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

2 and thus more vapor generation is needed one

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Unformatted text preview: the composition of the cumulative distillate would be 60.5 mol% ethanol. Exercise 13.12 (continued) Analysis: Distillation with 2 stages plus a reboiler (continued) Exercise 13.13 Subject: Batch distillation of a mixture of acetone (A) and ethanol (E) at 101 kPa in a column with equilibrium stages and a reboiler. Given: Feed of 50 mol% A and 50 mol% E. Vapor-liquid equilibrium data at 101 kPa. Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser. Find: (a) Number of equilibrium stages if the instantaneous distillate composition is 90 mol% acetone when the residue is 10 mol% acetone for a reflux ratio, L/D = 1.5 times the minimum value. (b) Plot of reflux ratio versus the still pot composition and amount of residue left if the column has 8 stages (9 including the reboiler) and the reflux rate is varied to maintain the distillate composition constant at 90 mol% acetone. (c) Total vapor generated for distillation with 8 stages (9 including the reboiler) at constant reflux ratio to obtain a residue of 10 mol% acetone and a cumulative distillate composition of 90 mol% acetone Which method of operation requires more energy. A more efficient operating policy to produce a distillate of 90 mol% acetone. Analysis: (a) First compute the minimum L/V. The slope of the operating line for this condition is given by: L V = min ( y at 0.9) - ( y in equilibrium with x at 0.10) 0.9 − 0.25 = = 0.8125 ( x at 0.9) - ( x at 0.10) 0.9 − 01 . From Eq. (7-27), Rmin = 0.8125/(1 - 0.8125) = 4.33. Therefore, R = 1.5Rmin = 1.5(4.33) = 6.5. Operating L/V = R/(R+1) = 6.5/7.5 = 0.867. On the next page, the number of equilibrium stages are stepped off by the McCabe-Thiele method, where the given vapor-liquid equilibrium data are plotted, operating line has a slope of 0.867, and the steps start at x = 0.9 and end at x = 0.1. Almost 9 equilibrium stages are stepped off. Analysis: (continued) Exercise 13.13 (continued) Analysis: (continued) Exercise 13.13 (continued) (b) For constant acetone mole fraction in the distillate = 0.90, and 8 equilibrium stages plus a reboiler, the McCabe-Thiele diagram can be used to determine the residue composition and amount of residue as a function of reflux ratio. For example, as shown at the bottom of the following page, when L/V = 0.90, R = 0.9/(1-0.9) = 9, and xW = 0.065. From Eq. (13-6), x − (W / W0 ) xW W 0.4 Solving, = For xW = 0.065, W / W0 = 0.48 ( yD )avg = 0.90 = 0 1 − (W / W0 ) W0 0.9 − xW For other reflux ratios, the following results are obtained and plotted on the next page. R=L/D 2.33 3.0 4.0 5.67 9.0 19 ∞ xW 0.48 0.35 0.22 0.13 0.065 0.035 0.006 W/W0 0.95 0.73 0.59 0.52 0.48 0.46 0.45 (c) The case of constant reflux ratio is a tedious trial and error problem. A guess of the reflux ratio is made and a series of McCabe-Thiele constructions is carried out as in Fig. 13.4 or 13.5 with 9 total stages, such that the initial xW = 0.5 and the final xW = 0.1. This gives a series of values of instantaneous yD and xW. Using these values, Eq. (13-2) is integrated to obtain W when the final xW = 0.1is reached. Then, Eq. (13-6) is solved for y D avg . If the value is not 0.9, the procedure is repeated with different values of the reflux ratio until the va...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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