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Unformatted text preview: aterial balance on P gives Eq. (136) for the mole fraction of C7 in the
cumulative distillate, Therefore, for a basis of W0 = 1 mole, yD avg = W0 x0 − Wx
W0 − W ( yD )avg = (3) 1(0.4) − 0.3(0.0575)
= 0.547
1 − 0.3 Exercise 13.2 (continued)
x
0.40
0.39
0.38
0.37
0.36
0.35
0.34
0.33
0.32
0.31
0.30
0.29
0.28
0.27
0.26
0.25
0.24
0.23
0.22
0.21
0.20
0.19
0.18
0.17
0.16
0.15
0.14
0.13
0.12
0.11
0.10
0.09
0.08
0.07
0.06
0.05
0.04 y f =1/(yx) 0.580
0.578
0.575
0.573
0.571
0.569
0.566
0.564
0.562
0.559
0.557
0.554
0.552
0.549
0.546
0.543
0.540
0.536
0.533
0.529
0.525
0.521
0.517
0.512
0.507
0.502
0.497
0.491
0.486
0.479
0.473
0.466
0.459
0.452
0.444
0.436
0.427 5.558
5.329
5.117
4.922
4.741
4.573
4.417
4.272
4.137
4.011
3.894
3.785
3.683
3.588
3.499
3.416
3.338
3.266
3.198
3.135
3.076
3.021
2.970
2.923
2.879
2.838
2.801
2.767
2.735
2.707
2.681
2.658
2.638
2.621
2.606
2.594
2.584 Value of
Integral
0.054
0.107
0.157
0.205
0.252
0.297
0.340
0.382
0.423
0.462
0.501
0.538
0.575
0.610
0.645
0.678
0.711
0.744
0.775
0.806
0.837
0.867
0.896
0.925
0.954
0.982
1.010
1.037
1.065
1.092
1.118
1.145
1.171
1.197
1.223
1.249 Exercise 13.3
Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene.
Given: Charge of 100 moles of 30 mol% benzene (B) and 70 mol% toluene (T). Distillation
until the cumulative distillate contains 45 mol% benzene.
Assumptions: Relative volatility = α = αBT = 2.5. Perfect mixing in the still. Exiting vapor in
equilibrium with the liquid.
Find: Moles of residue.
Analysis: For constant relative volatility, Eq. (135) applies with α = 2.5, W0 = 100, x0 = 0.30.
Thus, ln 100
2
0.30
1− x
= ln
+ 2.5 ln
W
3
x
0.70 (1) From Eq. (136), the mole fraction of benzene in the cumulative distillate is given by:
yD avg = 0.45 = 30 − Wx
100 − W (2) Eqs. (1) and (2) are solved by first solving Eq. (2) for x in terms of W to obtain
x = 0.45 − 15
W (3) Substituting Eq. (3) into Eq. (1) and solving the resulting nonlinear equation gives W = 57.9
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 readily done by setting up a table in W,
working in increments from W = 100 moles down to a low value of W. Exercise 13.4
Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene.
Given: Charge of 250 lb of 70 mol% benzene (B) and 30 mol% toluene (T). Distillation at 1
atm until 1/3 of the original charge is distilled.
Assumptions: Ideal Kvalues. Perfect mixing in still. Exiting vapor in equilibrium with liquid.
Find: Compositions of the residue and distillate.
Analysis: Using molecular weights of 78.11 for benzene and 92.14 for toluene, the average
molecular weight of the charge = 0.7(78.11) + 0.3(92.14) = 82.32. Therefore, the charge contains
250/82.32 = 3.037 lbmol. Final mass of t...
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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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