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

# The range of the corresponding xb values depends on

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online