Unformatted text preview: equation to determine
thermodynamic properties of the liquid phases. The liquid density of the feed = 8.319 lb/gal
Feed rate = 100,000(8.319)/24 = 34,664 lb/h. The permeate and retentate flow rates are obtained
by material balance:
Total material balance:
34,664 = R + P
(1)
Ethyl acetate material balance:
0.02(34,664) = 0.002 R + 0.457 P
(2)
Solving Eqs. (1) and (2), R = 33,293 lb/h and P = 1,371 lb/h
Using a MW of 80.10 for ethyl acetate, the following component material balance is obtained:
Component
Ethyl acetate
Water
Total lb/h:
Feed
693
33,971
34,664 Retentate
66.5
33,226.5
33,293.0 Permeate
626.5
744.5
1,371.0 lbmol/h:
Feed
7.87
1,885.73
1,893.60 Retentate
0.76
1,844.40
1,845.16 Permeate
7.11
41.33
48.44 Assume that the membrane flux is based on a liquid phase of permeate at 30oC. From the
Chemcad program, the liquid density at 30oC is 58.82 lb/ft3 or 2.077 lb/L. Therefore, the volume
flow of the permeate is 1,371/2.077 = 660 L/h. Therefore, since the flux is 1.0 L/m2h,
the membrane area = 660/1.0 = 660 m2.
As discussed in Section 14.7, assume the permeate vapor leaves at the dewpoint temperature at
the permeate pressure of 3 cmHg. From the Chemcad program the permeate temperature =
26.2oC.
As discussed in Section 14.7, compute the exit temperature of the liquid retentate by an adiabatic
enthalpy balance. Using Chemcad, the retentate temperature = 15.5oC. Therefore, temperature
drop of the feed = 30 â€“ 15.5 = 14.5oC. Exercise 14.23
Subject: Calculation of permeances from laboratory data on pervaporation.
Given: Permeation flux of 1.6 kg/m2h for a 17 wt% ethanol in water feed, giving a permeate of
12 wt% ethanol at 60oC and a permeate pressure of 15.2 mmHg. Other conditions in Example
14.12.
Find: Permeances for ethyl alcohol (1) and water (2). Selectivity for water.
Analysis: Molecular weights for ethyl alcohol and water are 46.07 and 18.02, respectively.
From Example 14.12, the vapor pressure of ethyl alcohol = 352 mmHg, and that of water = 149
mmHg. The mole fractions in the feed are:
x1 = 0.17 / 46.07
= 0.0742 and
.
017 (1.0 âˆ’ 0.17)
+
46.07
18.02 x2 = 1 âˆ’ 0.0742 = 0.9258 The mole fractions in the permeate are:
012 / 46.07
.
y1 =
= 0.0506 and y2 = 1 âˆ’ 0.0506 = 0.9494
012 (1.0 âˆ’ 0.12)
.
+
46.07
18.02
From Eq. (1439), with no sweep, the component fluxes are proportional to the weight fractions
in the permeate. Therefore,
N1 = 16(0.12)
.
16(0.88)
.
= 0.00417 kmol / h  m2 and N 2 =
= 0.07814 kmol / h  m2
46.07
18.02 For Eq. (1480), we need the activity coefficients in the feed mixture. Obtain these from
the van Laar equations given in Example 14.12:
0.9232(0.9258)
Î³ 1 = exp 16276
.
1.6276(0.0742) + 0.9232(0.9258)
Î³ 2 = exp 0.9232 16276(0.0742)
.
16276(0.0742) + 0.9232(0.9258)
. 2 = 3.488
2 = 1.014 Exercise 14.23 (continued)
Analysis (continued)
From Eq. (1480),
N1
0.00417
PM1 =
=
= 4.62 Ã—10 âˆ’5 kmol/hm 2 âˆ’ mmHg
s
Î³1 x1 P âˆ’ y1 p1 3.488(0.0742)(352) âˆ’ 0.0506(15.2)
1
PM 2 = N2
0.07814
=
= 6.23 Ã— 10âˆ’4 kmol/hm 2 âˆ’ mmHg
s
Î...
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 Spring '11
 Levicky
 The Land

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