Unformatted text preview: i
Therefore, the permeances are:
PM O = 150,000 1.70 × 1010 = 2.55 × 105 lbmol / ft 2  h  psi
2 Using the given selectivity,
PM N = 7.29 × 106 lbmol / ft 2  h  psi
2 Assume that α = αO2N2 = a constant = 3.28 from above. Then, for crossflow with a binary
mixture, we have the following relationships between oxygen mole fractions in the permeate and
retentate as a function of the cut, θ = nP/nF.
x − x R (1 − θ) 0.21 − x R (1 − θ)
From material balance Eq. (1), Example 14.6, y P = F
=
(5)
θ
θ
From Eq. (1449), where the y and x pertain to O2, yP = x 1
1− α
R 1− θ
θ 1 − xR α
α −1 xF
1 − xF α
α −1 − xR Combine this equation with Eq. (5) to obtain:
0.21
1− θ =
−
x R + x R0.4386 0.1486 1 − x R α
α −1 1.439 −
= x R0.4386 1− θ
01486 1 − x R
.
θ 1.439 − x1.439
R (6) − x 1.439
R For O2, xR begins at 0.21 at the feed end and decreases toward the retentate end. Use Eq. (6) to
compute values of θ as xR is stepped toward 0. Use Eq. (5) to compute the corresponding value
of yP.
For a differential molar transfer rate, dn, the differential membrane area required is given by Eq.
(1450), which is also given in incremental form for numerical calculation. If we base this
calculation on O2,
d AM = PM O 2 ydn
,
xPF − yPP ∆AM = 2.55 × 10 −5 y∆n
x (500) − y (20) (9) where y is the local value given by Eq. (3) and ∆n is the incremental decrease in the feed flow. Exercise 14.20 (continued)
Analysis: Company A (continued)
Using the above equations, the calculations are carried out on a spreadsheet with results below.
From the spreadsheet results, the membrane area is, by interpolation, 2,388 ft2 when the oxygen
mole fraction in the permeate = 0.40. However, at that condition, the calculations show a
retentate with an oxygen mole fraction of 0.128 and, therefore, a nitrogen mole fraction of 0.872.
Thus, the ideal of 90 mol% N2 is not quite achieved. At that value, the oxygen content of the
permeate drops to 37.4 mol% and the membrane area increases to 3,262 ft2. Exercise 14.21
Subject: Removal of CO2 and H2S from highpressure sour natural gas by gas permeation to
produce a pipeline gas.
Given: Feed and product conditions:
Stream
Feed gas Pipeline gas
Pressure, psia
1000
980
Composition, mol%:
C H4
70
97.96
H2 S
10
0.04
C O2
20
2.00
Hollowfiber membrane of 0.5 µm skin thickness with a permeability for CO2 of 13.3 barrer, and
selectivities of CO2 to CH4 and of H2S to CH4 , both of 50.
Assumptions: T = 60oF and permeateside pressure = 20 psia. Crossflow.
Find: Membrane area for feed flow rate of 10,000 scfm (0oC, and 1 atm).
Analysis: The feed molar flow rate = 10,000 scfm/[(379 scf/lbmol)(60 s/min)] = 0.44 lbmol/s
The membrane thickness = lM = 0.5 microns = 0.5 x 104 cm From Eq. (141),
PM
13.3
PM CO2 = CO2 =
= 266,000 barrer / cm
lM
0.5 × 10 −4 = 2.66 × 10−5 cm3 (STP) / cm2  s  cmHg
Convert permeance to American Engineering units of lbmol/ft2spsi:
1 barrer/cm = 1 x 1010 (30.48 cm/ft)2 (76 cmHg/14.696...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details