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

Mass transfer resistance of the water film does not

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Unformatted text preview: 7(42.3) 50,000 From Eq. (2), k c = 2/3 0.948 2.41 = 0121 m/s . Using properties in Example 3.14, PAs 10 cA s = yA s c = c= (0.0327) = 4.3 × 10−4 kmol / m3 and cA ∞ = 0 P 760 ( ) From Eq. (1), N A = kc cA s − cA∞ = (0.121)(4.3 × 10−4 − 0) = 5.20 × 10−5 kmol/s-m2 From Eq. (3-171), packed bed correlation can't used with NRe = 50,000. Exercise 3.35 Subject: Experimental data for stripping of CO2 (A) from water by air (B) in a wetted-wall tube. Given: Location in the tube where P = 10 atm, T =25oC (298 K), CO2 flux from water into air = NA = 1.62 lbmol/h-ft2, partial pressure of CO2 at gas-liquid interface = pA i = 8.2 atm, and partial pressure of CO2 in bulk air = pA b = 0.1 atm. Diffusivity of CO2 in air = DAB = 1.6 x 10-2 cm2/s. Assumptions: Neglect stripping of water, but account for the bulk flow effect. Mass-transfer resistance of the water film does not need to be considered. Ideal gas law. Find: Using the film theory, find the mass-transfer coefficient, kc , for the gas phase and the film thickness, δ. Analysis: Solve the problem in cgs units. From Eq. (3-105) and a rearrangement of Eq. (3-189), kc = N A 1 − yA LM c yA i − yA b = DAB δ (1) NA = 1.62 lbmol/h-ft2 = 0.00022 mol/s-cm2 From the ideal gas law, c =P/RT = (10)/(82.06)(298) = 0.000409 mol/cm3 yA i = pA i (1 − yA ) LM = From Eq. (1), kc = P = 8.2 = 0.82 10 yA b = pA b P = 0.1 = 0.01 10 (1 − yA b ) − (1 − yA i ) (1 − 0.01) − (1 − 0.82) = = 0.475 (1 − yA b ) (1 − 0.01) ln ln (1 − 0.82) (1 − yA i ) (0.00022)(0.475) = 0.315 cm/s (0.000409) ( 0.82 − 0.01) From Eq. (1), δ = DAB / kc = 1.6 x 10-2/0.315 = 0.051 cm Exercise 3.36 Subject: Experimental data for absorption of CO2 (A) from air into water (B) using a packed column of Pall rings. Given: Location in a packed column where pA i = 150 psia, cA b = 0 , and NA = 0.017 lbmol/h-ft2 (2.31 x 10-6 mol/s-cm2). Diffusivity of CO2 in water = DAB = 2.0 x 10-5 cm2/s. Henry's law for CO2 between air and water is p, psia = 9,000 x. Assumptions: Negligible bulk flow effect. Only the mass-transfer resistance in the water need be considered. Neglect stripping of water. Find: (a) Liquid-phase mass-transfer coefficient, kc , and film thickness, δ, for the film theory. (b) Contact time for penetration theory. (c) Average eddy residence time and probability distribution for surface renewal theory. Analysis: Solve the problem in cgs units. (a) From Eq. (3-105) and a rearrangement of Eq. (3-189), kc = NA D = AB δ c xA i − xA b (1) Total liquid-phase concentration, since almost pure water, = c = ρ/MB = 1.0/18 = 0.0556 mol/cm3 From given Henry's law, xA i = pA i /9,000 = 150/9,000 = 0.0167 and xA b = 0.0. From Eq. (1), kc = 2.31× 10−6 = 0.0025 cm/s (0.0556) ( 0.0167 − 0 ) From Eq. (1), δ = DAB /kc = 2 x 10-5/0.0025 = 0.0080 cm (b) From a rearrangement of Eq. (3-194), tc = 4 DAB 4(2 × 10 −5 ) = = 4.08 s (3.14)(0.0025) 2 πkc2 1 DAB 2 × 105 (c) From Eq. (3-201), t = = 2 = = 3.2 s s kc (0.0025) 2 From Eq. (3-196), the probability distribution is: φ{t} = s exp(− st ) = 0.313 exp(−0.313t ) Exercise 3.36 (...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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