Unformatted text preview: Pa:
From Section 14.1, 1 barrer/cm = 1010 cm3(STP)/cm2scmHg
Conversion factors: 22.42 x 106 cm3(STP)/kmol, 104 m2/cm2, 3600 s/h, and 1.333 kPa/cmHg
1 × 10−10 (3600)
kmol
Therefore, 1 barrer / cm =
= 1205 × 10−10 2
.
6
−4
m  h  kPa
22.42 × 10 10 1333
.
kmol
N2
m  h  kPa
kmol
PM
= 10,000(1.205 × 10 −10 ) = 1.21 × 10 −6 2
CH 4
m  h  kPa
Let x = kmol/h of CH4 in permeate. Then 800  x = kmol/h of CH4 in retentate.
nN
180
N N2 = 2 =
= 6.03 × 10−6 ∆PN 2
(1)
avg
AM A M PM = 50,000(1205 × 10 −10 ) = 6.03 × 10−6
. N CH 4 = nCH 4
AM = x
= 121 × 10 −6 ∆PCH 4
.
AM 0.2(5500) + 5450 ∆PN 2 avg = 2 20
820 − x − 100 2
0.8(5500) + 5450 (2) avg 800 − x
820 − x 180
180 + x (3) x
(4)
avg
2
180 + x
Combining Eqs. (1) and (3), and (2) and (4) and solving the resulting two nonlinear equations,
x = 281 kmol/h of CH4 in the permeate and the membrane area = 48,800 m2.
∆PN 2
= 612 kPa and ∆PCH 4
= 4760 kPa
∆PCH 4 = avg avg − 100 Exercise 14.4
Subject: Hollowfiber module
Given: Hollow fibers, each 42 microns i.d., 85 microns o.d., and 1.2 m long in a membrane
module with 4,000 ft2 of membrane surface area based on the i.d.
Assumptions: Module is a right circular cylinder in shape.
Find: (a) Number of hollow fibers in the module.
(b) Diameter of the module if fibers on a square spacing of 120 microns center to center.
(c) Membrane surface area per unit volume of module (packing density) in m2/m3.
Compare results to Table 14.4.
Analysis:
(a) For one fiber, the inside surface area = Ai = πDiL = 3.14(42/106)(1.2) = 0.000158 m2
or 0.000158/(9.29 x 102) = 0.001704 ft2. Therefore, number of fibers = N = 4000/0.001704 =
2,350,000 fibers.
(b) For square spacing of 120 microns center to center, each fiber occupies a crosssectional area of 120 by 120 microns or 120(120)/1012 = 14.4 x 109 m2. For 2,350,000 fibers,
crosssectional area of the membrane module = 2,350,000(14.4 x 109) = 0.0338 m2 = 338 cm2.
If the area is circular, the diameter of the module = [4(338)/3.14]1/2 = 20.8 cm.
(c) The volume of the module for a length of 1.2 m = 0.0338(1.2) = 0.0406 m3.
The packing density = membrane surface area based on the inside diameter/unit module volume.
Thus, the packing density = 0.000158(2,350,000)/0.0406 = 9,150 m2/m3. Table 14.4 gives a
range of 500 to 9,000 m2/m3 for a hollowfiber module. Exercise 14.5
Subject: Geometry of a spiralwound membrane module.
Given: Module of 0.3 m diameter (D) and 3 m length (L). Packing density = 500 m2/m3.
Assumptions: Collection tube of 1 cm in diameter.
Find: Center to center spacing of membrane in the spiral.
Analysis: Membrane volume = V = πD2L/4 = 3.14(0.3)23/4 = 0.212 m3.
Membrane area = membrane volume x packing density = 0.212(500) = 106 m2.
Length around the spiral = 106/3 = 35.3 m
Circumference of outer circle = πD = 3.14(0.3) = 0.943 m
Circumference of inner circle = 3.14(0.01) = 0.0314 m
Average circumference = (0.943 + 0.0314)/2 = 0.487 m
Number of turns of the spiral = length around spiral/aver. cir...
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 Spring '11
 Levicky
 The Land

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