Exercise 13.1
Subject:
Evaporation of a small amount of nheptane (C7) from a drum of toluene (T).
Given:
A drum of toluene containing 2 mol% C7. Simple batch differential (Rayleigh)
distillation at 1 atm. Vaporliquid equilibrium data for the C7toluene system at 1 atm.
Assumptions:
Perfect mixing in the still. Exiting vapor in equilibrium with the liquid.
Find:
(a) Composition of the residue in the pot as a function of wt% distilled. Composition of
residue and the cumulative distillate at 50 wt% distilled.
(b) Composition of the cumulative distillate and the residue when 50% of the C7 has been
distilled. Weight percent of the original charge distilled.
Analysis:
The vaporliquid equilibrium data indicate that C7 is more volatile than T. Therefore,
make the calculations in terms of C7. Because the initial concentration of C7 in the charge is
low (2 mol%), assume that the distillation takes place at constant relative volatility,
α = α
C7T
.
Alternatively, the
y  x
data in the low concentration region could be fitted to a quadratic equation
and used with Eq. (132). The lowest concentration in the data table is 2.5 mol% C7, where
x
=
0.025 and
y
= 0.048. Therefore, for the binary system, using Eqs. (219) and (221),
K
C7
=
y/x
= 0.048/0.025 = 1.92
K
T
= (1
y
)/(1
x
) = (1  0.048)/(1  0.025) = 0.976
α
C7T
=
K
C7
/
K
T
= 1.92/0.976 = 1.97
A similar calculation at the next point in the data table,
x
= 0.062 and
y
= 0.107, gives
α
C7T
=
1.81. Therefore the relative volatility increases as
x
drops to zero. Take
α
C7T
= 2 for the region
of interest.
From Eq. (135),
ln
ln
ln
ln
.
ln
.
W
W
x
x
x
x
x
x
0
0
0
1
1
1
1
0 025
2
1
1
0 025
°
±
²
³
´
µ
=

°
±
²
³
´
µ
+


°
±
²
³
´
µ
¶
·
¸
¹
º
»
=
°
±
²
³
´
µ
+


°
±
²
³
´
µ
α
α
(1)
Taking a basis of
W
0
= 1 lbmole, this equation can be solved with a spreadsheet for
W
for a
series of values of
x
from 0.020 to 0.0. For each value of
x
, a material balance on C7 gives Eq.
(136) for the mole fraction of C7 in the cumulative distillate,
y
W x
Wx
W
W
D
¼ ½
avg
=


0
0
0
(2)
With values of
W
,
x
, and
y
D
¼ ½
avg
, together with the molecular weights of 100.21 for C7 and
92.14 for T, amounts in mass units can be computed, using Eqs. (1) and (2). The spreadsheet is
given on the following page, from which the following results are obtained.
(a) Initial residue = 92.29 lb. At 50 wt% distilled, residue = 46.145 lb. From the spreadsheet,
the
mole fraction of C7 in the residue = 0.0102, the mass fraction of C7 in the residue = 0.0110
.
The
mass fraction of C7 in the cumulative distillate = 0.0324
.
(b) When 50% of the C7 has distilled, the
mole fraction of C7 in the residue = 0.0142, the mass
fraction of C7 in the cumulative distillate = 0.0365, and the wt% of charge distilled = 29.9%.
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Exercise 13.1 (continued)
x
ln(
W
o/
W
)
W
mols C7
lb C7
mols T
lb T
Total lb
wt%
C7 mass
lb cum.
C7 mass
in
in
in
in
residue
distilled
fraction
distillate
fraction
residue
residue
residue
residue
in residue
in cum.
distillate
0.0200
0.000
1.000
0.0200
2.0040
0.980
90.29
92.29
0.000
0.0217
0.00
0.0196
0.021
0.979
0.0192
1.9231
0.960
88.45
90.37
2.083
0.0213
1.92
0.0421
0.0192
0.042
0.958
0.0184
1.8439
0.940
86.61
88.45
4.163
0.0208
3.84
0.0417
0.0188
0.064
0.938
0.0176
1.7664
0.920
84.77
86.53
6.240
0.0204
5.76
0.0413
0.0184
0.087
0.917
0.0169
1.6907
0.900
82.93
84.62
8.313
0.0200
7.67
0.0408
0.0180
0.109
0.896
0.0161
1.6166
0.880
81.09
82.71
10.382
0.0195
9.58
0.0404
0.0176
0.133
0.876
0.0154
1.5443
0.860
79.26
80.80
12.448
0.0191
11.49
0.0400
0.0172
0.157
0.855
0.0147
1.4737
0.840
77.43
78.90
14.510
0.0187
13.39
0.0396
0.0168
0.181
0.835
0.0140
1.4048
0.821
75.59
77.00
16.569
0.0182
15.29
0.0392
0.0164
0.206
0.814
0.0133
1.3376
0.801
73.76
75.10
18.625
0.0178
17.19
0.0388
0.0160
0.231
0.794
0.0127
1.2722
0.781
71.94
73.21
20.677
0.0174
19.08
0.0384
0.0156
0.257
0.773
0.0121
1.2084
0.761
70.11
71.32
22.725
0.0169
20.97
0.0379
0.0152
0.284
0.753
0.0114
1.1463
0.741
68.28
69.43
24.771
0.0165
22.86
0.0375
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 Spring '08
 Ofoli
 Equilibrium, Distillation, reflux, cumulative distillate

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