{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

# 327 90766 c4 71926e 6 31340e 6 83303e 18 c5 2 2 6

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}