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Unformatted text preview: Ratebased modeling must be used for membrane separations because an equilibrium
model does not apply.
Compared to distillation and extractive distillation, membranes usually can not achieve
sharp separations
Compared to liquidliquid extraction, absorption, and stripping, membranes produce
products that are usually miscible, rather than immiscible. Exercise 14.2
Subject: Calculation of permeabilities of hydrogen and methane in barrer units.
Given: Feed gas at 450 psia and 200oF, with the following component flow rates in lbmol/h:
H2 1,872.3, CH4 193.1, C2 11.4, Benzene 4.8, Toluene 4.2, and pXylene 0.6. Membrane
system has a total area of 16,000 ft2, with a thickness of 0.3 microns. The permeate exits at 50
psia with 1,685.1 lbmol/h of H2 and 98.8 lbmol/h of CH4. The retentate exits at 450 psia.
Assumptions: Mass transfer driving force based on the difference between the average partial
pressure of the feed and retentate, minus the partial pressure of the permeate.
Find: The permeabilities of H2 and CH4 in barrer units.
Analysis: Use Eq. (141) and the definition of the barrer,
1 barrer = 1010 cm3 gas (STP of 0oC and 1 atm)cm/(cm2scm2 Hg pressure head)
The total feed flow rate = 2086.3 lbmol/h. The permeate rate = 1783.9 lbmol/h.
The retentate flow rate = 2086.3  1783.9 = 302.4 lbmol/h.
Hydrogen: Flux through the membrane = NH2 = 1,685.1/16,000 = 0.1053 lbmol/hft2
Partial pressure driving force =
1872.3 187.2
+
.
2086.3 302.4 − 50 16851 = 3412 − 47.2 = 294 psi
∆pH 2
= 450
.
avg
2
1783.9
From Eq. (141), permeability = PM H =
2 N H2 l M
∆pH 2 avg = 0.1053(0.3)
lbmol  µm
= 1075 × 10 −4
.
294
h  ft 2 − psi To convert to barrer, use: 0.4536 kmol/lbmol, 22.42 x 106 cm3 (STP)/kmol, 104 cm/µm,
3600 s/h, 9.29 x 102 cm2/ft2, and 5.17 cm Hg/psi. Thus,
( 0.4536 ) ( 22.42 ×106 ) (10−4 )
−4
PM H2 = 1.075 × 10
= 1.075 × 10−4 5.882 × 10−5 =
2
( 9.29 ×10 ) ( 3600 )( 5.17 ) cm3 (STP)cm
6.32 × 10
= 63.2 barrer
cm 2 scm Hg
Methane: Flux through the membrane = NCH4 = 98.8/16,000 = 0.00618 lbmol/hft2
1931
.
94.3
+
98.8
∆pCH 4
= 450 2086.3 302.4 − 50
= 910 − 2.8 = 88.2 psi
.
avg
2
1783.9
9 PM CH4 = N CH4 lM ( ∆p )
CH 4 avg = 0.00618(0.3) 5.882 ×10 −5
= 12.4 barrer
88.2
10−10 Exercise 14.3
Subject: Separation of N2 from CH4 by an asymmetric polyimide polymer membrane.
Given: Feed gas of 200 kmol/h of N2 and 800 kmol/h of CH4 at 5500 kPa and 30oC.
Permeances are 50,000 barrer/cm for N2 and 10,000 barrer/cm for CH4. Retentate leaves at 5,450
kPa and 30oC, containing 20 kmol/h of N2, while permeate leaves at 100 kPa and 30oC,
containing 180 kmol/h of N2.
Assumptions: Diffusion driving force based on arithmetic average partial pressures of the feed
and retentate on the feed side, and based on the permeate product on the permeate side.
Find: Membrane surface are in m2. CH4 flow rate in the permeate in kmol/h.
Analysis: From Eq. (141),
nN 2 = 180 kmol / h = AM PM N2 ∆PN 2 avg and nCH 4 = AM PM 2 CH 4 ∆PCH 4 avg Convert permeances in barrer/cm to kmol/m hk...
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 Spring '11
 Levicky
 The Land

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