Unformatted text preview: arithmetic average =
[(1  0.0306) + (1  0.00578)]/2 = 0.982.
t= ρ L 1 − xW LM M W cDW yWI − yW 2
z2 − z12 = 62.4 0.982
2
z2 − z12
(18.02)(0.00255)(0.992)(0.0306 − 0.00578) 2
2
= 54,200 z2 − z12 = 54,200 z2 − 0.0417 2 (1) When z2 = 3.5 inches = 0.2917 ft, Eq. (1) gives t = 4,520 h Exercise 3.4 (continued)
Analysis: (continued)
(b) For other values of z2 , the following results are obtained using a spreadsheet with
Eq. (1) above:
z2 , inches
0.5
1.0
1.5
2.0
2.5
3.0
3.5 z2 , feet
0.0417
0.0833
0.1250
0.1667
0.2083
0.2500
0.2917 time, hours
0
282
753
1,412
2,259
3,294
4,520 Exercise 3.5
Subject: Mixing of argon (A) and xenon (X) by molecular diffusion.
Given: Bulb 1 containing argon. Bulb 2 containing xenon. Bulbs connected by a 0.002 m (0.2
cm) inside diameter by 0.2 m (20 cm) long tube. 105oC (378 K) and 1 atm are maintained. At
time, t, equal 0, diffusion is allowed to occur through the connecting tube. Diffusivity = 0.180
cm2/s.
Assumptions: Gas in each bulb is perfectly mixed. The only mass transfer resistance is in the
connecting tube. Ideal gas law. Equimolar, countercurrent diffusion.
Find: At a time when the argon mole fraction = 0.75 in one bulb and 0.20 in the other bulb,
determine,
(a) Rates and directions of mass transfer for A and X.
(b) Transport velocities of A and X.
(c) Molar average velocity of the gas mixture.
Analysis: This exercise is similar to Example 3.1.
Area normal to diffusion = A = 3.14(0.2)2/4 = 0.0314 cm2
From the ideal gas law, total concentration of gas =
c = P/RT = 1/(82.06)(378) = 0.0000322 mol/cm3
(a) From form of Eq. (318),
nA = cDA,X A ( yA1 − yA 2 ) = 5 × 10 −9 = ( 0.0000322 )( 0.180 )( 0.0314 )( 0.75 − 0.20 ) ∆z
mol/s from bulb 1 to 2 20 nX = 5 × 10−9 mol/s from bulb 2 to 1
(b) Because of equimolar, countercurrent diffusion, species velocities relative to
stationary coordinates are equal to diffusion velocities. From Eq. (39), for argon (A), υA = υA
υX = D JA
nA
5 × 10−9
0.00495
=
=
=
=
cA AcyA (0.314)(0.0000322)( yA )
yA 0.00495
yX (2) Thus, the transport velocities depend on the mole fractions. (1) Exercise 3.5 (continued)
Analysis: (b) (continued)
Using Eqs. (1) and (2) over the range of mole fractions, noting that mole
fractions are linear with distance because of equimolar, countercurrent diffusion,
Distance from
Bulb 1, cm
0 (Bulb 1)
5
10
15
20 (Bulb 2) yA yX υΑ, cm/s υΞ, cm/s 0.7500
0.6125
0.4750
0.3375
0.2000 0.2500
0.3875
0.5250
0.6625
0.8000 0.0066
0.0081
0.0104
0.0147
0.0248 0.7500
0.6111
0.4760
0.3367
0.1996 (c) Because we have equimolar, countercurrent diffusion, the molar average velocity of the
mixture is zero. Exercise 3.6
Subject: Measurement of diffusivity of toluene (T) in air (A), and comparison with prediction.
Given: Vertical tube, 3 mm in diameter and open at the top, containing toluene at 39.4oC (312.6
K). Initially, the toluene level, z1, is 1.9 cm below the top. It takes 960,000 s for the level to
drop to z2 equal to 7.9 cm below the top. The toluene v...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details