Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Given stagnant vapor film of 01 inch 000833 ft

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Unformatted text preview: 30 cm3 Benzene loss rate = 9.14 x 10-7 (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 0.1-inch (0.00833-ft) 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 vapor-liquid interface. Find: Mass transfer rate of toluene in lbmol/h-ft2. Analysis: At the vapor-liquid interface, use Raoult's law, Eq. (2-44). 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 finite-difference form of Fick's law, Eq. (3-16), for the diffusion of toluene from the bulk vapor to the vapor-liquid interface, NT = cDT,B ( yT − yTI ) = ( 0.00217 )( 0.2 )( 0.300 − 0.211) ∆z = 0.00464 lbmol/h-ft 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 gas-liquid 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 gas-liquid interface is obtained from Raoult's law, Eq. (2-44), 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 dew-point 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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