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

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Unformatted text preview: readily done by setting up a table in W, and working in increments from W = 100 moles down to a low value of W. Exercise 13.5 (continued) Analysis: (continued) Case 2: For 40 mol% distilled, W = 100 - 40 = 60 moles. Substitute this value into Eq. (1) ln 100 100 0.60 1− x = ln = 0.5108 = 0.6993 ln + 2.43 ln W x 60 0.40 This is a nonlinear equation in x. Solving with a spreadsheet or a nonlinear equation solver, x = 0.50. Substituting this value for x and W = 60 moles into Eq. (2) gives: ( yD )avg = 60 − Wx 60 − 60(0.5) = = 0.75 100 − W 100 − 60 Case 3: The original charge contains 60 moles of B per 100 moles of charge. If 60% of the benzene is distilled, then 36 moles of benzene are in the cumulative distillate and 60 - 36 = 24 moles of benzene are in the residue. To find the number of moles of distillate = 100 - W, and the residue and cumulative distillate benzene mole fractions, use a spreadsheet. Solve Eq. (1) for W using a series of decreasing values of x, starting from x0 = 0.6. For each value of x, compute the moles of benzene in the residue from Wx, until a value of 24 moles is obtained. Then compute the cumulative distillate mole fraction from Eq. (2). The spreadsheet result is: x W Wx 0.60 100.00 60.00 0.58 89.87 52.13 0.56 81.03 45.38 0.54 73.25 39.55 0.52 66.36 34.51 0.50 60.24 30.12 0.48 54.77 26.29 0.46 49.86 22.94 0.44 45.44 19.99 By interpolation, for Wx = 24 moles, x = 0.466 and W = 51.47 moles. From Eq. (2), ( yD )avg = 60 − Wx 60 − 24 = = 0.742 100 − W 100 − 51.47 Exercise 13.6 Subject: Simple differential batch (Rayleigh) distillation of a mixture of phenol and water. Given: Mixture of 15 mol% phenol (P) and 85 mol% water (W). Distillation at 260 torr at the conditions below. Vapor-liquid equilibrium data given in wt% water Assumptions: Perfect mixing in the still. Exiting vapor in equilibrium with the liquid. Find: Fraction of original charge that remains as residue and the residue concentration when the cumulative distillate contains 98 mol% water. Analysis: Make the calculations in terms of water because it is the more volatile component. First convert the given equilibrium data from wt% to mol% using molecular weights of 18.016 for water and 94.108 for phenol. Using a spreadsheet, the results are: wt% W wt% P wt% W wt% P x, water y, water in liquid in liquid in vapor in vapor 1.54 4.95 6.87 7.73 19.63 28.44 39.73 82.99 89.95 93.38 95.74 98.46 95.05 93.13 92.27 80.37 71.56 60.27 17.01 10.05 6.62 4.26 41.10 79.72 82.79 84.45 89.91 91.05 91.15 91.86 92.77 94.19 95.64 58.90 20.28 17.21 15.55 10.09 8.95 8.85 8.14 7.23 5.81 4.36 0.0755 0.2139 0.2782 0.3044 0.5606 0.6749 0.7749 0.9622 0.9791 0.9866 0.9916 0.7847 0.9536 0.9617 0.9659 0.9790 0.9815 0.9818 0.9833 0.9853 0.9883 0.9913 For 98 mol% W in the cumulative distillate, Eq. (13-6) becomes: yD avg = 0.98 = From Eq. (13-3), W0 x0 − Wx = W0 − W x x0 x0 − W x W0 W 1− W0 dx W = ln y−x W0 or 0.85 − = 0.85 x W x W0 W 1− W0 dx W = ln 0 y−x W (1) (2) Exercise 13.6 (continued) Analysis: Eqs. (1) and (2)...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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