Unformatted text preview: 30 cm3
Benzene loss rate = 9.14 x 107 (78.11)(72,930)(3600)(24)/454 = 991 lb/day Exercise 3.3
Subject: Countercurrent diffusion of toluene (T) and benzene (B) across vapor film at 170oF
(630oR) and 1 atm.
Given: Stagnant vapor film of 0.1inch (0.00833ft) thickness, containing 30 mol% toluene and
70 mol% benzene, in contact with liquid reflux containing 40 mol% toluene and 60 mol%
benzene. Diffusivity of toluene in benzene = 0.2 ft2/h. Vapor pressure of toluene = 400 torr.
Assumptions: Equal molar heats of vaporization for benzene and toluene, such that diffusion is
equimolar, countercurrent. Ideal gas law and Raoult's law apply. All mass transfer resistance is
in the vapor phase, i.e. liquid is assumed to be uniform in composition. Given vapor
composition is for bulk conditions. Phase equilibrium at the vaporliquid interface.
Find: Mass transfer rate of toluene in lbmol/hft2.
Analysis: At the vaporliquid interface, use Raoult's law, Eq. (244). Then, the mole fraction of
toluene at the interface is,
Ps
400
yTI = xT T = 0.4
= 0.211
P
760
From ideal gas law, total gas concentration, c, is P/RT = 1/(0.7302)(630) = 0.00217 lbmol/ft3
From the finitedifference form of Fick's law, Eq. (316), for the diffusion of toluene from the
bulk vapor to the vaporliquid interface,
NT = cDT,B ( yT − yTI ) = ( 0.00217 )( 0.2 )( 0.300 − 0.211) ∆z
= 0.00464 lbmol/hft 2 0.00833 Benzene diffuses at the same rate in the opposite direction. Exercise 3.4
Subject: Drop in level of water (W) contained in a vertical tube when evaporating into air at
25oC (537oR).
Given: Tube with an inside diameter of 0.83 inch. Initial liquid level of water in tube = 0.5 inch
from the top. Air above the tube has a dew point of 0oC. Diffusivity of water vapor in air =
0.256 cm2/s or 0.992 ft2/h.
Assumptions: Pressure = 1 atm. Ideal gas. Phase equilibrium at the gasliquid interface with
Raoult's law for mole fraction of water in the vapor adjacent to liquid water.
Find: (a) Time for the liquid level to drop from 0.5 inch to 3.5 inches.
(b) Plot of liquid level as a function of time.
Analysis: (a) The mole fraction of water in the air adjacent to the gasliquid interface is
obtained from Raoult's law, Eq. (244), with xW = 1 for pure liquid water, using a vapor pressure
of 0.45 psia for water at 25oC.
P s 0.45
yWI = W =
= 0.0306
P 14.7
In the bulk air, the mole fraction of water is obtained from the dewpoint condition. Thus, the
partial pressure of water vapor = vapor pressure of water at 0oC = 0.085 psia. Therefore,
p
P s 0.085
yW = W = W =
= 0.00578
P
P
14.7
The equation for the time, t, for the water level to drop from level z1 = 0.5 inch (0.0417 ft) to
level z2 equal to as large as 3.5 inches (0.2917 ft) is derived in Example 3.2, where the result is
given by Eq. (6). Applying that equation here, with ρΛ = 62.4 lb/ft3 for liquid water, total gas
concentration, c, by the ideal gas law to give c=P/RT = 1/(0.7302)(537) = 0.00255 lbmol/ft3, and
a bulk flow factor = (1  xW)LM = to a good approximation to the...
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 Spring '11
 Levicky
 The Land

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