Mass Transport
2
25 Example 4: An open tank is filled with 2ft of the top with pure methanol. The tank is tapered, as shown below. The air within the tank is stationary but circulation of air immediately above the tank is adequate to assure a negligible concentration of methanol at this point. The tank and air space are at 77 o F and 1 atm and the diffusivity of methanol in air in such conditions is 0.62 ft 2 /s. Calculate the rate of loss of methanol from the tank at steady – state. Given that the partial pressure of methanol at 77 o F is 135 mm Hg. Solution: Since air does not diffuse in methanol, so N B = 0 47 equation apply area, variable layer with stagnant through diffusing is case the ∴ P (P dP RT P D A dZ N 2 A 1 A 2 1 P P A) T A T AB Z Z A ∫ ∫ − − = Then we have to find the area (A) in terms of (Z) 4ft d check atm 0 P then 2 Z 6ft d check atm 0.178 Hg mm 135 P then 0 Z : conditions boundary following For the ) ( * * ) 6 ( * 4 ) 6 ( 4 ) ( 4 6 4 4 2 2 2 1 tan 2 2 1 2 1 A 2 1 A 1 2 2 2 = = = = = = = − − = − − = = ∴ − = + − = + = − = − = = ∫ ∫ A A P P A T A T AB Z Z A P P dP T R P D Z dZ N Then Z d A Z Z L d m d Z L m Z L m π π π θ Now we have to calculate all the parameters in the equation, so we can find A N , therefore:
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An example on Molecular Diffusion
Transcript of Mass Transport
25
Example 4: An open tank is filled with 2ft of the top with pure methanol. The tank is tapered, as shown below. The air within the tank is stationary but circulation of air immediately above the tank is adequate to assure a negligible concentration of methanol at this point. The tank and air space are at 77oF and 1 atm and the diffusivity of methanol in air in such conditions is 0.62 ft2/s. Calculate the rate of loss of methanol from the tank at steady – state. Given that the partial pressure of methanol at 77oF is 135 mm Hg. Solution: Since air does not diffuse in methanol, so NB = 0
47equation apply area, variablelayer withstagnant through diffusing is case the ∴
P(P
dPRT
P D AdZ N
2A
1A
2
1
P
P A)T
ATABZ
ZA ∫∫ −
−=
Then we have to find the area (A) in terms of (Z)
4ft dcheck atm 0P then 2Z6ftdcheck atm 0.178 Hg mm 135P then 0Z
:conditionsboundary following For the
)(**
)6(
*4
)6(4
)(4
6442
2
21tan
2
2
1
2
1
A2
1A1
2
22
=======
−−
=−
−==∴
−=+−=+=
−=
−==
∫∫A
A
P
P AT
ATABZ
Z
APP
dPTR
PD
Z
dZN
Then
ZdA
ZZLdmd
ZLm
ZLm
π
ππ
θ
Now we have to calculate all the parameters in the equation, so we can find AN ,
therefore:
26
sec10*722.162.0
R*lbmolft atm0.7302 R Use
atm 908.0
822.01ln
822.01
lnP
atm 1 Hg mm 760 0 - 760 P
atm 0.822 mmHg 625135760 P
24
2
3
BLM
B
B
1
2
22
2
1
fthrftD
PP
PP
AB
B
B
BB
−==
=
=−
=−
=
===
==−=
Now by integrating the above equation, we find:
[ ]
seclbmol 10*111.8
)61
41(
0178.0*908.0*537*7302.01*10*722.1
4N
**
61
61N*4
74
A
112
A2
−−
=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
−
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
−=⎥⎦
⎤⎢⎣
⎡−
−−
π
π AABLM
TAB PPPRT
PDZZ
Example 5: A small diameter tube closed at one end was filled with acetone to within 18 mm of the top and maintained at 290 K and 99.75 kPa with a gentle stream of air blowing across the top. After 15 ksec, the liquid level had fallen to 27.5 mm. The vapor pressure of acetone at that temperature is 21.95 kN/m2. Calculate the diffusivity of acetone in air, given the following data: Mwt of acetone = 58 The density of acetone (ρ) = 790 kg/m3 Solution: Since evaporation is occurred through constant area, then apply equation 39 and by integrating this equation we find:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−⎟
⎟⎠
⎞⎜⎜⎝
⎛
−=
22Z
)P P(.DP
)(P T RMwt
t 20
2f
AAABT 21
A ZBLMρ
To apply this equation we must first calculate each parameter in the equation as: PT = 99.75 kPa T = 290 K
K*kmolm*kPa 314.8
3=R
