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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online