Unformatted text preview: urated vapor line. The vapor composition at point M is
69 mol% nC6 and 31 mol% nC8.
(b) From Fig. 7.37, reproduced below, the enthalpy of the initial mixture of 20 mol%
nC6 at 100oF, shown at point G, is 6,500 Btu/lbmol = HG . A vertical line is drawn upward from
point G until it reaches the point E, located on the dashed tie line between the saturated liquid
and saturated vapor curves, at a point that is 60% of the way from the saturated liquid curve to
the saturated vapor curve. The enthalpy of the twophase mixture at point E is 23,000 Btu/lbmol
= HE. Therefore, the energy required = Q = HE  HG = 23,000  6,500 = 16,500 Btu/lbmol.
(c) From the reproduction of Fig. 7.37 below, the equilibrium liquid phase
composition is at point D, which is 7.5 mol% nC6 , while the equilibrium vapor composition is at
point F, which is 28.5 mol% nC6 . Checking this by material balance on nC6 , for a basis of 100
lbmoles of mixture:
0.20(100) = 20 lbmoles
0.075(40) + 0.285(60) = 20.1 lbmoles
This is a reasonably good check. Analysis: (continued) Exercise 7.54 (continued) Exercise 7.55
Subject:
Determining compositions and energy requirements for a mixture of nC6 and nC8
with an enthalpyconcentration diagram for 101 kPa.
Given: Enthalpyconcentration diagram in Fig. 7.37.
Find: (a) Temperature and compositions of liquid and vapor, where F1 = 950 lb/h of 30 mol%
nC6 at 180oF is adiabatically mixed with F2 = 1,125 lb/h of 80 mol% nC6 at 240oF.
(b) Energy and resulting phase compositions when a mixture of 60 mol% nC6 at 260oF is
cooled and partially condensed to 200oF.
(c) Compositions and relative amounts of the two phases when the equilibrium vapor
from part (b) is further cooled to 180oF.
Analysis: (a) In Fig. 7.37, reproduced on the next page, the two feeds before mixing are shown
at F1 and F2. When these two feeds are adiabatically mixed, the resulting twophase mixture
contains the following, where it is necessary to convert mass to moles.
Component
nC6
nC8
Total
MW
86.11
114.14
Feed F1 :
lb
232
718
950
lbmol
2.69
6.29
8.98
Feed F2 :
lb
845
280
1,125
lbmol
9.81
2.45
12.26
Total feeds:
lbmol
12.5
8.74
21.24
mole fraction 0.5885 0.4115
100
Based on the mole fractions of the combined feed, point M, the adiabatic mixing point is located
in Fig.7.37 on the next page. By coincidence the mixing line connecting the two feeds is also a
tie line for a temperature of 204oF. The equilibrium vapor is located at point V, while the
equilibrium liquid is located at point L. The equilibrium vapor has a mole fraction of 0.75 for
nC6 and 0.25 for nC8. The equilibrium liquid has a mole fraction of 0.35 for nC6 and 0.65 for
nC8.
Although not requested, the moles of equilibrium vapor and liquid can be determined by material
balances. By total material balance,
F1 + F2 = 21.24 = V + L
(1)
By material balance on nhexane,
12.5 = 0.75V + 0.35L
(2)
Solving Eqs. (1) and (2), V = 12.68 lbmol and L = 8.56 lbmol
The molar % vaporized = 12.69/21.24 x 100% = 59.7%. This is consistent with the value
obtained by the inverse lever arm rule on the diagra...
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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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