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Unformatted text preview: 2 x2 P2 − y2 p2 1.014(0.9258)(149) − 0.9496(15.2) The selectivity is defined in Example 14.12. For water (2) in the permeate, the selectivity
is given by: α 2,1 = (100 − w1 ) P ( w1 ) P
(100 − w1 ) F ( w1 ) F with w in wt% In this exercise, w1 = 12 wt% in the permeate and 17 wt% in the feed.
Therefore, the selectivity for water = α 2,1 = (100 − 12 ) P (12 ) P
(100 − 17 ) F (17 ) F = 1.50 Exercise 14.24
Subject: Laboratory data for separation of benzene (B) from cyclohexane (C) by a
pervaporation process. Consideration of Stage 2 in a 3stage process.
Given: Feed of 9,905 kg/h of 57.5 wt% B at 75oC to Stage 2. Retentate is 16.4 wt% B at
67.5oC. Permeate is 88.2 wt% B at 27.5oC. Total permeate flux = 1.43 kg/m2h. Selectivity for
B = 8.
Find: Flow rates of retentate and permeate in kg/h. Membrane area in m2.
Analysis: First, make a material balance to obtain the component flow rates in the feed,
retentate, and permeate. Let R = retentate flow rate in kg/h and P = permeate flow rate in kg/h.
By total overall material balance, R + P = 9905 (1) By an overall benzene material balance, 0.164R + 0.882P = 0.575(9905) = 5,695 (2) Solving Eqs. (1) and (2), P = 5,670 kg/h and R = 4,235 kg/h
Calculate the component flow rates in kg/h. Stream
Component:
B
C
Total Feed
kg/h
5,695
4,210
9,905 Retentate
kg/h
694
3,541
4,235 Permeate
kg/h
5,001
669
5,670 Membrane area = AM = total permeate flow rate/permeate flux = 5,670/1.43 = 3,965 m2 Exercise 14.25
Subject: Section 1 of a cheese whey ultrafiltration process to obtain a dry powder of combined
TP and NPN as in Example 14.13
Given: Conditions of Example 14.13, including a whey feed of 1,000,000 lb/.day with the same
composition and solute retention factors. Section 1 only with just two stages of continuous
bleedandfeed ultrafiltration to reach 55 wt% (dry basis) of combined TP and NPN.
Find: The component material balance in lb/day of operation, the % recovery (yield) from the
whey of the TP and NPN in the final concentrate, and the number of cartridges required.
Analysis: Section 1 is shown in Figure 14.31. The calculations are similar to those of Example
14.13 where two cases are considered for Section 1, a single stage and four stages. As is the case
in Example 14.13, the calculations assume equal membrane area for the two stages, and are
carried out by a double “trial and error” procedure using a spreadsheet with a Solver function.
First assume a membrane area per stage. In Example 14.13, the computed area was
12,500 ft2 for a single stage and 1,617 ft2 per stage in a fourstage system, giving a total of 6,468
total ft2. Assume a total area for two stages of 9,000 ft2 or 4,500 ft2 per stage.
Next, find by iteration, the overall concentration factor, R1 = P2 + R2, that gives the fresh
feed rate, F1, to the first stage, which is computed in Example 14.13 at 5,882 gal/h for 20 hr/day
operation. Use a spreadsheet, working backward from stage 2 to stage 1, using Equation (1) in
Example 14.13, where Jn = hourly membrane flux at 1/24 of that e...
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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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