Solutions_F11_ECH157_HW3

# Solutions_F11_ECH157_HW3 - d If r → ∞ there will a...

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Unformatted text preview: d) If r → ∞ , there will a large amount of mixing between the two tanks as a result of the very high internal circulation. Thus the process acts like ci q c2 q Total Volume = V1 + V2 Model : dc 2 = q (c i − c 2 ) dt c1 = c2 (complete internal mixing) (V1 +V2) Degrees of freedom analysis is same as part b) Homework Problem Set #3: Solution Key Problem 1: 5.19 a) For the original system, dh1 h = Cqi − 1 dt R1 dh h h A2 2 = 1 − 2 dt R1 R2 A1 where A1 = A2 = π(3)2/4 = 7.07 ft2 ft 3 /min gpm h 2.5 ft = 0.187 3 R1 = R2 = 1 = Cqi 0.1337 × 100 ft /min C = 0.1337 5-23 (9) (10) Using deviation variables and taking Laplace transforms, H 1′ ( s ) = Qi′ ( s ) C = CR1 0.025 = A1 R1 s + 1 1.32 s + 1 1 R1 ′ H 2 ( s) 1 / R1 R2 / R1 1 = = = 1 ′ H 1 (s) A2 R2 s + 1 1.32 s + 1 A2 s + R2 ′ H 2 ( s) 0.025 = Qi′( s ) (1.32s + 1) 2 A1 s + For step change in qi of magnitude M, h1′max = 0.025M ′ h2 max = 0.025M since the second-order transfer function 0.025 is critically damped (ζ=1), not underdamped (1.32 s + 1) 2 2.5 ft Hence Mmax = = 100 gpm 0.025 ft/gpm For the modified system, A dh h = Cq i − dt R A = π(4) 2 / 4 = 12.6 ft 2 V = V1 + V2 = 2 × 7.07ft 2 × 5ft = 70.7ft3 hmax = V/A = 5.62 ft R= h Cq i H ′( s ) = Qi′ ( s ) = 0.5 × 5.62 ft = 0.21 3 0.1337 × 100 ft /min C As + 1 R = CR 0.0281 = ARs + 1 2.64 s + 1 ′ hmax = 0.0281M 2.81 ft Mmax = = 100 gpm 0.0281 ft/gpm 5-24 Hence, both systems can handle the same maximum step disturbance in qi. b) For step change of magnitude M, Qi′( s ) = M s For original system, ′ Q2 ( s ) = 1 1 0.025 M ′ H 2 ( s) = R2 0.187 (1.32 s + 1) 2 s 1 1.32 1.32 = 0.134M − − 2 s (1.32s + 1) (1.32s + 1) t −t / 1.32 q ′ (t ) = 0.134 M 1 − 1 + e 2 1.32 For modified system, Q ′( s ) = 1 1 0.0281 M 2.64 1 H ′( s ) = = 0.134 M − R 0.21 (2.64 s + 1) s s 2.64 s + 1 [ q ′(t ) = 0.134 M 1 − e −t / 2.64 ′ Original system provides better damping since q 2 (t ) < q ′(t ) for t < 3.4. Problem 2: 5.20 a) Caustic balance for the tank, ρV dC = w1c1 + w2 c 2 − wc dt Since V is constant, w = w1 + w2 = 10 lb/min For constant flows, ′ ρVsC ′( s ) = w1C1′ ( s ) + w2 C 2 ( s ) − wC ′( s ) w1 C ′( s ) 5 0.5 = = = ′ C1 ( s ) ρVs + w (70)(7) s + 10 49s + 1 5-25 ′ C m (s) K , = C ′( s ) τs + 1 K = (3-0)/3 = 1 τ ≈ 6 sec = 0.1 min , (from the graph) ′ C m ( s) 1 0.5 0.5 = = C1′ ( s ) (0.1s + 1) (49s + 1) (0.1s + 1)(49s + 1) b) 3 s C1′ ( s ) = 1.5 s (0.1s + 1)(49 s + 1) 1 c ′ (t ) = 1.51 + (0.1e −t / 0.1 − 49e −t / 49 ) m (49 − 0.1) ′ C m ( s) = c) ′ C m ( s) = 0.5 3 1.5 = (49 s + 1) s s (49 s + 1) ( c ′ (t ) = 1.5 1 − e − t / 49 m The responses in b) and c) are nearly the same. Hence the dynamics of the conductivity cell are negligible. 