Unformatted text preview: vapor phase. Analysis: (a) For the WilkeChang equation (339), use φB = 2.6 and MB =18.
From Perry's Handbook, µB = 0.78 cP.
From Table 3.3, υΑ = 3(14.8) + 8(3.7) + 7.4 = 81.4
From Eq. (339), 7.4 × 10−8 ( 2.6 × 18 )
=
0.78(81.4)0.6 1/ 2 DA,B 308 = 1.43 × 10−5 cm 2 /s (b) For the FullerSchettlerGiddings Eq. (336), use T = 308K and P = 2 atm, with,
2
M A,B =
= 38.3
1
1
+
60.4 28
From Table 3.1, VA = 3(15.9) + 8(2.31) + 611 = 72.3
. DA,B = VB = 18.5 0.00143(308)1.75
= 0.056 cm 2 / s
1/ 2
1/ 3
1/ 3 2
(2)(38.3) [72.3 + 18.5 ] Exercise 3.15 (continued)
Analysis: (continued)
(c) The molar density of liquid water = ρ/M = 1/18 = 0.056 mol/cm3.
Therefore, DABρΜ = 1.43 x 105 (0.056) = 8.01 x 107 mol/scm
(d) From the ideal gas law, the molar density of gaseous nitrogen =
P/RT = 2/(82.06)(308) = 7.91 x 105 mol/cm3.
Therefore, DABρΜ = 0.056 (7.91 x 105) = 4.43 x 106 mol/scm
(e) The diffusivity in the gas is about 3 orders of magnitude more than in the liquid.
(f) The product of diffusivity and molar density in the gas is less than one order of
magnitude more than in the liquid.
(g) For equal mole fraction gradients, diffusion through the liquid phase is comparable to
that in the gas phase. Exercise 3.16
Subject: Liquid diffusivities for the ethanol (A) benzene (B) system at 45oC (318 K) over the
entire composition range.
Given: Experimental activity coefficients in Exercise 2.23.
Find: Effect of composition on diffusivities of both ethanol and benzene.
Analysis: Use Eq. (342) of HaydukMinhas to estimate the infinitedilution liquid diffusivities,
with liquid viscosities from Perry's Handbook. From Table 3.5, the parachor of benzene is
given. The parachor for ethanol is estimated from Table 3.6. Table 3.3 is used to estimate
molecular volumes. The resulting parameters are as follows:
P
Component
µ , cP
υ
Benzene
205.3
96
0.48
Ethanol
125.3 59.2
0.79
For benzene at infinite dilution in ethanol, values of molecular volume and the parachor for
ethanol must be multiplied by 8 times the viscosity of ethanol. Thus, from Eq. (342), using
P = 8(0.79)(125.3) = 792 and υ = 8(0.79)(59.2) = 374 for ethanol, the solvent in this case, DB,A ∞ 3181.29 792 0.5 / 205.30.42 = 155 × 10−8
. 0.79 0.92 374 0.23 = 2.51 × 10 −5 cm2 / s For ethanol at infinite dilution in benzene, Eq. (342) gives, DA,B ∞ = 155 × 10
. −8 3181.29 205.30.5 / 125.30.42
0.48 0.92 96 0.23 = 3.4 × 10 −5 cm2 / s Use Eqs. (345) and (346) of Vignes to compute DA,B and DB,A as a function of composition, DA,B = DA,B
DB,A = DB,A xB
∞ xA
∞ DB,A
DA,B xA
∞ xB
∞ 1+
1+ ∂ ln γ A
∂ ln xA
∂ ln γ B
∂ ln xB (1)
T ,P (2)
T ,P The partial derivatives are evaluated numerically with,
∂ ln γ A ∆ ln γ A
=
∂ ln xA ∆ ln xA and ∂ ln γ B ∆ ln γ B
=
∂ ln xB ∆ ln xB Exercise 3.16 (continued)
Analysis: (continued)
Using the data from Exercise 2.23,
xA
DA,B ,
DB,A ,
∆ ln γ A
∆ ln γ B
ln γ Α
l n γB
cm2/s
cm2/s
∆ ln xA
∆ ln xB
0.0000
3.4x1...
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 Spring '11
 Levicky
 The Land

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