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

0000 251x10 5 these results give the following plot

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Unformatted text preview: 0-5 0.0374 2.0937 0.0220 0.0673 -0.501 -0.466 1.66x10-5 1.78x10-5 0.0972 1.6153 0.0519 0.2057 -0.773 -0.757 0.727x10-5 0.777x10-5 0.3141 0.7090 0.2599 0.4170 -0.785 -0.783 0.645x10-5 0.651x10-5 0.5199 0.3136 0.5392 0.949x10-5 0.985x10-5 0.6143 -0.664 -0.651 0.1079 0.8645 0.7087 -0.414 -0.353 1.56x10-5 1.72x10-5 0.8140 0.9193 0.0002 1.3177 0.9392 -0.186 -0.121 2.08x10-5 2.25x10-5 0.9591 -0.0077 1.3999 1.0000 2.51x10-5 These results give the following plot showing that the liquid diffusivities do not vary linearly with composition. Exercise 3.17 Subject: Estimation of the diffusivity of an electrolyte. Given: 1-M aqueous NaOH at 25oC (298 K). Assumptions: Dilute solution. Find: Diffusivity of the Na+ (A) and OH- (B) ions. Analysis: Eq. (3-47) of Nernst and Haskell applies, with n+ = 1 and n- = 1. From Table 3.7, λ+ = 50.1 for Na+ , and λ− = 197.6 for OH-. From Eq. (3-47), RT DA,B = F2 1 1 + n+ n− 1 λ+ + 1 λ− = 1 1 + 1 1 1 1 + 50.1 197.6 8.314 ( 298 ) 96,500 2 = 2.1× 10−5 cm 2 /s Note that DA,B may be 10 to 20% higher for 1-M solution. Exercise 3.18 Subject: Estimation of the diffusivity of an electrolyte and comparison with experiment. Given: Experimental value of 1.28 x 10-5 cm2/s for 2 M aqueous NaCl at 18oC (291 K). Assumptions: Dilute solution Find: Diffusivity of Na+ (A) and Cl- (B) ions. Analysis: Eq. (3-47) of Nernst and Haskell applies, with n+ = 1 and n- = 1. From Table 3.7, for 25oC, λ+ = 50.1 for Na+ , and λ− = 76.3 for Cl-. Correction to ionic conductances for 18oC = T/334 µwater = 291/(334)(1.05) = 0.83 1 1 1 1 RT + 8.314 ( 291) + n+ n− 1 1 DA,B = = = 1.3 × 10−5 cm 2 /s 1 1 1 1 96,500 2 + F2 + 50.1(0.83) 76.3(0.83) λ+ λ− This value compares very well with the experimental value. Exercise 3.19 Subject: Estimation of diffusivity of N2 (A) in H2 (B) in pores of a solid catalyst. Given: Catalyst at 300oC (573 K) and 20 atm, with porosity of 0.45 and tortuosity of 2.5. Assumptions: Only mass transfer mechanism is ordinary molecular diffusion. Correction for high pressure is not necessary because of high temperature. Find:. Diffusivity Analysis: Use Eq. (3-49), with ε = 0.45 and τ =2.5. Estimate diffusivity of nitrogen in hydrogen at 573 K and 20 atm from Eq. (3-36) of Fuller, Schettler, and Giddings, with, M A,B = From Table 3.1, 2 1 1 + 28 2.016 VA = 18.5, DA,B = = 3.76 VB = 612 . 0.00143(573)1.75 = 0124 cm2 / s . 1/ 2 1/ 3 1/ 3 2 (20)(3.76) [18.5 + 612 ] . From Eq. (3-49), Deff = ε DA,B (0.45)(0.124) = = 0.022 cm 2 / s τ 2.5 Exercise 3.20 Subject: Diffusion of hydrogen through the steel wall of a spherical pressure vessel. Given: Gaseous hydrogen (A) stored at 150 psia and 80oF in a 4-inch inside diameter spherical pressure vessel of steel, with a 0.125-inch wall thickness. Solubility of hydrogen in steel at these conditions = 0.094 lbmol/ft3. Diffusivity of hydrogen in steel at these conditions = 3.0 x 10-9 cm2/s. Air outside the vessel with zero partial pressure of hydrogen. Assumptions: Henry's law for solubility of hydrogen i...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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