Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Analysis from the results of example 1413 the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 3-stage 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/m2-h. 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 bleed-and-feed 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 four-stage 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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online