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

# 5 microns 05 x 10 4 cm from eq 14 1 pm 133 pm co2 co2

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Unformatted text preview: Analysis: (continued) d AM = PM O 2 ydn , xPF − yPP ∆AM = 170 × 10 . −2 y∆n x (17.64) − y (0.9671) (9) where y is the local value given by Eq. (3) and ∆n is the incremental decrease in the feed flow. Using the equations, the calculations are carried out on a spreadsheet with the results below. From the spreadsheet results, the membrane area is, by interpolation, 314,500 ft2 when the acetone mole fraction in the permeate = 0.05. However, at that condition, the calculations show a retentate with an acetone mole fraction of 0.00136, where 0.0005 is desired. At that value, the permeate acetone content drops to 3.55 mol% and the membrane area increases to 548,400 ft2. Analysis: (continued) Exercise 14.19 (continued) Exercise 14.20 Subject: Separation of air by gas permeation. Given: Permeate of 5 ton/day (2,000 lb/ton) of 40 mol% O2 and retentate of 90 mol% N2 from a feed of air. 500 psia on feed side and 20 psia on permeate side. Assumptions: Crossflow. No pressure drop on feed side. Find: Membrane area in m2 for the following membranes: Company A Module type Hollow-fiber 15 Permeance for O2, barrer/µm Ratio of O2 to N2 permeances 3.5 B Spiral-wound 35 1.9 Analysis: First calculate the overall material balance. P = 5(2,000) = 10,000 lb/day The average molecular weight of the permeate = [0.4(32) + 0.6(28)] = 29.6 Therefore, P = 10,000/29.6 = 337.8 lbmol/day. Total molar balance: F = R + 337.8 (1) O2 molar balance: 0.21F = 0.1R + 0.4(337.8) = 0.1R + 135.1 (2) Solving Eqs. (1) and (2), F = 921.2 lbmol/h and R = 583.4 lbmol/h On a component basis, the overall material balance is: lbmol/day: Component: Feed Retentate Permeate O2 193.5 58.3 135.2 N2 727.7 525.1 202.6 Total 921.2 583.4 337.8 Now check each module type to see if it is even possible to obtain a permeate of 40 mol% O2. For crossflow, the initial oxygen content of the local permeate at the feed end must be greater than 40 mol% because the mol% will decrease as the feed moves to the other end. PF = PR = 500 psia, PP = 20 psia, r = PP/PF = 20/500 = 0.04 αx Rearranging Eq. (14-36), y = (3) ,where, y and x refer to O2 and α is for O2 to N2. 1 + (α − 1) x From Eq. (14-43), α = α * 2 -N 2 O x (α − 1) + 1 − rα 0.21(α − 1) + 1 − 0.04α = α * 2 -N 2 O x (α − 1) + 1 − r 0.21(α − 1) + 1 − 0.04 For Company A, α * 2 -N 2 = 3.5. Substituting this into Eq. (4) and solving, α = 3.28. O Substituting this α into Eq. (3), y = 0.4658. Therefore, consider Company A. For Company B, α * 2 -N 2 = 1.9. Substituting this into Eq. (4) and solving, α = 1.84. O Substituting this α into Eq. (3), y = 0.3285. Therefore, Company B is dropped. (4) Exercise 14.20 (continued) Analysis: Company A (continued) 15 PM O = −4 = 150,000 barrer / cm 2 10 = 150 × 10 −5 cm3 (STP) / cm2 - s - cmHg . Convert permeance to American Engineering units of lbmol/ft2-h-psi: 1 barrer/cm = 1 x 10-10 (30.48 cm/ft)2 (76 cmHg/14.696 psi) / (22,400 x 454 cm3 (STP)/lbmol) = 1.70 x 10-10 lbmol/ft2-h-ps...
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