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1.60
0.563
6
411
0.287
0.383
0.189
0.510
0.729
8
113
0.045
0.120
0.033
0.179
0.844
10
37
0.008
0.042
0.006
0.068
0.913
For these conditions, it takes 10 batch stages of absorption and flashing to obtain a vapor of at
least 90 mol% O2. However, the yield of O2 is very low, 0.068/21 x 100% = 0.32%. It is
doubtful that any processing conditions exist that can deliver 90 mol% O2 with a high yield by
absorption with water, followed by flash desorption. Exercise 4.64
Subject: Absorption of ammonia from nitrogen into water at 200C and 1 atm (760 torr).
Given: 14 m3 of gas containing equimolar NH3 (A) and N2 (N)to be contacted in one
equilibrium stage with 10 m3 of water (W). Equilibrium data for partial pressure of NH3 over
water at 20oC.
Assumptions: No mass transfer of water and nitrogen.
Find: % absorption of NH3.
Analysis: First compute the kmol of water. Density = 1000 kg/m3. M of water = 18.02.
Therefore, we have (10)(1000)/18.02 = 555 kmol = L. Next, compute the kmol of gas. M of
ammonia = 17.03. Assume the 14 m3 are at 1 atm and 20oC. At 1 atm and 0oC, have 22.4
m3/kmol. At 20oC and 1 atm, have 22.4(293/273) = 24.04 m3/kmol or 14/24.04 = 0.582 kmol.
Because NH3 is the only component being transferred use mole ratios, with:
X = moles NH3 / mole H2O and Y = moles NH3 / mole N2. Convert equilibrium data to these
variables, using:
Y = pA/(760 torr  pA) , where pA is in torr. X= grams NH 3
100 grams H 2 O 18.02 1
grams NH 3
= 0.0106
17.03 100
100 grams H 2 O p of
g NH3 /
NH3 ,torr 100 g H2O
470
40
298
30
227
25
166
20
114
15
69.6
10
50.0
7.5
31.7
5
24.9
4
18.2
3
15
2.5
12
2
0
0 Y X 1.6207
0.6450
0.4259
0.2795
0.1765
0.1008
0.0704
0.0435
0.0339
0.0245
0.0201
0.0160
0.0000 0.4240
0.3180
0.2650
0.2120
0.1590
0.1060
0.0795
0.0530
0.0424
0.0318
0.0265
0.0212
0.0000 Exercise 4.64 (continued)
Analysis: (continued) The kmol of N2 = 0.5(0.582) = 0.291 kmol = G.
Material balance for NH3 , where 0 = inlet and 1 = outlet, with Y0 = 1.0 , and X0 = 0.0:
(1)
LX 0 + GY0 = G = 0.291 = LX 1 + GY1 = 555 X 1 + 0.291Y1
Rearranging Eq. (1), Y1 = 1 − 1907 X 1
(2)
Because Y1 at the outlet must be less than the inlet value of 1, X1 must be a very small number.
Assume that at small values of X, the equilibrium curve is linear. Using the lowest point
at Y = 0.0160 and X = 0.0212, then:
Y1 = (0.0160/0.0212)X1 = 0.755X1
(3)
Solving Eqs. (2) and (3) simultaneously, Y1 = 3.96 x 104 mol NH3/mol N2
X1 = 5.25 x 104 mol NH3/mol H2O Therefore, in the final gas, have Y1G = (3.96 x 104)(0.291) = 0.000115 kmol NH3
Therefore, NH3 not absorbed = 0.000115/0.291 x 100% = 0.04%.
Therefore, % NH3 absorbed = 100  0.04 = 99.96. Exercise 4.65
Subject: Desublimation of phthalic anhydride (PA) from a gas by cooling.
Given: 8,000 lbmol/h of a gas containing 67 lbmol/h of PA cooled at 770 torr.
s
Assumptions: Only PA desublimes. Dalton's law. At equilibrium, pPA = PPA . Find: % recovery of PA for temperatures where the vapor pressure of PA, are (a) 0.7 torr,
(b) 0.4 torr, and (c) 0.1 torr.
Analysis: Let nPA = lbmol/h of PA still in the gas at...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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