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

29 x 10 2 0001704 ft2 therefore number of fibers n

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Unformatted text preview: Rate-based modeling must be used for membrane separations because an equilibrium model does not apply. Compared to distillation and extractive distillation, membranes usually can not achieve sharp separations Compared to liquid-liquid extraction, absorption, and stripping, membranes produce products that are usually miscible, rather than immiscible. Exercise 14.2 Subject: Calculation of permeabilities of hydrogen and methane in barrer units. Given: Feed gas at 450 psia and 200oF, with the following component flow rates in lbmol/h: H2 1,872.3, CH4 193.1, C2 11.4, Benzene 4.8, Toluene 4.2, and p-Xylene 0.6. Membrane system has a total area of 16,000 ft2, with a thickness of 0.3 microns. The permeate exits at 50 psia with 1,685.1 lbmol/h of H2 and 98.8 lbmol/h of CH4. The retentate exits at 450 psia. Assumptions: Mass transfer driving force based on the difference between the average partial pressure of the feed and retentate, minus the partial pressure of the permeate. Find: The permeabilities of H2 and CH4 in barrer units. Analysis: Use Eq. (14-1) and the definition of the barrer, 1 barrer = 10-10 cm3 gas (STP of 0oC and 1 atm)-cm/(cm2-s-cm2 Hg pressure head) The total feed flow rate = 2086.3 lbmol/h. The permeate rate = 1783.9 lbmol/h. The retentate flow rate = 2086.3 - 1783.9 = 302.4 lbmol/h. Hydrogen: Flux through the membrane = NH2 = 1,685.1/16,000 = 0.1053 lbmol/h-ft2 Partial pressure driving force = 1872.3 187.2 + . 2086.3 302.4 − 50 16851 = 3412 − 47.2 = 294 psi ∆pH 2 = 450 . avg 2 1783.9 From Eq. (14-1), permeability = PM H = 2 N H2 l M ∆pH 2 avg = 0.1053(0.3) lbmol - µm = 1075 × 10 −4 . 294 h - ft 2 − psi To convert to barrer, use: 0.4536 kmol/lbmol, 22.42 x 106 cm3 (STP)/kmol, 10-4 cm/µm, 3600 s/h, 9.29 x 102 cm2/ft2, and 5.17 cm Hg/psi. Thus, ( 0.4536 ) ( 22.42 ×106 ) (10−4 ) −4 PM H2 = 1.075 × 10 = 1.075 × 10−4 5.882 × 10−5 = 2 ( 9.29 ×10 ) ( 3600 )( 5.17 ) cm3 (STP)-cm 6.32 × 10 = 63.2 barrer cm 2 -s-cm Hg Methane: Flux through the membrane = NCH4 = 98.8/16,000 = 0.00618 lbmol/h-ft2 1931 . 94.3 + 98.8 ∆pCH 4 = 450 2086.3 302.4 − 50 = 910 − 2.8 = 88.2 psi . avg 2 1783.9 -9 PM CH4 = N CH4 lM ( ∆p ) CH 4 avg = 0.00618(0.3) 5.882 ×10 −5 = 12.4 barrer 88.2 10−10 Exercise 14.3 Subject: Separation of N2 from CH4 by an asymmetric polyimide polymer membrane. Given: Feed gas of 200 kmol/h of N2 and 800 kmol/h of CH4 at 5500 kPa and 30oC. Permeances are 50,000 barrer/cm for N2 and 10,000 barrer/cm for CH4. Retentate leaves at 5,450 kPa and 30oC, containing 20 kmol/h of N2, while permeate leaves at 100 kPa and 30oC, containing 180 kmol/h of N2. Assumptions: Diffusion driving force based on arithmetic average partial pressures of the feed and retentate on the feed side, and based on the permeate product on the permeate side. Find: Membrane surface are in m2. CH4 flow rate in the permeate in kmol/h. Analysis: From Eq. (14-1), nN 2 = 180 kmol / h = AM PM N2 ∆PN 2 avg and nCH 4 = AM PM 2 CH 4 ∆PCH 4 avg Convert permeances in barrer/cm to kmol/m -h-k...
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