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

13 at 5882 galh for 20 hrday operation use a

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Unformatted text preview: equation to determine thermodynamic properties of the liquid phases. The liquid density of the feed = 8.319 lb/gal Feed rate = 100,000(8.319)/24 = 34,664 lb/h. The permeate and retentate flow rates are obtained by material balance: Total material balance: 34,664 = R + P (1) Ethyl acetate material balance: 0.02(34,664) = 0.002 R + 0.457 P (2) Solving Eqs. (1) and (2), R = 33,293 lb/h and P = 1,371 lb/h Using a MW of 80.10 for ethyl acetate, the following component material balance is obtained: Component Ethyl acetate Water Total lb/h: Feed 693 33,971 34,664 Retentate 66.5 33,226.5 33,293.0 Permeate 626.5 744.5 1,371.0 lbmol/h: Feed 7.87 1,885.73 1,893.60 Retentate 0.76 1,844.40 1,845.16 Permeate 7.11 41.33 48.44 Assume that the membrane flux is based on a liquid phase of permeate at 30oC. From the Chemcad program, the liquid density at 30oC is 58.82 lb/ft3 or 2.077 lb/L. Therefore, the volume flow of the permeate is 1,371/2.077 = 660 L/h. Therefore, since the flux is 1.0 L/m2-h, the membrane area = 660/1.0 = 660 m2. As discussed in Section 14.7, assume the permeate vapor leaves at the dew-point temperature at the permeate pressure of 3 cmHg. From the Chemcad program the permeate temperature = 26.2oC. As discussed in Section 14.7, compute the exit temperature of the liquid retentate by an adiabatic enthalpy balance. Using Chemcad, the retentate temperature = 15.5oC. Therefore, temperature drop of the feed = 30 – 15.5 = 14.5oC. Exercise 14.23 Subject: Calculation of permeances from laboratory data on pervaporation. Given: Permeation flux of 1.6 kg/m2-h for a 17 wt% ethanol in water feed, giving a permeate of 12 wt% ethanol at 60oC and a permeate pressure of 15.2 mmHg. Other conditions in Example 14.12. Find: Permeances for ethyl alcohol (1) and water (2). Selectivity for water. Analysis: Molecular weights for ethyl alcohol and water are 46.07 and 18.02, respectively. From Example 14.12, the vapor pressure of ethyl alcohol = 352 mmHg, and that of water = 149 mmHg. The mole fractions in the feed are: x1 = 0.17 / 46.07 = 0.0742 and . 017 (1.0 − 0.17) + 46.07 18.02 x2 = 1 − 0.0742 = 0.9258 The mole fractions in the permeate are: 012 / 46.07 . y1 = = 0.0506 and y2 = 1 − 0.0506 = 0.9494 012 (1.0 − 0.12) . + 46.07 18.02 From Eq. (14-39), with no sweep, the component fluxes are proportional to the weight fractions in the permeate. Therefore, N1 = 16(0.12) . 16(0.88) . = 0.00417 kmol / h - m2 and N 2 = = 0.07814 kmol / h - m2 46.07 18.02 For Eq. (14-80), we need the activity coefficients in the feed mixture. Obtain these from the van Laar equations given in Example 14.12: 0.9232(0.9258) γ 1 = exp 16276 . 1.6276(0.0742) + 0.9232(0.9258) γ 2 = exp 0.9232 16276(0.0742) . 16276(0.0742) + 0.9232(0.9258) . 2 = 3.488 2 = 1.014 Exercise 14.23 (continued) Analysis (continued) From Eq. (14-80), N1 0.00417 PM1 = = = 4.62 ×10 −5 kmol/h-m 2 − mmHg s γ1 x1 P − y1 p1 3.488(0.0742)(352) − 0.0506(15.2) 1 PM 2 = N2 0.07814 = = 6.23 × 10−4 kmol/h-m 2 − mmHg s...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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