ch07 - 1234567898 7.1 In the absence of more accurate data,...

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Unformatted text preview: 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 ) 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 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 ...
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This note was uploaded on 04/10/2008 for the course CHE 242 taught by Professor Cummings during the Spring '08 term at Vanderbilt.

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