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

11 00321 exercise 132 subject simple differential

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Unformatted text preview: when 50% of the C7 has been distilled. Weight percent of the original charge distilled. Analysis: The vapor-liquid equilibrium data indicate that C7 is more volatile than T. Therefore, make the calculations in terms of C7. Because the initial concentration of C7 in the charge is low (2 mol%), assume that the distillation takes place at constant relative volatility, α = αC7-T. Alternatively, the y - x data in the low concentration region could be fitted to a quadratic equation and used with Eq. (13-2). The lowest concentration in the data table is 2.5 mol% C7, where x = 0.025 and y = 0.048. Therefore, for the binary system, using Eqs. (2-19) and (2-21), KC7 = y/x = 0.048/0.025 = 1.92 KT = (1-y)/(1-x) = (1 - 0.048)/(1 - 0.025) = 0.976 αC7-T = KC7/KT = 1.92/0.976 = 1.97 A similar calculation at the next point in the data table, x = 0.062 and y = 0.107, gives αC7-T = 1.81. Therefore the relative volatility increases as x drops to zero. Take αC7-T = 2 for the region of interest. From Eq. (13-5), ln W0 1 x 1− x = ln 0 + α ln W α −1 x 1 − x0 = ln 0.025 1− x + 2 ln x 1 − 0.025 (1) Taking a basis of W0 = 1 lbmole, this equation can be solved with a spreadsheet for W for a series of values of x from 0.020 to 0.0. For each value of x, a material balance on C7 gives Eq. W x − Wx (13-6) for the mole fraction of C7 in the cumulative distillate, y D avg = 0 0 (2) W0 − W With values of W, x, and y D avg , together with the molecular weights of 100.21 for C7 and 92.14 for T, amounts in mass units can be computed, using Eqs. (1) and (2). The spreadsheet is given on the following page, from which the following results are obtained. (a) Initial residue = 92.29 lb. At 50 wt% distilled, residue = 46.145 lb. From the spreadsheet, the mole fraction of C7 in the residue = 0.0102, the mass fraction of C7 in the residue = 0.0110. The mass fraction of C7 in the cumulative distillate = 0.0324. (b) When 50% of the C7 has distilled, the mole fraction of C7 in the residue = 0.0142, the mass fraction of C7 in the cumulative distillate = 0.0365, and the wt% of charge distilled = 29.9%. Exercise 13.1 (continued) W lb C7 mols T lb T Total lb wt% C7 mass lb cum. C7 mass in in in residue distilled fraction distillate fraction residue ln(Wo/W) mols C7 in x residue residue residue in residue in cum. distillate 0.0200 0.000 1.000 0.0200 2.0040 0.980 90.29 92.29 0.000 0.0217 0.00 0.0196 0.021 0.979 0.0192 1.9231 0.960 88.45 90.37 2.083 0.0213 1.92 0.0421 0.0192 0.042 0.958 0.0184 1.8439 0.940 86.61 88.45 4.163 0.0208 3.84 0.0417 0.0188 0.064 0.938 0.0176 1.7664 0.920 84.77 86.53 6.240 0.0204 5.76 0.0413 0.0184 0.087 0.917 0.0169 1.6907 0.900 82.93 84.62 8.313 0.0200 7.67 0.0408 0.0180 0.109 0.896 0.0161 1.6166 0.880 81.09 82.71 10.382 0.0195 9.58 0.0404 0.0176 0.133 0.876 0.0154 1.5443 0.860 79.26 80.80 12.448 0.0191 11.49 0.0400 0.0172 0.157 0.855 0.0147 1.4737 0.840 77.43 78.90 14.510 0.0187 13.39 0.0396 0.0168 0.181 0.835 0.0140 1.4048 0.821 75.59 77.00 16.569 0.0182 15.29 0.0392 0.0164 0.206...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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