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Unformatted text preview: Analysis: (continued)
d AM = PM O 2 ydn
,
xPF − yPP ∆AM = 170 × 10
. −2 y∆n
x (17.64) − y (0.9671) (9) where y is the local value given by Eq. (3) and ∆n is the incremental decrease in the feed flow.
Using the equations, the calculations are carried out on a spreadsheet with the results below.
From the spreadsheet results, the membrane area is, by interpolation, 314,500 ft2 when the
acetone mole fraction in the permeate = 0.05. However, at that condition, the calculations show
a retentate with an acetone mole fraction of 0.00136, where 0.0005 is desired. At that value, the
permeate acetone content drops to 3.55 mol% and the membrane area increases to 548,400 ft2. Analysis: (continued) Exercise 14.19 (continued) Exercise 14.20
Subject: Separation of air by gas permeation.
Given: Permeate of 5 ton/day (2,000 lb/ton) of 40 mol% O2 and retentate of 90 mol% N2 from a
feed of air. 500 psia on feed side and 20 psia on permeate side.
Assumptions: Crossflow. No pressure drop on feed side.
Find: Membrane area in m2 for the following membranes:
Company
A
Module type
Hollowfiber
15
Permeance for O2, barrer/µm
Ratio of O2 to N2 permeances
3.5 B
Spiralwound
35
1.9 Analysis: First calculate the overall material balance. P = 5(2,000) = 10,000 lb/day
The average molecular weight of the permeate = [0.4(32) + 0.6(28)] = 29.6
Therefore, P = 10,000/29.6 = 337.8 lbmol/day.
Total molar balance: F = R + 337.8
(1)
O2 molar balance: 0.21F = 0.1R + 0.4(337.8) = 0.1R + 135.1
(2)
Solving Eqs. (1) and (2), F = 921.2 lbmol/h and R = 583.4 lbmol/h
On a component basis, the overall material balance is:
lbmol/day:
Component:
Feed Retentate Permeate
O2
193.5
58.3
135.2
N2
727.7
525.1
202.6
Total
921.2
583.4
337.8
Now check each module type to see if it is even possible to obtain a permeate of 40 mol% O2.
For crossflow, the initial oxygen content of the local permeate at the feed end must be greater
than 40 mol% because the mol% will decrease as the feed moves to the other end.
PF = PR = 500 psia, PP = 20 psia, r = PP/PF = 20/500 = 0.04
αx
Rearranging Eq. (1436), y =
(3) ,where, y and x refer to O2 and α is for O2 to N2.
1 + (α − 1) x
From Eq. (1443), α = α * 2 N 2
O x (α − 1) + 1 − rα
0.21(α − 1) + 1 − 0.04α
= α * 2 N 2
O
x (α − 1) + 1 − r
0.21(α − 1) + 1 − 0.04 For Company A, α * 2 N 2 = 3.5. Substituting this into Eq. (4) and solving, α = 3.28.
O
Substituting this α into Eq. (3), y = 0.4658. Therefore, consider Company A.
For Company B, α * 2 N 2 = 1.9. Substituting this into Eq. (4) and solving, α = 1.84.
O
Substituting this α into Eq. (3), y = 0.3285. Therefore, Company B is dropped. (4) Exercise 14.20 (continued)
Analysis: Company A (continued)
15
PM O = −4 = 150,000 barrer / cm
2
10
= 150 × 10 −5 cm3 (STP) / cm2  s  cmHg
.
Convert permeance to American Engineering units of lbmol/ft2hpsi:
1 barrer/cm = 1 x 1010 (30.48 cm/ft)2 (76 cmHg/14.696 psi) / (22,400 x 454 cm3 (STP)/lbmol)
= 1.70 x 1010 lbmol/ft2hps...
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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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