let the first-order continuous transfer function be
( )
( ) 1
Y s K
X s = τ +s
From Eq. 7-34, the discrete model should be
/ /
( ) t ( 1) [1 t ] ( 1)
y k =e−∆ τy k− + −e−∆ τ Kx k− (2)
Comparing Eqs. 1 and 2, for ∆t=1, gives
τ = 5 and K = 6
0 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 time,t y (t ) actual data fitted model
Figure S7.12. Response of the fitted model and the actual data
7.13
To fit a first-order discrete model ) 1 ( ) 1 ( ) (k =a1′y k− +b1x k− y
Using the expressions for a1′ and b1 from the solutions to Exercise 7.12,
with the data in Table E7.12 gives 918
. 0
1′ =
a , b1 =0.133
Using the graphical (tangent) method of Fig.7.5 . 1
=
K , θ =0.68, and τ =6.8
The response to unit step change for the first-order model given by
0.68 6.8 1 s e s − + is 8 . 6 / ) 68 . 0 ( 1 ) ( = − −t− e t y
Figure S7.13- Response of the fitted model, actual data and graphical method 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 0 2 4 time,t 6 8 10 y (t ) actual data fitted model graphical method
