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

The relationship between yd and xw in eq 1 with x xw

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Unformatted text preview: 0.083 0.301 0.565 0.688 0.793 Exercise 13.11 (continued) Analysis: (continued) From the above table, ln(W0/W) = 0.793 when the instantaneous distillate reaches a mole fraction of 0.55 for benzene. Solving, W = 0.452 kmole. The composition of the cumulative distillate is given by Eq. (13-6), which becomes: ( yD )avg = 1(0.5) − Wx 0.5 − 0.452(0.125) = = 0.809 1−W 1 − 0.452 Exercise 13.12 Subject: Batch distillation of a mixture of ethanol (E) and water (W) by both simple differential (Rayleigh) distillation and in a column with 2 equilibrium stages and a reboiler (3 total stages), under conditions of a constant reflux ratio. Given: Feed containing 3.3 mol% E and 96.7 mol% W. Vapor-liquid equilibrium data from Exercise 7.29. 20 mol% of charge is distilled. Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser. Find: Plot of instantaneous vapor composition as a function of mole percent distilled for Rayleigh distillation. Maximum ethanol purity that can be obtained when using the column with the equivalent of 3 stages. Analysis: Rayleigh distillation: Because the vapor-liquid equilibrium data indicate that E is the more volatile component below 89 mol% E in the liquid, base the calculations on E. Use Eq. (13-3), with x0 = 0.033 and W/W0 = 1 - 0.2 = 0.8 (20 mol% vaporized). Thus, Eq. (13-3) becomes: x 0.033 dx dx = ln 0.8 = −0.223 or = 0.223 (1) 0.033 y − x x y−x where x in the lower integration limit is the mole fraction of E in the residue at 20 mol% distilled and y is in equilibrium with x. Equilibrium data from Exercise 7.29 that are in the lower mole fraction range are: y 0.1700 0.3891 0.4375 0.4704 x 0.0190 0.0721 0.0966 0.1238 A fit of these data to a cubic equation that goes through the origin gives: y = 10.4016 x - 89.8114 x2 + 295.358 x3 (2) Substitute Eq. (2) into Eq. (1) and, using a spreadsheet, integrate the resulting equation using the trapezoidal rule with a ∆x increment of 0.001 starting from x = 0.033 with decreasing values of x until the integral equals 0.223 or W/W0 = 0.8. As an example, using this method for the first four increments, the integral is equal to ∆x[ 0.5fx=0.40 + fx=0.39 + fx=0.38 + fx=0.37 + 0.5fx=0.36 ], where f = 1/(y-x). The complete spreadsheet is given on the following page. From it, the following result is obtained: Mole fraction of E in the residue at 20 mol% distilled = 0.0055. A plot of instantaneous vapor composition as a function of mole percent distilled is given on a following page. For x = 0.0055, Eq. (13-6) for the mole fraction of E in the cumulative distillate gives, yD avg = W0 x0 − Wx x0 − W / W0 x 0.033 − 0.8(0.0055) = = = 0143 . W0 − W 1 − W / W0 1 − 0.8 (3) Exercise 13.12 (continued) Analysis: Rayleigh Distillation (continued) x 0.033 0.032 0.031 0.030 0.029 0.028 0.027 0.026 0.025 0.024 0.023 0.022 0.021 0.020 0.019 0.018 0.017 0.016 0.015 0.014 0.013 0.012 0.011 0.010 0.009 0.008 0.007 0.006 0.005 y f=1/(y-x) Increment Cumulative of...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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