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

The range of the corresponding xb values depends on

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Unformatted text preview: 8 mole fraction of 1 in instantaneous distillate. Second specification of 100 kmol/h of boilup Stop operation when mole fraction of 1 in the bottoms (residue) = 0.05 Step size = 0.02 h recorded every 5 increments with stop tolerance of 0.001 Results are as follows: (a) 7.28 h for operation step 1. (b) kmol of cumulative distillate (from material balance) = 200 kmol at 78oC. (c) Initial (minimum) reflux ratio = 2.35 Final (maximum) reflux ratio = 3.7 (d) The variation of the distillate rate, computed from the reflux ratios, is as follows: 0 2 4 6 7.28 Time, h 29.9 28.6 26.9 24.1 21.3 Distillate, kmol/hr Exercise 13.16 Subject: Batch rectification with variable reflux ratio and constant distillation composition. Given: 500 lbmol charge of 48.8 mol% A and 51.2 mol% B. αA,B = 2.0. Total condenser, 7 theoretical trays in column, and partial reboiler. Constant 95 mol% A in distillate. Boilup rate = 213.5 lbmol/h. Assumptions: Negligible holdup on trays and in condensing system. Perfect mixing in reboiler. No pressure drop. Constant molar overflow. Find: When mole fraction of A in boiler drops to 0.192: (a) Time in h. (b) Total amount of distillate, lbmol. Analysis: (a) Using the procedure outlined in Section 13.3, a McCabe-Thiele construction is applied, as in Fig. 13.7, to obtain the relationship between the reflux ratio and the mole fraction of A in the residue for xD = constant = 0.95 and 7 + 1 = 8 equilibrium stages. The construction for the initial residue composition of xW = x0 = 0.488 is shown on the following page, where the correct slope of the operating line was obtained by trial and error to be L/V = 0.706. From Eq. (7-7), the corresponding reflux ratio = R = (L/V)/[1 - (L/V)] = 0.706/(1 - 0.706) = 2.40. In the table below, other values of increasing L/V by the McCabe-Thiele construction give the values of xW listed below until the final value of 0.192 is reached. These values are used to obtain time, t, for distillation starting from a bubble-point charge, with Eq. (13-16): t= W0 x D − x0 V dxW x0 xWt 1 − L / V ( x D − xW L/V 0.706 0.744 0.783 0.828 0.870 0.898 xW 0.488 0.425 0.365 0.290 0.230 0.192 2 R 2.4 2.9 3.6 4.8 6.7 8.8 = 500 0.95 − 0.488 2135 . I= dxW 0.488 0.192 1 − L / V 0.95 − xW 1 1 − L / V ( x D − xW 15.9 14.2 13.5 13.4 14.8 17.1 2 2 ∆ t, h 1.03 0.90 1.09 0.92 0.66 The above table includes the numerical integration of Eq. (1) by the trapezoidal rule. For example, the ∆t of 0.95 in the 3rd row is obtained from: 500 0.95 − 0.488 ∆t = Iavg (∆xW) = 1.08[(15.9 + 14.2)/2](0.488 - 0.425) = 1.03 h 213.5 (1) Exercise 13.16 (continued) Analysis: (continued) The total time is the sum of ∆t values from the above table: t = 1.03 + 0.90 + 1.09 + 0.92 + 0.66 = 4.60 h This is close to the value of 4.75 h obtained using the batch distillation program of Chemcad. (b) The total distillate is obtained by material balance on component A, giving an equation like (13-13): D = W0 1 − xD − x0 xD − xW = 500 1 − 0.95 − 0.48...
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