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

# Permeance for water 00134 lbft2 h psi osmotic

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Unformatted text preview: + 622 ∆παϖγ = − 5.6 = 514.5 − 5.6 = 508.9 psi 2 (∆P - ∆π)avg = 1,950 - 508.9 = 1,441 psi nH 2O 10, 000, 000 gal From Eq. (1), PM H O = = = 3.47 × 10−3 2 AM ( ∆P − ∆π )avg 2, 000, 000(1, 441) day-ft 2 − psi Now consider the actual crossflow case. The (P - π) on the permeate side is very small (less than 4% of that on the feed-permeate side). Therefore, the change in (∆P - ∆π) along the length of the membrane is little affected by conditions on the permeate side. Therefore, the case computed above well approximates crossflow. From Eq. (14-70), SP = ( csalt )Permeate / ( csalt )Feed = 0.00855 / 0.620 = 0.0138 Exercise 14.17 Subject: Reverse osmosis with multiple stages. Given: 100 gal/min of feed. Assumptions: All units of same size. Find: Design 1 - Drawing for single stage of 4 units in parallel to obtain 50% recovery. Design 2 - Drawing and % recovery for 2 stages in series with respect to retentate (4 units in parallel followed by 2 units in parallel). Design 3 - Drawing and % recovery for 3 stages in series with respect to retentate (4 units in parallel, then 2 units in parallel, then 1 unit). Analysis: Design 1: The design is shown in a sketch below. Consider the following conditions for the single stage of 4 units in parallel. To simplify calculations and comparison of designs, assume: 1. Water feed dilute in dissolved salts. Take the 100 gal/min of feed = F, containing 1 lb salt/100 lb of water = XF1. 2. Density of all streams = 8.33 lb water/gal = ρ. 3. Driving force for water transfer across membrane = ∆P - ∆π = 1,000 psi for all units. 4. Water flux across all membrane units = 4.5 gal/day-ft2. 5. Salt concentration in permeate from the stage = 0.01 lb salt/100 lb of water = Y1. Therefore for the single stage of 4 units in parallel, the water transfer rate for 50% recovery of potable water = 0.5(100) = 50 gal/min. Therefore, the total membrane area is, AM = (50)(60)(24)/4.5 = 16,000 ft2 or 4,000 ft2 per unit. Compute the permeance for water transport from a modification of Eq. (14-69): mwater 8.33(50) = = 1.666 × 10 −5 lb / min - ft 2 - psi AM ( ∆P − ∆π ) 16,000(1,000) Compute the permeance for salt transport from a modification of Eq. (14-52), where the transport of salt across the membrane = 50(8.33)(0.0001) = 0.0416 lb/min, and a concentration difference in terms of mass ratios is used, based on an arithmetic average on the feed side and perfect mixing on the permeate side. Therefore, PM water = PM salt = AM msalt 0.0416 = = 175 × 10−4 lb / min - ft 2 - ( ∆ mass ratio) . 0.01 + 0.0199 XF + XR 16,000 − 0.0001 − YP 2 2 Exercise 14.17 (continued) Analysis: (continued) Design 2: The design is shown in a sketch below. Retentate, R1 from Stage 1 is feed to the Stage 2. Stage 1 consists of 4 membrane modules in parallel, each with 4,000 ft2 for a total of 16,000 ft2, and Stage 2 consists of 2 membrane modules in parallel, each with 4,000 ft2 for a total of 8,000 ft2. Permeates from Stage 1 and...
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