Unformatted text preview: a specification on y2. The
following results were obtained for Cases 2a, 2b, and 2c:
y2
F1
R2
P1
T1
T2
xF1
x1
y1
x2
a
0.74 126.58
26.58 86.58 0.684 0.693 0.5754 0.3900 0.6610 0.4827
b
0.761 164.64
64.64 124.64 0.757 0.481 0.5648 0.3590 0.6308 0.5103
c
0.78 271.64 171.64 231.64 0.853 0.259 0.5605 0.3300 0.6004 0.5376
For Case 2b, which corresponds to the same y2 as for Case 1, the membrane areas are as follows:
0.6308(0.757)(124.64)(2820)
A M 1 in m 2 =
= 254.5 m2
−5
9.0 × 10 0.510(1,551) − 0.761(76.6) (10,000)
0.76064(0.481)(60)(2820)
= 93.9 m2
9.0 × 10 0.510(1,551) − 0.761(76.6) (10,000)
The total membrane area for Case 2b is 348.4 m2, compared to a total of 390 m2 for Case 1.
AM 2 in m2 = −5 Exercise 14.12
Subject: Importance of concentration polarization in dialysis.
Given: From Example 14.7: At a certain location in a tubular membrane separator, for the
dialysis of solute A from solvent B, concentrations for A of 5 x 102 kmol/m3 on the feed side and
1.5 x 102 kmol/m3 on the permeate side. Permeance for A = 7.3 x 105 m/s and feed passes
through the tube of inside diameter = D = 0.4 cm. Permeateside masstransfer coefficient = 0.06
cm/s or 6 x 104 m/s For the feed side masstransfer coefficient, Reynolds number = 25,000 and
Schmidt number for solute A = 500, with a molecular diffusivity for A of 6.5 x 105 cm2/s.
Find: Importance of concentration polarization.
Analysis: The flux of A through the membrane, in terms of the three resistances is given by a
modification of Eq. (1453), with :
NA = cA F − cA P
5.0 × 10 −2 − 15 × 10 −2
.
=
1
1
1
1
1
1
+
+
+
+
−5
6 × 10 −4
k A F PM A kA P
k A F 7.3 × 10 (1) Since the Reynolds number is greater than 10,000, turbulent flow exists inside the tube and if
fully developed flow is assumed, the masstransfer coefficient is given by a modification of Eq.
(1454), where, if the analogy to heat transfer is applied with Eqs. (3165) and (3166) for a
circular tube, a = 0.023, dH = D, and the (dH/L)d correction for the entry effect is not necessary.
In Example 14.7, the feedside masstransfer coefficient for NRe = 15,000 and D = 0.5 cm is
computed to be 5.1 x 104 m/s. Therefore, for this exercise, by ratios, using Eq. (1454), k A F = 51 × 10
. −4 0.5
0.4 25,000
15,000 0.8 = 9.6 × 10 −4 m / s Substituting into Eq. (1), 5.0 × 10 −2 − 15 × 10 −2
.
2.5 × 10 −2
2.5 × 10 −2
NA =
=
=
= 2.1 × 10 −6 kmol / s  m2
1
1
1
1,042 + 13,700 + 1,667
16,409
+
+
−4
−5
−4
9.6 × 10
7.3 × 10
6 × 10
The sum of the feedside and permeateside masstransfer resistances is 2,709 of a total of
16,409. Therefore 2,709/16,409 or 16.5% of the total resistance is due to the two external
resistances. Therefore, concentration polarization is important, but not to a major extent. Exercise 14.13
Subject: Dialysis of an aqueous stream to separate Na2SO4 from a highmolecularweight
substance (A).
Given: Aqueous feed of 100 gal/h at 20oC, containing 8 wt% Na2SO4 and 6 wt% A.
Continuous countercurrent...
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 Spring '11
 Levicky
 The Land

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