Unformatted text preview: ble 3.5,
and are listed below. The parachors for ethyl acetate, obtained by structural contributions from
Table 3.6, is 2(55.5) + 40.0 + 63.8 = 214.8. Molecular volumes are obtained from Table 3.3, and
are listed below. Solvent viscosities are obtained from Perry's Handbook. The results of
applying Eq. (1) are as follows
Solvent, B
Benzene
Acetone
Ethyl acetate PB
205.3
161.5
214.4 υB
96
74
106.1 µB , cP
0.60
0.32
0.45 DA,B (HM)
1.5 x 105
2.5 x 105
1.93 x 105 DA,B (Expt.)
2.09 x 105
2.92 x 105
2.18 x 105 The predictions by the HaydukMinhas equation are low, but quite good for ethyl acetate.
For water, the WilkeChang equation, Eq. (339), is most applicable,
DA,B 7.4 × 10 −8 φ B M B
=
0
µ Bυ A.6 1/ 2 T (2) From Table 3.3, υA = (2)14.8 + 4(3.7) + 7.4 + 12 = 63.8 cm3/mol.
Solvent viscosity = 0.95 cP, MB = 18, and φB = 2.6. Substitution into Eq. (1) gives, 7.4 × 10−8 ( 2.6 × 18 )
=
0.95(63.8)0.6 1/ 2 DA,B 298 = 1.3 × 10 −5 cm 2 /s This is close to the experimental value of 1.19 x 105 cm2/s. Exercise 3.14
Subject: Vapor diffusion through an effective film thickness.
Given: Water (A) at 11oC (284 K) evaporating into dry air (B) at 25oC (298 K) at the rate of
0.04 g/hcm2.
Assumptions: 1 atm pressure. Ideal gas law. Raoult's law for water vapor mole fraction.
Find: Effective stagnant air film thickness, assuming molecular diffusion of the water vapor
through the air film.
Analysis: Apply Eq. (331) for Fick's law with the bulk flow effect. Solving for film thickness,
cDA,B ln
∆z = 1 − yA b 1 − yA I (1) NA From the ideal gas law, using T = (284 + 298)/2 = 291 K, c = P/RT = 1/(82.06)(284) =
4.29 x 105 mol/cm3.
Molar flux of water vapor = NA = 0.04/MA = 0.04/(18)(3600) = 6.17 x 107 mol/scm2.
Mole fraction of water vapor in the bulk dry air = yAb = 0 Mole fraction of water vapor at the interface, yA I = Ps at 11oC/P = 0.191/14.7 = 0.013.
Estimate diffusivity of water vapor in air at 291 K and 1 atm from Eq. (336) of Fuller, Schettler,
and Giddings, with,
2
M A,B =
= 22.2
1
1
+
18 29
From Table 3.1,
VA = 131,
.
VB = 19.7 DA,B = 0.00143(291)1.75
= 0.24 cm2 / s
(1)(22.2)1/ 2 [1311/ 3 + 19.71/ 3 ]2
. Substitution into Eq. (1),
∆z = 4.29 × 10−5 ( 0.24 ) ln
6.17 × 10−7 Note that the bulk flow has little effect here. 1− 0
1 − 0.013 = 0.218 cm Exercise 3.15
Subject: Mass transfer of isopropyl alcohol (A) by molecular diffusion through liquid water and
gaseous nitrogen at 35oC (308 K) and 2 atm
Given: Critical conditions for nitrogen and isopropyl alcohol and molar liquid volume of
isopropyl alcohol.
Assumptions: Ideal gas law.
Find: (a)
(b)
(c)
(d)
(e)
(f)
(g) Diffusivity of A in liquid water by WilkeChang equation.
Diffusivity of A in gaseous nitrogen by the Fuller, Schettler, Giddings equation.
The product, DABρΜ = DAB c for part (a).
The product, DABρΜ = DAB c for part (b).
Comparison of diffusivities in parts (a) and (b).
Comparison of results from parts (c) and (d).
Conclusions about diffusion in the liquid versus the...
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 Spring '11
 Levicky
 The Land

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