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Unformatted text preview: aporizes into 1atm (760 torr) air, which
is stagnant inside the tube. The density of toluene is 0.852 g/cm2. Its vapor pressure is 57.3 torr.
Assumptions: Isothermal vaporization. Mass transfer resistance only in the air in the tube.
Neglect counterdiffusion of air. Molecular diffusion of toluene through the air in the tube.
Toluene mole fraction of zero in the air at the top of the tube. Phase equilibrium at the gasliquid
interface, given by Raoult's law. Ideal gas law.
Find: Experimental and predicted values of diffusivity of toluene in air at 39.4oC and 1 atm.
Analysis: Applying Raoult's law, Eq. (244), to the gasliquid interface, yT = xT PTs
57.3
= 10
.
= 0.0754
P
760 Therefore, the driving force for molecular diffusion, ∆yT = 0.0754  0.0 = 0.0754
Total gas concentration from ideal gas law, c = P/RT = 1/(82.06)(312.6) = 3.9 x 105 mol/cm3
Molecular weight of toluene = ML = 92.1
The correction for bulk flow = (1xA)LM = approximately [(10) + (10.0754)]/2 = 0.962
Integrating and solving Eq. (6) in Example 3.2 for the experimental gas diffusivity, DT,A = 2
ρ L (1 − xA ) LM z2 − z12 tM L c ( ∆yT ) 2 = (0.852)(0.962)
7.92 − 1.92
= 0.0927 cm 2 /s
−5
(960, 000)(92.1)(3.9 × 10 )(0.0754)
2 Compute the predicted value from Eq. (336) of Fuller, Schettler, and Giddings, with:
M T,A = From Table 3.1, DT,A = VA = 19.7, 2
= 44.1
1
1
+
92.1 29
VT = 7(15.9) + 8(2.31) − 18.3 = 1115
. 0.00143(312.6)1.75
= 0.0887 cm 2 /s
(1)(44.1)1/ 2 [(19.7)1/ 3 + (111.5)1/ 3 ]2 The predicted value is 4.3% less than the measured value. This is good agreement. Exercise 3.7
Subject: Countercurrent molecular diffusion of hydrogen (H) and nitrogen (N) in a tube.
Given: Tube of 1 mm (0.1 cm) inside diameter and 6 inches (15.24 cm) long. At one end of the
tube (1) is pure hydrogen (H) blowing past. At the other end of the tube (2) is pure nitrogen
blowing past. The temperature is 75oC (348 K) and pressure is 1 atm.
Assumptions: Ideal gas law.
Find: (a) Estimate the diffusivity and the rate of diffusion of nitrogen in mol/s for equimolar,
countercurrent diffusion.
(c) Plot of hydrogen mole fraction with distance.
Analysis: (a) Estimate diffusivity from (336), with T =348 K, P = 1 atm,
M H,N = From Table 3.1, VH = 612,
. DH,N = 2
1
1
+
2.02 28
VN = 3.765 = 18.5 0.00143(348)1.75
= 1.033 cm 2 /s
1/ 2
1/ 3
1/ 3 2
(1)(3.765) [(6.12) + (18.5) ] From Eq. (318), Fick's law for equimolar, countercurrent diffusion, with,
c = P/RT = 1/(82.06)(348) = 0.000035 mol/cm3
∆yN = mole fraction driving force for nitrogen = 1  0 = 1
∆z = distance for diffusion = 15.24 cm
A = crosssectional area for diffusion = (3.14)(0.1)2/4 = 0.00785 cm2
Nitrogen flow rate,
nN = J N A = cDH,N (∆yN )
∆z A= (0.000035)(1.033)(1.0)
(0.00785) = 1.86 × 10−8 mol/s
15.24 (c) The mole fraction of hydrogen varies linearly through the tube because of equimolar,
countercurrent diffusion. Thus, Exercise 3.7 (continued)
Distance from End 1, inches Hydrogen mole fraction
0
1
1
0.833
2
0.667
3
0.500...
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 Spring '11
 Levicky
 The Land

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