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first increment of n, which takes n to 0.44  0.005 = 0.435, with an navg = (0.44 + 0.435)/2
=0.4375
Because the permeate side pressure is very low compared to the feedside pressure, the initial
composition of the permeate is given by simplifications of Eqs. (4) and (5):
yA/yC = 50(xA/xC) = 50(0.1/0.7) = 7.14 and yB/yC = 50(xB/xC) = 50(0.2/0.7) = 14.29
(0)
(0)
(0)
Substitution into Eqs. (6) to (8) gives yA = 0.3183, yB = 0.6372, yC = 0.0445
From Eqs. (1), (2), and (3),
(
(
xA1) = xA0) + x (1)
B =x ( 0)
B (
(
yA0) − xA0)
0.3183 − 010
.
∆n = 0.10 +
(0.005) = 0.0975
navg
0.4375 (
(
yB0) − xB0)
0.6372 − 0.20
+
∆n = 0.20 +
(0.005) = 0.1950
navg
0.4375 (
(
(
xC1) = 1 − xA1) − xB1) = 1 0.0975  0.1950 = 0.7075 Substitution into Eqs. (4) and (5) gives:
yA
yC
yB
yC (1) = α*
AC (
(
xA1) PF − yA0) PP
0.0975(1000)  0.3183(20)
= 50
= 6.449
(1)
( 0)
xC PF − yC PP
0.7075(1000)  0.0445(20) =α (
(
xB1) PF − yB0) PP
0.1950(1000)  0.6372(20)
= 50
= 12.897
(1)
( 0)
xC PF − yC PP
0.7075(1000)  0.0445(20) (1)
*
BC Using Eqs. (6) to (8),
1
= 0.0491
1 + 6.449 + 12.897
= 0.04915(6.449) = 0.3170 (
yC1) =
(
yA1) (
yB1) = 0.04915(12.897) = 0.6339 Exercise 14.21 (continued)
Analysis: (continued)
From Eq. (9), ∆AM = PM A (
yA1) ∆n
0.3170(0.005)
=
= 1,380 ft 2
8
(1)
(1)
1.26 × 10 [0.0975(1000) − 0.3170(20)]
xA PF − yA PP Calculations for succeeding increments of n are given below from spreadsheet calculations.
From the spreadsheet results, it is seen that when the mole fraction of H2S in the retentate reaches
0.0004, the mole fraction of CO2 is well below the specified 0.02. Thus, H2S controls. The mole
fractions of the permeate in the spreadsheet table are the local values. The average composition
of the final combined permeate is obtained by material balance from the feed gas and calculated
pipeline gas (final retentate from the spreadsheet) with the following results: Component mol%:
C H4
H2 S
C O2
Total flow,lbmol/s Feed gas Pipeline gas
70
10
20
0.440 99.89
0.04
0.07
0.255 Permeate
28.80
23.73
47.47
0.185 From 10,000 scfm of feed gas, 1,000(0.255/0.44) = 580 scfm of pipeline gas is obtained.
From the spreadsheet table, the final retentate pressure is very close to the required 980 psia.
The required membrane area = 197,000 ft2. If each module contained 4,000 ft2, as mentioned for
one case in the industrial example at the beginning of this chapter, then the number of modules in
parallel = 197,000/4,000 = say 50 modules.
This is not unreasonable. Analysis: (continued) Exercise 14.21 (continued) Exercise 14.22
Subject: Separation of ethyl acetate (EA) from water by pervaporation.
Given: 100,000 gal/day of water containing 2.0 wt% EA at 30oC and 20 psia. Dense
polydimethylsiloxane with a 1 µmthick skin in a spiralwound module. Permeate pressure = 3
cmHg. Membrane flux = 1.0 L/m2h with a separation factor of 100 for EA with respect to H2O.
Assumptions: Crossflow.
Find: Membrane area in m2 and temperature drop of feed for a retentate of 0.2 wt% EA and a
permeate of 45.7 wt% EA as a vapor at 3 cmHg pressure.
Analysis: Use the Chemcad process simulator with the Wilson...
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 Spring '11
 Levicky
 The Land

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