Unformatted text preview: quation, with CF2 = CF ,
Raffinate, R2 = F1/CF2, P2 = AJ2, R1 = P2 + R2, and CF1 = F1/R1, P1 = AJ1, and backcalculated
F1 = P1 + R1. If the backcalculated 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 (1481), with the following form
of (1492):
( 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 4stage 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.
 Spring '11
 Levicky
 The Land

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