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

# 5 57 508 lbh of a and 753 565 188 lbh of na2so4

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Unformatted text preview: a specification on y2. The following results were obtained for Cases 2a, 2b, and 2c: y2 F1 R2 P1 T1 T2 xF1 x1 y1 x2 a 0.74 126.58 26.58 86.58 0.684 0.693 0.5754 0.3900 0.6610 0.4827 b 0.761 164.64 64.64 124.64 0.757 0.481 0.5648 0.3590 0.6308 0.5103 c 0.78 271.64 171.64 231.64 0.853 0.259 0.5605 0.3300 0.6004 0.5376 For Case 2b, which corresponds to the same y2 as for Case 1, the membrane areas are as follows: 0.6308(0.757)(124.64)(2820) A M 1 in m 2 = = 254.5 m2 −5 9.0 × 10 0.510(1,551) − 0.761(76.6) (10,000) 0.76064(0.481)(60)(2820) = 93.9 m2 9.0 × 10 0.510(1,551) − 0.761(76.6) (10,000) The total membrane area for Case 2b is 348.4 m2, compared to a total of 390 m2 for Case 1. AM 2 in m2 = −5 Exercise 14.12 Subject: Importance of concentration polarization in dialysis. Given: From Example 14.7: At a certain location in a tubular membrane separator, for the dialysis of solute A from solvent B, concentrations for A of 5 x 10-2 kmol/m3 on the feed side and 1.5 x 10-2 kmol/m3 on the permeate side. Permeance for A = 7.3 x 10-5 m/s and feed passes through the tube of inside diameter = D = 0.4 cm. Permeate-side mass-transfer coefficient = 0.06 cm/s or 6 x 10-4 m/s For the feed side mass-transfer coefficient, Reynolds number = 25,000 and Schmidt number for solute A = 500, with a molecular diffusivity for A of 6.5 x 10-5 cm2/s. Find: Importance of concentration polarization. Analysis: The flux of A through the membrane, in terms of the three resistances is given by a modification of Eq. (14-53), with : NA = cA F − cA P 5.0 × 10 −2 − 15 × 10 −2 . = 1 1 1 1 1 1 + + + + −5 6 × 10 −4 k A F PM A kA P k A F 7.3 × 10 (1) Since the Reynolds number is greater than 10,000, turbulent flow exists inside the tube and if fully developed flow is assumed, the mass-transfer coefficient is given by a modification of Eq. (14-54), where, if the analogy to heat transfer is applied with Eqs. (3-165) and (3-166) for a circular tube, a = 0.023, dH = D, and the (dH/L)d correction for the entry effect is not necessary. In Example 14.7, the feed-side mass-transfer coefficient for NRe = 15,000 and D = 0.5 cm is computed to be 5.1 x 10-4 m/s. Therefore, for this exercise, by ratios, using Eq. (14-54), k A F = 51 × 10 . −4 0.5 0.4 25,000 15,000 0.8 = 9.6 × 10 −4 m / s Substituting into Eq. (1), 5.0 × 10 −2 − 15 × 10 −2 . 2.5 × 10 −2 2.5 × 10 −2 NA = = = = 2.1 × 10 −6 kmol / s - m2 1 1 1 1,042 + 13,700 + 1,667 16,409 + + −4 −5 −4 9.6 × 10 7.3 × 10 6 × 10 The sum of the feed-side and permeate-side mass-transfer resistances is 2,709 of a total of 16,409. Therefore 2,709/16,409 or 16.5% of the total resistance is due to the two external resistances. Therefore, concentration polarization is important, but not to a major extent. Exercise 14.13 Subject: Dialysis of an aqueous stream to separate Na2SO4 from a high-molecular-weight substance (A). Given: Aqueous feed of 100 gal/h at 20oC, containing 8 wt% Na2SO4 and 6 wt% A. Continuous countercurrent...
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