Unformatted text preview: 5 = 1.16 lb1/2s1ft1/2 or 2.83 kg1/2s1m1/2
and uL = [(526)(17.9)/3,600)]/(2.35)(62.4) = 0.018 ft/s = 0.00547 m3/m2s
Substitution of these values into Eq. (3) gives, kGa = 43.4(2.83)0.7(0.00547)0.45 = 8.6 s1
Note that this kGa is in concentration units for the driving force.
Therefore, from Table 6.7 by analogy to the liquid phase case,
HOG = HG = VMV/kGaSρV = (100/3,600)(24.2)/(8.6)(2.35)(0.0618) = 0.54 ft
Now, we must correct this value to 1.5inch metal Pall rings.
From Table 6.8, using interpolation when necessary, we have the following characteristics:
Packing
1.5inch Pall rings
2.0inch Pall rings a, m2/m3
139.4
112.6 Ch
0.644
0.784 ε
0.965
0.951 CV
0.373
0.410 From Eqs. (6136) to (6140), aPh/a is proportional to ε1/2/a . Therefore, aPh / a for 1.5inch Rings
0.965
=
aPh / a for 2.0inch Rings
0.951 1/ 2 112.6
= 0.814
139.4 From Eq. (6133), if we ignore holdup in the term (ε  hL), then,
1ε
a
H G is proportional to
1.25
CV a
aPh
1
0.965
H G for 1.5inch Rings 0.341 139.41.25
1
Therefore,
=
= 1.15
1
0.951
H G for 2.0inch Rings
0.814
0.410 112.61.25
This ratio should be about the same for HOG .
Therefore, for the 1.5inch rings, HOG = 0.54(1.15) = 0.62 ft.
From Eq. (2), column height = 3.46(0.62) = 2.1 ft. Exercise 6.37
Subject: Absorption of acetone from air with water in a packed column.
Given: Column operates at 60oF and 1 atm. Entering air is 50 ft3/min at 60oF and 1 atm,
containing 3 mol% acetone. 97% of the acetone is to be absorbed. Maximum allowable gas
superficial velocity is 2.4 ft/s. Equilibrium equation is Y = 1.75 X
(1)
Assumptions: No stripping of water. No absorption of air.
Find: (a)
(b)
(c)
(d)
(e)
(f) Minimum watertoair ratio.
Maximum acetone concentration possible in the water.
Nt for absorbent flow rate equal to 1.4 times the minimum.
Number of overall gas transfer units.
Height of packing for Kya = 12.0 lbmol/hft3mole ratio difference.
Height of packing as a function of molar flow ratio for constant V and HTU. Analysis: First compute material balance. Total gas in = 50(60)/379 = 7.916 lbmol/h
Acetone in entering gas = 0.03(7.916) = 0.237 lbmol/h
Air in entering gas = 7.916  0.237 = 7.679 lbmol/h
Acetone in exiting gas = 0.03(0.237) = 0.007 lbmol/h
Acetone in entering liquid = 0.0 lbmol/h
Acetone in exiting liquid = 0.237  0.007 = 0.230 lbmol/h
Xin = 0.0
Yin = 0.03/0.97 = 0.03093
Yout = 0.007/7.679 = 0.000912
(a) For minimum water rate, the exiting liquid will be in equilibrium with the entering air.
Therefore, using the equilibrium equation, Eq. (1), Xout = Yin /1.75 = 0.03093/1.75 = 0.01767
Therefore, water rate out and in = 0.230/0.01767 = 13.02 lbmol/h
Minimum watertoair molar ratio = 13.02/7.679 = 1.70
(b) Maximum possible acetone concentration in the liquid is at min. water rate, X = 0.01767
This corresponds to an acetone mole fraction of 0.01767/(1 + 0.01767) = 0.01736.
(c) Use Lin = L' = 1.4 Lmin = 1.4(13.02) = 18.23 lbmol/h
Using mole ratios with no acetone in entering liquid, and K' = constant...
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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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