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Unformatted text preview: flow dialyzer, using a pure water sweep of 100 gal/h. Membrane of
microporous cellophane with ε = 0.5, wet thickness, lM = 0.0051 cm, τ = 4.1, and pore diameter =
dp = 31 Angstroms. Properties of Na2SO4 and A:
Component
Na2SO4
A
Molecular weight
142
1000
Molecular diameter, Angstroms
5.5
15.0
Molecular diffusivity, cm2/s
0.77 x 105 0.25 x 105
Assumptions: Logmean driving forces. Masstransfer resistances on each side of the
membrane are each 25% of the total masstransfer resistance for each component. No transfer of
water through the membrane.
Find: Membrane area in m2 for a 10% transfer of A through the membrane. Corresponding
percent transfer of Na2SO4.
Analysis: For a concentration driving force, using Eqs. (1413), (1414), and (1415) and
taking into account restrictive diffusion because of the small pore size,
the permeances are given by: P Mi
For Na2SO4, PM Na SO =
2 4 εD
d
= i 1− M
lM τ
dp 0.5 0.77 × 10−5
0.0051(4.1) .
55
1−
31 4 (1) 4 = 8.43 × 10−5 cm / s 4 0.5(0.25 × 10 −5 )
15.0
1−
= 0.424 × 10 −5 cm / s
0.0051(4.1)
310
.
Using Eq. (1456), with the two film resistances together being equal to half the total resistance,
the overall mass transfer coefficients are onehalf of the permeances. Therefore,
KNa 2SO 4 = 4.22 × 10 −5 cm / s or 1.52 × 103 m / h For A, PM A = KA = 0.212 × 10 −5 cm / s or 7.63 × 105 m / h
Assume that the density of the feed is that of 14 wt% aqueous Na2SO4 at 20oC. From Perry's
Handbook, the density = 1.13 g/cm3 or 1130 kg/m3 or 9.41 lb/gal.
Therefore, the mass flowrate of the feed = 100(9.41) = 941 lb/h, comprised of 0.08(941) = 75.3
lb/h of Na2SO4, 56.5 lb/h of A, and 941  75.3  56.5 = 809.2 lb/h of water.
The density of the sweep is 8.33 lb/gal. Exercise 14.13 (continued)
Analysis: (continued)
The sweep liquid enters at a flow rate of 100(8.33) = 833 lb/h of water.
The permeate contains 10% of A in the feed or 0.10(56.5) = 5.7 lb/h.
To obtain the concentrations, assume that 75% of the Na2SO4 in the feed is transferred through
the membrane. Thus, the permeate contains 0.75(75.3) = 56.5 lb/h of Na2SO4. The retentate
contains 56.5  5.7 = 50.8 lb/h of A and 75.3  56.5 = 18.8 lb/h of Na2SO4. Compute the
concentrations in kg/m3. For example, the feed is 941 lb/h or 427 kg/h. Its volume is 427/1130
= 0.378 m3/h. Therefore, the concentration of Na2SO4 in the feed = 0.08(427)/0.378 = 90.4
kg/m3.
Assume the density of the permeate is 1.07 g/cm3 or 1070 kg/m3.
Assume the density of the retentate is 1.07 g/cm3 or 1070 kg/m3.
The total permeate = 895.2 lb/h or 406.1 kg/h with a volume of 0.394 m3/h.
The total retentate = 878.8 lb/h or 398.6 kg/h with a volume of 0.373 m3/h.
The results of the concentration calculations are:
Concentrations in kg/m3:
Component
Feed
Sweep Retentate Permeate
Na2SO4
90.4
0.0
22.9
65.0
A
67.8
0.0
61.8
6.6
67.8 − 6.6 − 618 − 0.0
.
= 615 kg / m3
.
67.8 − 6.6
ln
618 − 0.0
.
mA
5.7(0.4536)
From a modification of Eq. (1455), AM =
=
= 551 m2
−5
KA ∆cA 7.63 × 10 615
.
The logmean driving force for A is ∆cA = The logmean driving force for Na2SO4 is ∆cA =...
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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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