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Unformatted text preview: e base conditions is: α C 2 , nC4 = KC 2
KnC4 = yC 2 / xC 2
ynC4 / xnC 4 = (nC 2 )V / (nC2 ) L
(nnC 4 )V / (nnC4 ) L = (19.0 / 6.0)
= 7.39
(7.5 / 17.5) But we need a relative volatility of: α C 2 , nC4 = (nC2 )V / (nC 2 ) L
(nnC4 )V / (nnC 4 ) L = (17.5 / 7.5)
= 44.3
(1.25 / 23.75) For ideal solutions, where the Raoult's law Kvalue applies, Eq. (221) combined with Eq. (3 in
Table 2.3, gives relative volatility as independent of pressure and equal to the ratio of vapor
pressures, which depend only on temperature. In general, as the temperature is reduced, the
relative volatility increases. Assume that the hydrocarbon mixture, although not an ideal
solution, Exercise 4.23 (continued)
Analysis: (continued)
follows these same trends. Thus, to increase the relative volatility, the pressure has little effect.
We must decrease the temperature to obtain the desired relative volatility, and then adjust the
pressure to obtain the required compositions. The base case and calculations leading to the
desired separation are summarized in the following table: T, oF
P, psia
Kvalues: C2
C3
nC4
nC5
α of C2 to nC4
% C2 to vapor
% nC4 to vapor Base Case
150
205
4.1
1.5
0.56
0.215
7.3
75.8
30 70
14.7
4.0
0.46
0.055
0.077
73 40
14.7
7.5
1.11
0.165
0.028
45 Desired Case
40
16.4
6.7
1.00
0.148
0.025
45
70
4.9 Thus, at 40oF and 16.4 psia, the desired 70% of the ethane is found in the vapor product, with
only 5% of the nbutane.
The compositions of the vapor and liquid products for the base case and the desired case are as
follows:
Base Case:
Desired Case:
y
l, lbmol/h
x
y
l, lbmol/h
x
Component
υ, lbmol/h
υ, lbmol/h
C2
19.0
0.44
6.0
0.11
17.5
0.69
7.5
0.10
C3
13.4
0.31
11.6
0.20
6.4
0.25
18.6
0.25
7.5
0.17
17.5
0.31
1.22
0.05
23.78
0.32
nC4
nC5
3.5
0.08
21.5
0.38
0.22
0.01
24.78
0.33
Total: 43.4 1.00 56.6 1.00 25.34 1.00 74.66 1.00 Exercise 4.24
Subject: Cooling of a reactor effluent with recycle liquid from a partial condensation.
Given: Reactor effluent temperature of 1000oF and composition in lbmol/h of 2000 H2, 2000
CH4, 500 benzene, and 100 toluene. Partial condensation conditions or 100oF and 500 psia, and
component Kvalues at these conditions. Two heat exchangers in a recycle loop.
Find: (a) Composition and flow rate of vapor leaving flash drum in Fig. 4.38.
(b) Proof that vapor flow rate is independent of quench rate.
Analysis: (a) Assume that vapor rate is independent of quench rate. Therefore, conduct the
flash calculation on just the reactor effluent at the flash drum conditions of temperature and
pressure. Use the RachfordRice equations (Eqs. (3) and (6), Table 4.4):
f {Ψ} = yi = zi 1 − Ki
=0
i =1 1 + Ψ Ki − 1
C (1) zi Ki
1 + Ψ Ki − 1 (2) Nonlinear Eq. (1) is solved for Ψ = V/F , followed by calculation of V = ΨF, and then
calculations of vapor mole fractions from Eq. (2). The given input for Eq. (1) is:
Component
Hydrogen
Methane
Benzene
Toluene f, lbmol/h
2,000
2,000
500
100 zi
0.4348
0.4348
0.1087
0.0217 Ki
80
10
0.010
0.004 One method f...
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 Spring '11
 Levicky
 The Land

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