1.5 1 Cm'(t) d) ) 0.5 Part b) Part c) 0 0 20 40 60 80 100 time 120 140 160 Fig S5.20. Step responses for parts b) and c) 5-26 180 200 Two possibilities: 1. K1<0 and K1τ + K2 >0 2. K1 > 0 and K1τ + K2 < 0 e) Gain is negative if K1 < 0 Then zero is RHP if K1τ + K2 > 0 This is the only possibility. f) Constant term and e-t/τ term. g) If input is M/s, the output will contain a t term, that is, it is not bounded. Problem 3: 6.7 a) 2 s −3 −3 2 Q ′( s ) = P ′( s ) = 20 s + 1 20 s + 1 s p ′(t ) = (4 − 2) S (t ) ,P ′( s ) = Q ′(t ) = −6(1 − e −t / 20 ) b) ′ R ′( s ) + Q ′( s ) = Pm ( s ) r ′(t ) + q ′(t ) = p ′ (t ) = p m (t ) − p m (0) m r ′(t ) = p m (t ) − 12 + 6(1 − e − t / 20 ) K= r ′(t = ∞) 18 − 12 + 6(1 − 0) = =6 p (t = ∞) − p (t = 0) 4−2 Overshoot, OS = r ′(t = 15) − r ′(t = ∞) 27 − 12 + 6(1 − e −15 / 20 ) − 12 = = 0.514 r ′(t = ∞) 12 6-7 − πζ OS = exp 1− ζ2 = 0.514 , ζ = 0 .2 Period, T, for r ′(t ) is equal to the period for pm(t) since e-t/20 decreases monotonically. Thus, and c) T = 50 − 15 = 35 τ= T 1 − ζ 2 = 5 .4 6 2π ′ Pm ( s ) K K′ = 22 + P ′( s ) τ s + 2ζτs + 1 τ′s + 1 (K ′τ )s = 2 d) + ( Kτ′ + 2 K ′ζτ) s + ( K + K ′) (τ s 2 + 2ζτs + 1)(τ′s + 1) 2 2 Overall process gain = ′ Pm ( s ) P ′( s ) = K + K′ = 6 −3 = 3 s =0 % psi 6.8 a) Transfer Function for blending tank: Gbt ( s ) = K bt τ bt s + 1 where K bt = τ bt = qin ≠1 ∑ qi 2m 3 = 2 min 1m 3 / min Transfer Function for transfer line Gtl ( s ) = K tl (τ tl s + 1)5 where K tl = 1 τ tl = 6-8 0.1m 3 = 0.02 min 5 × 1m 3 / min t 1 t 2 1 t 3 1 t 4 −t / 6 c5 (t ) = 0.60 − 0.15 1 − e 1 + + + + 6 2! 6 3! 6 4! 6 Using Simulink, b) 0.6 c5 c4 0.58 c3 c2 Concentration 0.56 c1 0.54 0.52 0.5 0.48 0.46 0.44 0 5 10 15 20 25 time 30 35 40 45 50 Figure S6.10. Concentration step responses of the stirred tank. The value of the expression for c5(t) verifies the simulation results above: 52 53 54 c5 (30) = 0.60 − 0.15 1 − e −5 1 + 5 + + + = 0.5161 2! 3! 4! Problem 4: 6.11 a) Y (s) = − τa s + 1 E A B C = + 2+ 2 τ1 s + 1 s ss τ1 s + 1 We only need to calculate the coefficients A and B because Ce − t / τ1 → 0 for t >> τ1. However, there is a repeated pole at zero. 6-13 E (− τ a s + 1) B = lim =E s →0 τ1 s + 1 Now look at E (− τ a s + 1) = As (τ1 s + 1) + B (τ1 s + 1) + Cs 2 − Eτ a s + E = Aτ1 s 2 + As + Bτ1 s + B + Cs 2 Equate coefficients on s: − Eτ a = A + Bτ1 A = − E ( τ a + τ1 ) Then the long-time solution is y (t ) ≈ Et − E (τ a + τ1 ) Plotting (τa+τ1) y y(t)=Et =Et-E(τa+τ1) actual response -E(τa+τ1) time b) For a LHP zero, the apparent lag would be τ1 − τa c) For no zero, the apparent lag would be τ1 6-14 Practice Problems ...
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