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Diafiltration
5,245
353
3,476
603
500
49,897
60,074 Total Rejection,
σ
0.970
0.320
0.085
0.115
1.000 As in Example 14.13, assume the same membrane area for each of the 4 stages. Also, assume
the same flow rate of dilution water, Wn , to each stage, and set the permeate rate for each stage
equal to Wn . Therefore, the concentrate rate leaving each stage equals the feed rate above of
60,074 lb/day. This simplification leads to Eq. (2) of Example 14.13, which here for each stage
is:
W + 60, 074
(1)
60, 074
Equation (3) of Example 14.13 then gives the flow rate of each component in the concentrate
leaving stage n:
CF = ( mi ) R n = ( mi ) F 1
CF (1 − σi ) + σi n where, here, n = 4, and values of ( mi ) F and σi are given in the above table. (2) Exercise 14.26 (continued)
Analysis: (continued)
This exercise is solved by assuming a dilution water flow rate, W, for each stage, calculating CF
from (1), and calculating the flow rates of the components in the concentrate from the fourth
diafiltration stage using (2). If the resulting wt% (dry basis) of NP + NPN is not 85 wt%, the
procedure is repeated with a new assumed value of W. The procedure is carried out with a
spreadsheet using the Solver function.
The result is a dilution water flow rate of 57,240 lb/day for each of the four diafiltration stages.
The corresponding value of CF = 1.953. The component flow rates in the retentate leaving the
fourth stage are: Component
TP
NPN
Lactose
Ash
BF
Water Total Final concentrate
from Diafiltration,
Lb/day
4,686
48
283
52
500
54,505
60,074 This water flow rate per diafiltration stage is 2.45 times that in Example 14.13. The yield from
the feed to diafiltration is (4,686 + 48)/(5,245 + 353) x 100% = 84.6%, which is less than the
91.9% in Exercise 14.13. The membrane flux is the same as in Example 14.13, namely 0.5415
gal/hft2, but the volumetric permeate flow rate per stage is 57,240/[(20)(8.5)] = 337 gal/h. The
membrane area per stage is, therefore, 337/0.5415 = 622 ft2 and the number of cartridges per
stage is 622/26.5 = 24. Exercise 14.27
Subject: Deadend microfiltration of diluted skim milk using a combined operation of constant
permeate rate followed by constant pressure drop.
Given: Diluted skim milk with a protein concentration of 4.3 g/L. A constant permeate rate of
10 mL/min in stage 1 to a pressure drop of 25 psi, followed by stage 2 at this pressure drop until
the permeate rate drops to 5 mL/min. Membrane area of 0.00173 m2. Viscosity of 1 cP.
Assumption: Values of Rm = 1.43 x 1010 m1 and K2 = 3.78 x 1012 m/kg from Example 14.14.
Find: Permeate flux and cumulative permeate volume as a function of time.
Analysis: In stage 1, the permeate rate in m3/s for a given rate of 10 mL/min is,
−6
dV 10 (10 )
=
= 0.167 × 10−6 m3 /s
dt
60 The permeate flux in stage 1 is,
J= 1 dV 0.167 × 10−6
=
= 9.63 ×10 −5 m/s
AM dt
0.00173 cF = solid matter/unit volume of feed = 4.3 g/L = 4.3 kg/m3
From Example 14.14, viscosity = µ = 1 cP = 0.001 Pas
Pressure...
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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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