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Unformatted text preview: he residue = 0.667(250) = 166.7 lb.
Benzene is the more volatile component. Therefore, base the calculations on B. Use Eq. (133),
with x0 = 0.7 and W0 = 3.037 lbmol. Thus, Eq. (133) becomes:
x
0.7 dx
dx
W
3.037
(1)
= ln
or
= ln
x
0.7 y − x
3.037
y−x
W
where x is the mole fraction of B in the final residue and y is in equilibrium with x.
Equilibrium data at 1 atm are conveniently obtained from a simulation program. For example,
data from the Chemcad program are included below for the range of interest, based on Raoult's
law (ideal K values).
Using a spreadsheet with the equilibrium yx data, integrate Eq. (1) using the trapezoidal rule
with a ∆x increment of 0.02 starting from x = 0.70 with decreasing values of x. As an example,
using this method for the first four increments, the integral is equal to ∆x[ 0.5fx=0.40 + fx=0.39 +
fx=0.38 + fx=0.37 + 0.5fx=0.36 ], where f = 1/(yx). At each step , W is computed and then the mass in
pounds
until 166.7 lb of residue is reached.
The complete spreadsheet is given below From it, the following result is obtained:
By interpolation, mole fraction of B in the residue = 0.630, corresponding to W = 2.005 lbmol.
Using these values, a material balance on B gives for Eq. (136),
3.037(0.7) − 2.005(0.63)
= 0.836
3.037 − 2.005
y f =1/(y x) Value of W, lbmol
Integral ( yD )avg =
x
0.70
0.68
0.66
0.64
62 0.8550
0.8428
0.8302
0.8172
0.8037 6.452
6.143
5.875
5.643
5.444 0.1259
0.2461
0.3613
0.4722 3.307
2.678
2.374
2.116
1.894 W, lb
250.0
221.2
196.8
176.0
158.0 Exercise 13.5
Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene.
Given: Mixture of 60 mol% benzene (B) and 40 mol% toluene (T). Distillation at 1 atm at the
conditions below.
Assumptions: Relative volatility = α = αBT = 2.43, a constant. Perfect mixing in the still.
Exiting vapor in equilibrium with the liquid.
Find: (a)
(b)
for:
(1)
(2)
(3) Number of moles in the distillate for 100 moles of feed
Compositions of distillate and residue.
A (cumulative) distillate of 70 mol% B.
40 mol% of the feed distilled.
60 percent of B distilled. Analysis: Calculations are made in terms of benzene, the more volatile component. Because the
relative volatility is assumed constant, Eq. (135) applies, which for x0 = 0.60, W0 = 100 moles,
and α = 2.43 is:
ln 100
0.60
1− x
= 0.6993 ln
+ 2.43 ln
W
x
0.40 (1) The composition of the cumulative distillate is given by Eq. (136), which becomes:
yD Case 1: avg From Eq. (2), y D = avg 60 − Wx
100 − W
= 0.70 = (2)
60 − Wx
100 − W or W = 10
0.7 − x (3) Substituting Eq. (3) into Eq. (1) and solving the resulting nonlinear equation gives x = 0.33.
Then from Eq. (3), W = 27.0 moles. The distillate = W0  W = 100  27 = 73 moles.
In lieu of a nonlinear equation solver, a spreadsheet can be used by assuming a value of W and
solving Eq. (3) for x. Then substitute this value into Eq. (1) and solve for W. Repeat until the
assumed value of W equals the calculated value. This is...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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