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Unformatted text preview: diameter, need uo at flooding = 17(0.18/0.44)0.5 = 10.9 ft/s
At 50% of flooding, u = 5.45 ft/s. By ratio, column diameter = 2.5(17/5.45)0.5 = 4.4 ft
Column cross sectional area = S = 3.14(4.4)2/4 = 15.2 ft2.
Therefore, HOG = (792)/(2.5)(1)(15.2) = 20.8 ft, which is very high.
From Eq. (689), lT = HOG NOG = 20.8(8.5) = 177 ft. Given value of KGa seems very low. Exercise 6.32
Subject: Absorption of NH3 from air into water with a packed column.
Given: Conditions at a point in the column where P = 101.3 kPa, T = 20oC, 10 mol% NH3 in
bulk gas, 1 wt NH3 in bulk liquid, NH3 partial pressure of 2.26 kPa at the interface, and ammonia
absorption rate of 0.05 kmol/hm2.
Find: (a) X, Y, Yi , Xi , X*, Y*, KY, KX, kY, and kX in Fig. 6.49.
(b) % of mass transfer resistance in each phase.
1
1 H′
=
+
(c) Verify that
KY kY k X
Analysis: (a) At the point, as shown in Fig. 6.49, the mole ratios in the bulk are
X = (1/17)/(99/18) = 0.0107 and Y = 0.1/0.9 = 0.111.
At the interface, Yi = 2.26/(101.3  2.26) = 0.0229.
Then, using the equilibrium curve in Fig. 6.49, Xi = 0.028
The values of the mole fractions in equilibrium with the bulk values are obtained from the
equilibrium curve in Fig. 6.49. Thus, in equilibrium with Y = 0.111 is X* = 0.114.
In equilibrium with X = 0.0107 is Y* = 0.007
The given NH3 masstransfer flux can be written with the following combinations of masstransfer coefficients and driving forces, using Table 6.7 as a guide:
N = 0.05 = KY Y − Y * = K X X * − X = kY Y − Yi = k X X i − X
(1)
From Eq. (1), KY = 0.05/(Y  Y*) = 0.05/(0.111  0.007) = 0.480 kmol/hft2mole ratio
From Eq. (1), KX = 0.05/(X*  X) = 0.05/(0.114  0.0107) = 0.484 kmol/hft2mole ratio
From Eq. (1), kY = 0.05/(Y  Yi) = 0.05/(0.111  0.0229) = 0.568 kmol/hft2mole ratio
From Eq. (1), kX = 0.05/(Xi  X) = 0.05/(0.028  0.0107) = 2.89 kmol/hft2mole ratio
(b) To determine the relative masstransfer resistances in each phase, Eq. (1) can be
rearranged as in Eq. (678):
kX
(Y − Yi )
2.89
=−
=−
= −5.08
kY
( X − Xi )
0.568
Thus, the lowest coefficient and the highest driving force is that associated with the gas phase.
From the ratio of 5.08, the largest masstransfer resistance is in the gas phase. That resistance is
about five times the resistance in the liquid phase, with 84% in the gas phase, 16% in the liquid.
(c) From a mole ratio form of Eq. (680),
1
1
1 Yi − Y *
=
+
(2)
KY kY k X X i − X
Therefore,
Therefore, H′ = Yi − Y *
0.0229 − 0.007
=
= 0.919
0.028 − 0.0107
Xi − X 1 H′
1
0.919
1
1
+
=
+
= 2.08 compared to
=
= 2.08
kY k X 0.568 2.89
KY 0.480 Exercise 6.33
Subject: Stripping of ammonia from an aqueous solution with air in a packed column.
Given: Aqueous solution of 20 wt% NH3. Exit liquid to contain no more than 1 wt% NH3.
Exit gas to be 1,000 ft3/h of a 10 mol% NH3 in air. Equilibrium data in Fig. 6.49.
Temperature = 20oC = 68oF.
Assumptions: No stripping of water. No absorption of air. Operation at 1 atm.
Find: Volume of packing for KGa = 4 lbmol/hft3atm.
Analysi...
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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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