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Unformatted text preview: s: First, determine the molar flow rates of components in the exit gas.
From the ideal gas law applied to the exiting gas at 1 atm and 68oF,
ρ = P / RT = 1 / (0.7302)(68 + 460) = 0.00259 lbmol/ft3
Flow rate of exit gas = 1,000(0.00259) = 2.59 lbmol/h
Gas contains 0.10(2.59) = 0.259 lbmol/h of NH3 and 0.90(2.59) = 2.331 lbmol/h of air.
Next, determine the mole fractions in the inlet and exit liquid streams.
Take as a basis, 100 lb of inlet liquid.
Component
H2O
NH3
Total Inlet liquid:
lb
lbmol
80
4.44
20
1.18
100
5.62 mole fraction
0.79
0.21
1.00 Outlet liquid:
lb
lbmol
80.000 4.4400
0.808 0.0475
80.808 4.4875 mole fraction
0.9894
0.0106
1.0000 Determine the water flow rate, L', in and out by an ammonia balance:
0.21
0.0106
(1)
L′ + 0 =
L ′ + 0.259
0.79
0.9894
Solving Eq. (1), L' = 1.015 lbmol/h of H2O.
NH3 in entering liquid = (0.21/0.79)1.015 = 0.270 lbmol/h
NH3 in leaving liquid = 0.270  0.259 = 0.011 lbmol/h
Because of the concentrated solution, take into account the bulkflow effect.
For a partial pressure driving force, can write the rate of mass transfer, V for volume, as, nNH 3 = 0.259 = Rearranging Eq. (1), V= KG aV
*
pNH 3 − pNH 3
( yair ) avg
0.259( yair ) avg p *
NH 3 − pNH 3 LM KG a LM (2) (3) Exercise 6.33 (continued)
Analysis: (continued)
First consider the bulkflow effect.
At the top of the column, in the bulk exiting air, yair = 0.90
To obtain the corresponding value of y in the vapor in equilibrium with the entering liquid at the
top of the column, use Fig. 6.49. In entering bulk liquid, moles NH3/moleH2O =1.18/4.44=0.266.
From Fig. 6.49, by a rough extrapolation, moles NH3/mole H2O in equilibrium gas = 0.266.
Therefore, y*air = 1  0.266/1.266 = 1  0.210 = 0.79. The average yair = (0.90 + 0.79)/2 = 0.845.
At the bottom of the column, in the bulk entering air, yair = 1.00
To obtain the corresponding value of y* in the vapor in equilibrium with the leaving liquid at the
column bottom, use Fig. 6.49.
In leaving bulk liquid, moles NH3/mol H2O = 0.0475/4.4 = 0.0108.
From Fig. 6.49, moles NH3/mole H2O in equilibrium gas = 0.007 = mole fraction of NH3
Therefore, y*air = 1  0.007 = 0.993. The average yair = (1.000 + 0.993)/2 = 0.997
The (yair)avg = (0.845 + 0.997)/2 = 0.921
Now, determine the log mean partial pressure driving force for ammonia, using Dalton's law
with the above mole fractions.
At the top of the column, for ammonia, (p* p)top = P (y* y)top = (1.0)(0.210  0.10) = 0.11 atm
At the column bottom, for ammonia, (p* p)bottom = P (y* y)bottom = (1.0)(0.007  0.0) = 0.007 atm
Therefore, for ammonia, (p* p)LM = 0.037 atm
From Eq. (3), packed volume = V = 0.259(0.921)
= 1.61 ft3
(0.037)(4) Exercise 6.34
Subject: Absorption of NH3 from air into water in a packed column.
Given: Entering gas is 2,000 lb/h of air (dry basis) containing NH3 with a partial pressure of 12
torr and saturated with water vapor at 68oF and 1 atm. Equilibrium data in Fig. 6.49. Absorb
99.6% of the NH3.
Assumptions: No absorption of air. No st...
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 Spring '11
 Levicky
 The Land

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