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ch14 - Exercise 14.1 Subject Differences of membrane...

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Exercise 14.1 Subject: Differences of membrane separations from some other separation operations. Find: How membrane separations differ from: (a) Absorption and stripping (b) Distillation (c) Liquid-liquid extraction (d) Extractive distillation Analysis: In general, membrane separations differ from absorption, stripping, distillation, liquid-liquid extraction, and extractive distillation in the following respects: 1. The separating agent for membrane separations is a semi-permeable membrane. 2. 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.
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Exercise 14.2 Subject: Calculation of permeabilities of hydrogen and methane in barrer units. Given: Feed gas at 450 psia and 200 o F, with the following component flow rates in lbmol/h: H 2 1,872.3, CH 4 193.1, C 2 11.4, Benzene 4.8, Toluene 4.2, and p-Xylene 0.6. Membrane system has a total area of 16,000 ft 2 , with a thickness of 0.3 microns. The permeate exits at 50 psia with 1,685.1 lbmol/h of H 2 and 98.8 lbmol/h of CH 4 . 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 H 2 and CH 4 in barrer units. Analysis: Use Eq. (14-1) and the definition of the barrer, 1 barrer = 10 -10 cm 3 gas (STP of 0 o C and 1 atm)-cm/(cm 2 -s-cm 2 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 = N H2 = 1,685.1/16,000 = 0.1053 lbmol/h-ft 2 Partial pressure driving force = Δ p H avg 2 ° ± = 450 1872 3 2086 3 187 2 302 4 2 50 16851 17839 3412 47 2 294 . . . . . . . . + ² ³ ´ ´ ´ µ · · · - ² ³ ´ µ · = - = psi From Eq. (14-1), permeability = P N l p M M H 2 2 2 H H avg 2 lbmol - m h - ft psi = = = × - - Δ ° ± 01053 0 3 294 1075 10 4 . ( . ) . μ To convert to barrer, use: 0.4536 kmol/lbmol, 22.42 x 10 6 cm 3 (STP)/kmol, 10 -4 cm/ μ m, 3600 s/h, 9.29 x 10 2 cm 2 /ft 2 , and 5.17 cm Hg/psi. Thus, ( ) ( )( ) ( ) ( )( ) H 2 6 4 4 4 5 2 3 -9 2 0.4536 22.42 10 10 1.075 10 1.075 10 5.882 10 9.29 10 3600 5.17 cm (STP)-cm 63.2 6.32 10 cm -s-cm barrer Hg - - - - ° ± × ° ± ² ³ = × = × × = ´ µ × ² ³ ´ µ × = M P Methane: Flux through the membrane = N CH4 = 98.8/16,000 = 0.00618 lbmol/h-ft 2 Δ p CH avg 4 ° ± = 450 1931 2086 3 94 3 302 4 2 50 988 17839 910 2 8 88 2 . . . . . . . . . + ² ³ ´ ´ ´ µ · · · - ² ³ ´ µ · = - = psi ( ) 4 CH 4 4 5 CH 10 CH avg 0.00618(0.3) 5.882 10 88. 12.4 bar r 2 10 re - - ° ± × = = = ² ³ Δ ´ µ M M N l P p
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Exercise 14.3 Subject: Separation of N 2 from CH 4 by an asymmetric polyimide polymer membrane. Given: Feed gas of 200 kmol/h of N 2 and 800 kmol/h of CH 4 at 5500 kPa and 30 o C. Permeances are 50,000 barrer/cm for N 2 and 10,000 barrer/cm for CH 4 . Retentate leaves at 5,450 kPa and 30 o C, containing 20 kmol/h of N 2 , while permeate leaves at 100 kPa and 30 o C, containing 180 kmol/h of N 2 .
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