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0.0374
2.0937
0.0220
0.0673
0.501
0.466
1.66x105
1.78x105
0.0972
1.6153
0.0519
0.2057
0.773
0.757
0.727x105 0.777x105
0.3141
0.7090
0.2599
0.4170
0.785
0.783
0.645x105 0.651x105
0.5199
0.3136
0.5392
0.949x105 0.985x105
0.6143
0.664
0.651
0.1079
0.8645
0.7087
0.414
0.353
1.56x105
1.72x105
0.8140
0.9193
0.0002
1.3177
0.9392
0.186
0.121
2.08x105
2.25x105
0.9591 0.0077 1.3999
1.0000
2.51x105
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: 1M aqueous NaOH at 25oC (298 K).
Assumptions: Dilute solution.
Find: Diffusivity of the Na+ (A) and OH (B) ions.
Analysis: Eq. (347) 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. (347),
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 1M solution. Exercise 3.18
Subject: Estimation of the diffusivity of an electrolyte and comparison with experiment.
Given: Experimental value of 1.28 x 105 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. (347) 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. (349), with ε = 0.45 and τ =2.5. Estimate diffusivity of nitrogen in hydrogen
at 573 K and 20 atm from Eq. (336) 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. (349),
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 4inch inside diameter spherical
pressure vessel of steel, with a 0.125inch 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 109
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.
 Spring '11
 Levicky
 The Land

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