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

For each step in x eq 4 is used to calculate use eq 6

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Unformatted text preview: eously gives, R1 = 599.5 lb/h and P2 = 400.5 lb/h Dissolved solids in purified water, P2, equals [300/(1,000,000 + 300)](400.5) = 0.12 lb/h Dissolved solids in the concentrate (retentate from Stage 1) = 150 - 0.12 = 149.88 lb/h Compute material balance for Stage 2: Let P1 = permeate flow rate from Stage 1 = feed flow rate to Stage 2. Dissolved solids material balance: 0.004 P1 = 0.12 + 0.026 (P1 - 400.5) (3) Solving Eq. (3), P1 = 467.9 lb/h, containing 0.004(467.9) = 1.87 lb/h dissolved solids Retentate flow rate from Stage 2 = 467.9 - 400.5 = 67.4 lb/h, containing 1.87 - 0.12 = 1.75 lb/h dissolved solids The complete material balance is as follows: Component Dissolved solids Water Total Flow rate, lb/h: Fresh Combined feed feed to Stage 1 150 151.75 850 1,000 915.65 1,067.40 Retentate from Stage 1 149.88 Permeate from Stage 1 1.87 Retentate from Stage 2 1.75 Permeate from Stage 2 0.12 449.62 599.50 466.03 467.90 65.65 67.40 400.38 400.50 Exercise 14.18 (continued) Analysis: (continued) Now calculate the membrane area of each stage using Eq. (14-69). Assume osmotic pressure = π = C (wt% solids), with π = 1,700 psia for 25 wt% solids. Therefore, C = 1,700/25 = 68. For the feed, π = 68(151.75/1,067.40)100 = 967 psia For the retentate from Stage 1, π = 68(25) = 1,700 psia For the permeate from Stage 1, π = 68(0.4) = 27.2 psia For the retentate from Stage 2, π = 68(2.6) = 176.8 psia For the permeate from Stage 2, π = 68(0.12/400.50)100 = 2.0 psia nH 2 O nH 2 O From Eq. (14-69), AM = = PM H O ∆P − ∆π avg 0.0134 ∆P − ∆π avg 2 Stage 1: ∆P − ∆π avg = (1,756 - 15) - [(967 + 1,700)/2 - 27.2] = 1,741 - 1,306 = 435 psi AM = Stage 2: ∆P − ∆π avg 466.03 = 80 ft 2 0.0134(435) = (518 - 15) - [(27.2 + 176.8)/2 - 2.0] = 503 - 100 = 403 psi AM = 400.38 = 74 ft 2 0.0134(403) Exercise 14.19 Subject: Recovery of VOCs from air by gas permeation. Given: Feed gas of 1500 scfm (0oC, 1 atm) of air at 40oC and 1.2 atm, containing 0.5 mol% acetone (A). Pressure of permeate side = 5 cmHg. Membrane of thin-composite with 2µm-thick silicone rubber skin gives 4 barrer for air and 20,000 barrer for acetone. Assumptions: Crossflow. Find: Membrane area for retentate containing 0.05 mol% acetone and permeate containing 5 mol% acetone. Analysis: First complete the overall material balance. Feed flow rate = 1,500(60)/359 = 250.7 lbmol/h Acetone in the feed = 0.005(250.7) = 1.25 lbmol/h Air in the feed = 250.7 - 1.25 = 249.45 lbmol/h Overall total molar balance: F = 250.7 = R + P (1) Overall acetone molar balance: 1.25 = 0.0005 R + 0.05 P (2) Solving Eqs. (1) and (2) simultaneously, R = 228.0 lbmol/h and P = 22.7 lbmol/h The resulting component material balance is as follows: lbmol/hr: Component: Feed Retentate Permeate Acetone 1.25 0.11 1.14 Air 249.45 227.89 21.56 Total 250.70 228.00 22.70 Convert the given permeabilities in barrer to permeances in American Engineering units. 20,000 PM A = = 10 × 108 barrer / cm . −4 2 × 10 = 10 × 10 −2 cm3 (STP...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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