Unformatted text preview: h a reflux ratio of 3.5, from Eq. (77), the slope of the rectifying section operating line is,
L/V = R/(1 + R) = 3.5/4.5 = 0.778
The qline is a horizontal line at y = 0.44. For 14 plates with 50% efficiency, the column has the
equivalent of 7 equilibrium stages + 1 for the partial reboiler.
The McCabeThiele construction is shown on the next page, where it is seen that it is possible to
obtain the desired distillate composition.
(b) and (c) From the McCabeThiele diagram, the mole fraction of benzene in the bottoms is
xB = 0.24. As a weight fraction, this corresponds to,
0.24(78.11)
= 0.211 weight fraction or 21.1 wt% benzene
0.24(7811) + 0.76(92.13)
.
Compute the distillate rate by overall molar material balances.
Overall total mass balance:
158.22 = D + B
(1)
Overall benzene mass balance:
69.65 = 0.974D + 0.240B
(2)
Solving Eqs. (1) and (2):
D = 43.16 kmol/h
B = 115.06 kmol/h
By weight, D = 43.16[0.974(78.11) + 0.026(92.13)] = 3,387 kg/h Analysis: (a) (continued) Exercise 7.15 (continued)
McCabeThiele Diagram Exercise 7.16
Subject:
Effect on the separation of A from B by distillation when 3 of 7 theoretical plates
rust and drop to the bottom of the column.
Given: . Column has 7 theoretical plates + partial reboiler. Saturated liquid feed of 100 kmol/h
of 50 mol% A is sent to plate 5 from the top. Distillate contains 90 mol% A. The L/V = 0.75 in
the rectifying section. Vaporliquid equilibrium data.
Assumptions: Constant molar overflow. Total condenser.
Find: Case 1: Column before the 3 plates rust and drop.
(a) Composition of the bottoms product.
(b) The L/V in the stripping section.
(c) The kmol/h of bottoms product.
Case 2: If plates 5, 6, and 7 counted down from the top are lost:
(a) Composition of bottoms product.
Case 3: Same as Case 2, except replace reflux with the same molar flow rate of product
containing 80 mol% A:
(a) Composition of distillate.
(b) Composition of bottoms.
Analysis:
Case 1: Apply the McCabeThiele method in terms of component A, which is
more volatile than B. The rectifying section operating line passes through [0.90, 0.90] with a
slope of 0.75. The qline is vertical through x = 0.50. Step off 4 stages in the rectifying section.
Then, by trial and error, find an xB with a corresponding stripping section operating line that
gives 4 equilibrium stages in the stripping section. The result is shown on the following page,
where:
(a) Bottoms contains 7 mol% A and 93 mol% B.
(b) The slope of the stripping section operating line from the coordinates of the line is:
L / V = {[0.90 − 0.75(0.90 − 0.50] − 0.07}/(0.50 − 0.07) = 1.23
(c) By material balances, F = D + B and FxF = 50 = 0.9D + 0.07B. Solving these two
equations, distillate flow rate = 51.8 kmol/h and bottoms flow rate = 48.2 kmol/h
Case 2: We now have 4 equilibrium stages and a partial reboiler, with the feed being
sent to the reboiler. Assume that utility rates are such that L/V and L / V are the same as in Case
1. Then, on the McCabeThiele diagram, the val...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details