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

# 3 c 884 1773 where c is in mg tcel solution and t is

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Unformatted text preview: : Effective diffusivity of benzene vapor in the pores of silica gel. Given: Silica gel with ρp = 1.15 g/cm3 = 71.8 lb/ft3 , εp = 0.48, dp = 30 Angstroms = 30 x 10-8 cm, and tortuosity = τ = 3.2. Diffusion of benzene vapor in air in the pores, where from Exercise 15.16, temperature = 70oF and pressure = 1 atm. Differential heat of adsorption = 11,000 cal/mol. Adsorption equilibrium constant from Eq. (1) of Example 15.10 = K = 5,120 (lb/ft3 particles)/(lb/ft3 gas). Find: Effective diffusivity of benzene vapor through air in the pores, accounting for bulk (molecular) diffusion, Knudsen diffusion, and surface diffusion. Analysis: From Eq. (15-75), for all three mechanisms of internal diffusion, De = εp 1 τ 1 1 + Di DK + Di ρpK s εp = 0.48 3.2 1 1 1 + Di DK + Di ρpK s 0.48 (1) From Perry's Handbook, at 32oF, the diffusivity of benzene in air at 1 atm = 0.077 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power. Therefore, Di = 0.077[(70 + 460)/(32 + 460)]1.75 = 0.0877 cm2/s. From Eq. (14-21), using 78 for the molecular weight of benzene and T = 70oF = 294 K , DK = 4850 dp(T/Mi)1/2 = 4850(30 x 10-8)(294/78)1/2 = 0.00282 cm2/s From Eq. (15-76), using m =1 because silica gel is an insulating adsorbent, −0.45 − ∆Hads −0.45 11,000 Di s = 0.016 exp = 0.016 exp = 3.34 × 10−6 cm2 / s mRT (1)(1987)(294) . In Eq. (1), the units of K are (g adsorbed/g adsorbent)/(g solute/cm3 of gas). Therefore, converting the units of the given K, K = (5,120/ρp)[62.4 (lb/ft3)/(g/cm3)] = 5,120(62.4)/71.8 = 4,450 (g adsorbed/g adsorbent)/(g solute/cm3 of gas). Substituting the above values of Di , DK, (Di)s, and K into Eq. (1) gives: De = 0.48 3.2 = 0.15 1 1 1 + 0.0877 0.00282 + ( 3.12 ×10 −6 ) (1.15)(4, 450) 0.48 1 + 0.0333 = 0.15(0.00273 + 0.0333) = 0.0054 cm 2 /s 11.40 + 355 Note that the internal diffusion process is controlled by surface diffusion. Exercise 15.20 Subject: Effective diffusivity of water vapor in the pores of activated alumina. Given: Activated alumina with ρp = 1.38 g/cm3, εp = 0.52, dp = 60 Angstroms = 60 x 10-8 cm, and tortuosity = τ = 2.3. Diffusion of water vapor in air in the pores, where from Exercise 15.17, temperature = 21oC and pressure = 653.3 kPa. Find: Effective diffusivity of water vapor through air in the pores, accounting for bulk (molecular) diffusion and Knudsen diffusion, but not surface diffusion. Analysis: From a modification of Eq. (15-75), De = εp 1 τ 1 1 + Di DK = 0.52 2.3 1 1 1 + Di DK (1) From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s. From Eq. (3-36), the diffusivity is proportional to T to the 1.75 power and inversely proportional to the pressure. Therefore, Di = 0.22(101.3/653.3)[(70 + 460)/(32 + 460)]1.75 = 0.0388 cm2/s From Eq. (14-21), using 18 for the molecular weight of water and T = 21oC = 294 K , DK = 4850 dp(T/Mi)1/2 = 4850(60 x 10-8)(294/18)1/2 = 0.0118 cm2/s Substituting the above values of Di and DK into Eq. (1) gives: De = 0.52 2.3...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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