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Unformatted text preview: egral Cumulative
0.986 Exercise 13.9
The results in the above table indicate that all of the isopropanol is evaporated, because at xW =
0.002, the value of ln(W0/W) = 0.986, which corresponds to W/W0 = 0.565 or only 43.5 mol%
distilled. By the time, 70 mol% of the isopropanol is evaporated, the mole fraction of
isopropanol in the residue will be essentially 0. The mole fraction of isopropanol in the
cumulative distillate = 40/70 = 0.571. Exercise 13.10
Subject: Batch distillation of a mixture of benzene (B) and toluene (T) in a column with 3
equilibrium stages and a reboiler (4 total stages), under conditions of a constant reflux ratio.
Given: Feed of 100 moles contains 30 mol% B and 70 mol% T. Distillation at 1 atm with a
reflux of L/V = 0.6. Constant relative volatility of 2.5 for B with respect to T.
Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser.
Find: Moles of residue when the cumulative distillate is 45 mol% B.
Analysis: Make calculations in terms of B, the more volatile component. Eq. (13-2) applies,
where yD is the mole fraction of B in the instantaneous vapor leaving the top stage, and xW is the
mole fraction of B in the liquid in the reboiler.
xW y − x
From Eq. (13-6), the mole fraction of benzene in the cumulative distillate is given by:
30 − WxW
y D avg = 0.45 =
100 − W
xW = 0.45 −
From Eq. (4-8), the vapor-liquid equilibrium curve is given by,
1 + x (α − 1) 1 + 15x
This is the equilibrium curve for applying the McCabe-Thiele method. The relationship between
yD and xW in Eq. (1) is obtained from a McCabe-Thiele diagram by drawing a series of operating
lines of slope = L/V = 0.6. For each operating line, starting from the intersection with the 45o
line, which is yD, 4 stages are stepped off to determine the corresponding xW. A typical
construction that starts from yD = 0.74 is shown on the next page, where the xW = 0.277. Other
sets of values are given in the
0.20 0.050 ln Exercise 13.10 (continued)
A curve fit of the above data gives:
yD = 0.0059054 + 4.24118 xW − 5.66817 xW (5) Substituting Eq. (5) into Eq. (1) gives: ln 100
W 0.30 xW dxW
0.0059054 + 4.24118 xW − 5.66817 xW (6) Exercise 13.10 (continued)
Eq. (6) can be solved analytically or numerically. A numerical solution is used here with a
spreadsheet. The integral is evaluated by the trapezoidal rule. The following table starts at the
upper limit of 0.30 and moves in increments downward. For each increment, the increment of
the integral from x1 to x2 is (x1 - x2)(f1/2 + f2/2), where f is the integrand. For the addition of each
integral increment, W is computed, and then the benzene mole fraction in the cumulative
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land