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

B fraction of a monolayer adsorbed if the adsorption

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Unformatted text preview: o 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/h-ft2, 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: Dead-end 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 m-1 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 Pa-s Pressure...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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