Separation Process Principles- 2n - Seader & Henley - Solutions Manual

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Unformatted text preview: 0 0.5237 0.8866 0.9943 0.9999873 1.0000 x2 5.000 4.3911 4.0569 4.0032 4.0000 4.0000 (b) Starting guess is x1 = 4, x2 = 5 The Newton-Raphson method fails. After only one iteration, x1 = -6.2135, x2 = 10.7719 (c) Starting guess is x1 = 1, x2 = 1 The Newton-Raphson method fails. After only one iteration, x1 = -15.4910, x2 = 24.9880 (d) Starting guess is x1 = 8, x2 = 1 The Newton-Raphson method fails. After only one iteration, x1 = 5.1320, x2 = -0.4449 Note that by a homotopy continuation method, two roots in the positive domain are obtained: x1 = 1, x2 = 4 x1 = 4.0715, x2 = 0.65027 If the Newton-Raphson method is damped by using a line search, the method fails less often. Exercise 10.10 Subject: Solving a set of two nonlinear equations by the Newton-Raphson method. Given: Two nonlinear equations in x1 and x2: x2 − x1 = 0 2 1 1 1− + exp 1 − 1 − 2 x1 + x2 = 0 4π 4π sin πx1 x2 − exp 2 x1 Find: Solutions for three different starting guesses Analysis: (a) Starting guess is x1 = 0.4, x2 = 0.9 The Newton-Raphson method converges in 4 iterations: Iteration x1 x2 Start 0.4000 0.9000 1 0.1415 0.8394 2 0.2806 0.9073 3 0.2984 0.9026 4 0.2994 0.9030 (b) Starting guess is x1 = 0.6, x2 = 0.9 The Newton-Raphson method converges to a different solution in just 3 iterations. Iteration x1 x2 Start 0.6000 0.9000 1 0.4901 1.0235 2 0.4999 1.0002 3 0.5000 1.0000 (c) Starting guess is x1 = 1, x2 = 1 The Newton-Raphson method converges with some oscillations in 14 iterations. Iteration x1 x2 Iteration x1 x2 Start 1.0000 1.0000 8 1.0042 0.4451 1 0.9091 0.6914 9 0.7160 1.2837 2 1.0720 0.2987 10 0.6514 0.9886 3 0.9727 0.5500 11 0.5135 1.0452 4 1.1633 -0.0252 12 0.5045 1.0008 5 1.0511 0.3321 13 0.5001 1.0001 6 0.9512 0.6053 14 0.5000 1.0000 7 1.0876 0.2072 Exercise 10.11 Subject: Calculations for the first iteration of the Bubble-Point method Given: Feed of 1000 kmol/h of a bubble-point mixture of 60 mol% methanol (M), 20 mol% ethanol (E), and 20 mol% n-propanol (P) is sent to the middle stage of a column with a total condenser, 3 equilibrium stages in the column, and a partial reboiler, and operating at 1 atm. The distillate rate = 600 kmol/h and the reflux rate = 2,000 kmol/h. Assumptions: Ideal solutions with Raoult's law K-values. Constant molar overflow. Find: Liquid-phase component mole fractions and stage temperatures from the first iteration of the bubble-point method. Analysis: By material balance, the bottoms flow rate = 1000 - 600 = 400 kmol/h. The liquid rate above the feed = 2000 kmol/h. The liquid rate below the feed = 3000 kmol/h. The boilup rate over the entire column = 3000 - 400 = 2600 kmol/h. Based on the distillate rate, methanol will be mainly in the distillate, with ethanol and n-propanol mainly in the bottoms. The boiling points of these three components are, respectively, 64.7oC, 78.4oC, and 97.8oC. As a first approximation, assume a distillate temperature of 64.7oC and a bottoms temperature of (78.4 + 97.8)/2 = 88.1oC. Therefore, the initial guesses, assuming a linear temperature pro...
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