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

# 167 106 963 10 5 ms am dt 000173 cf solid matterunit

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Unformatted text preview: quation, with CF2 = CF , Raffinate, R2 = F1/CF2, P2 = AJ2, R1 = P2 + R2, and CF1 = F1/R1, P1 = AJ1, and back-calculated F1 = P1 + R1. If the back-calculated F1 is not equal to the actual value of 5,882 gal/h, then a new value of CF is assumed. The following values are calculated with the assumptions of A = 4,500 ft2 and CF = 30: CF2 = CF = 30, R2 = 5,882/30 = 196 gal/h, P2 = 4,500(0.411) = 1,851gal/h, R1 = 1,851 + 196 = 2,047 gal/h, and CF1 = 5,882/2,047 = 2.873, P1 = 4,500(0.929) = 4,182 gal/h, and backcalculated F1 = 4,182 + 2,047 = 6,229 gal/h. Since this is not the required value of 5,882, the Solver function, used with a spreadsheet gives CF = 36.487. However, the assumed area of 4,500 ft2 for each stage is not correct if it does not achieve the required value of 55 wt% (dry basis) for TP + NPN in the retentate from stage 2. To check this, the following calculations are made using the given values of the rejection factor, defined by (14-81), with the following form of (14-92): ( wt% i ) Fn CFn (1) ( wt% i ) Rn = CFn (1 − σi ) + σi For example, for TP in stage 1, with a wt% of 0.6 in the feed, σ = 0.97, and for A = 4,500 ft2 and CF1 = 2.873, Exercise 14.25 (continued) Analysis: (continued) ( wt% i )R n = 0.6 ( 2.873) 2.873 (1 − 0.97 ) + 0.97 = 1.63 The resulting wt% TP + NPN (dry basis) is not 55 wt%, even for CF = 36.487. Therefore, the assumed area is not correct. From the double trial and error, the following results are obtained in an effort to obtain the required value of 55 wt%: Assumed A, ft2 CF from Solver Wt% TP+NPN (dry basis) 4,500 3,800 3,750 3,730 36.487 18.847 17.878 17.502 65.79 56.22 55.35 55.00 The last result in the table gives the following material balance in lb/day, and % overall yield: Component TP NPN Lactose Ash B Fat Water Total Feed 6,000 3,000 49,000 8,000 500 933,500 1,000,000 Stage 1 Concentrate 5,733 1,464 20,327 3,384 500 362,064 393,472 Stage 2 Concentrate 4,874 293 3,183 545 500 47,741 57,136 % Overall Yield 81.2 9.8 6.5 6.8 100.0 The area per stage = 3,730 ft2 or a total of 7,460 ft2. The number of cartridges per stage = 141 or a total of 282 cartridges. Exercise 14.26 Subject: Section 2, involving diafiltration, of a cheese whey filtration process to obtain a dry powder of combined TP and NPN as in Example 14.13, wherein section 3 is eliminated. Given: Conditions of Example 14.13, including a whey feed of 1,000,000 lb/.day with the same composition and solute retention factors. Feed to the diafiltration section 2 is the 55 wt% concentrate from the 4-stage ultrafiltration section 1 of Example 14.13. The diafiltration section is to achieve 85 wt% concentrate (dry basis) of combined TP and NPN. Assumption: Assume 4 stages of diafiltration, as in Example 14.13. Find: A suitable set of operating conditions for section 2. Analysis: From the results of Example 14.13, the component flow rates in the concentrate leaving section 1 and feeding the diafiltration section is as follows in lb/day: Component TP NPN Lactose Ash BF Water Feed rate t...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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