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Unformatted text preview: Eq. (652), EOV = 1 − exp(− N OG ) = 1 − exp(−0.066) = 0.064 or 6.4%
(e) From Example 6.2, the separation requires 3 equilibrium stages.
If we assume the liquid on a tray is well mixed. Then, from Eq. (631), EMV = EOV = 0.064.
From Eq. (621), Na =Nt /Eo = 3/0.064 = 47 trays.
If we assume plug flow of liquid on a tray, then, if we take, below Eq. (6.33), λ=KV/L=9.89,
from Eq. (637), log 1 + E MV ( λ − 1) log 1 + 0.064(9.89 − 1)
=
= 0197
.
log λ
log 9.89
Then we need from Eq. (621), Na =Nt /Eo = 3/0.197 = 15 trays. Sufficient information is not
given to establish the partial mixing prediction. Therefore, the number of trays required ranges
widely from 15 to 47.
Eo = Exercise 6.24
Subject: Absorption of SO2 from air into water in an existing packed column.
Given: Feed gas flow rate of 0.062 kmol/s containing 1.6 mol% SO2. Absorbent is 2.2 kmol/s
of pure water. Packed column is 1.5 m2 in cross sectional area and packed with No. 2 plastic
super Intalox saddles to a 3.5m height. Exit gas contains an SO2 mole fraction of 0.004.
Operating pressure is 1 atm. At operating temperature, equilibrium curve for SO2 is y = Kx =
40x
Assumptions: No stripping of water. No absorption of air.
Find: (a)
(b)
(c)
(d) L/Lmin
NOG and Nt
HOG and HETP
KGa Analysis: Compute material balance. SO2free inlet air rate = 0.062(10.016) = 0.061 kmol/s
SO2 inlet rate in feed gas = 0.062(0.016) = 0.001 kmol/s = V'
SO2 outlet rate in gas = 0.061(0.004/0.996) = 0.00025 kmol/s
SO2 rate in outlet water = 0.001  0.00025 = 0.00075 kmol/s
Fraction absorbed = 0.00075/0.001 = 0.75 or 75%
(a) From Eq. (611), L'min = V'K(fraction absorbed) = 0.061(40)(0.75) = 1.83 kmol/s
Therefore, L/L'min = 2.2/1.83 = 1.20
0.887 N +1 − 0.887
(b) Take A = L/KV = 2.2/[(40)(0.062)] = 0.887, From Eq. (613), 0.75 =
0.887 N +1 − 1
Solving, Nt = 4. For NOG, use Eq. (689) with yin = 0.016, yout = 0.004, xin =0.0,
ln
N OG = 0.887 − 1 0.016
1
+
0.887
0.004 0.887
= 3.78
(0.887 − 1) / 0.887 (c) Given height of packing = 3.5 m = lT
From Eq. (673), HETP = lT /Nt = 3.5/4 = 0.875 m
From Eq. (689), HOG = lT /NOG = 3.5/3.78 = 0.926 m
V
K G aPS
3
= 0.045 kmol/sm atm (d) From Table 6.7, HOG = Therefore, K G a = V
H OG PS = 0.062
(0.926)(1)(1.5) Exercise 6.25
Subject: Operating data for absorption of SO2 from air into water in a packed column.
Given: Column operates at 1 atm (760 torr) and 20oC. Solutefree water enters at 1,000 lb/h.
Mole ratio of water to air is 25. Liquid leaves with 0.6 lb SO2/100 lb of solutefree water.
Partial pressure of SO2 in exit gas is 23 torr (0.0303 atm). Equilibrium data are given as partial
pressures of SO2 in air as a function of lb SO2 dissolved/ 100 lb H2O.
Assumptions: No stripping of water. No absorption of air. Density of liquid taken as water.
Find: (a) % of SO2 absorbed.
(b) Concentration of SO2 in the liquid at the gasliquid interface in lbmol/ft3 at a point
where the bulk liquid concentration is 0.001 lbmol SO2/lbmol of water and:
kL = 1.3 ft/h
kp = 0.195 lbmol/hft2atm
Analysis: (a) Inlet water rate = 1,000/18.02 = 55.5 lbmol/h
Inlet air rate = water rate/25 = 55.5/25 = 2.22 lbmol/h
Partial pressure of air in exit gas = 760  23 = 737 torr
By partial pressure ratio, SO...
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 Spring '11
 Levicky
 The Land

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