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Unformatted text preview: 2 flow rate in exit gas = 2.22(23/737) = 0.0693 lbmol/h
Molecular weight of SO2 = 64.06
SO2 flow rate in exit liquid = 0.6(1,000/(100)(64.06) = 0.0937 lbmol/h
By material balance, SO2 flow rate in entering air = 0.0693 + 0.0937 = 0.1630 lbmol/h
Partial pressure of SO2 in entering gas = 0.1630/(0.1630 + 2.22) = 0.0690 atm
% of SO2 absorbed = 0.0937/0.1630 x 100% = 57.5%
(b)
At the point, the rate of mass transfer for SO2 as a flux across the gasliquid interface, can
be written by the twofilm theory as,
kp(pb  pi) = kL(ci  cb) (1) As shown in Fig. 6.31, the equilibrium interface composition can be determined in terms of the
ratio of masstransfer coefficients. However, here, instead of compositions in mole fractions, the
gas composition is in partial pressure and the liquid is in concentration. From Eq. (1), kL
1.3
p − pi
=
= 6.67 = b
k p 0.195
ci − cb (2) At the point (column height location), bulk liquid concentration = 0.001 lbmol SO2/lbmol H2O
and the flow rate is 0.001(55.5) = 0.0555 lbmol/h for SO2 in the liquid phase. The flow of SO2 in
the gas phase at that location is obtained by a material balance around the top of the column:
SO2 in entering liquid + SO2 in gas at the point = SO2 in liquid at the point + SO2 in exit gas
Therefore, 0 + SO2 flow rate in gas at the point = 0.0693 + 0.0555 Analysis: (b) (continued) Exercise 6.25 (continued) SO2 flow rate in gas at the point = 0.1248 lbmol/h. Therefore, the partial pressure of SO2 in the
bulk gas at the point = 0.1248/(0.1248 + 2.22) x 1 atm = 0.0532 atm = pb .
The density of water is 62.4 lb/ft3 or 62.4/18.02 = 3.46 lbmol/ft3. Therefore,
the concentration of SO2 in the bulk liquid at the point = 0.001(3.46) = 0.00346 lbmol/ft3 = cb .
We now need an algebraic equilibrium relationship between pi and ci .
Convert the given equilibrium data to partial pressures and concentrations in the vicinity of the
values at the point:
The SO2 concentration in the liquid is obtained from, c, lbmol SO 2
lb SO 2
=
3
100 lb H 2 O
ft lb SO2 / 100 lb H2O
0.30
0.50
0.70
1.00 ci ,
lbmol SO2 / ft3
0.00292
0.00487
0.00682
0.00974 62.4
64.06 1
100 Partial pressure SO2 ,
torr
14.1
26.0
39.0
59.0 pi of SO2 ,
atm
0.01855
0.03421
0.05132
0.07763 These equilibrium data fit the curve, pi = 5.04536ci + 511.783ci2  21714.4ci3 (3) From Eq. (2), (4) Solving Eqs. (3) and (4), 0.0532  pi = 6.67(ci  0.00346)
ci = 0.00550 lbmol/ft3 and pi = 0.0396 atm The same result can be obtained by constructing a plot of SO2 partial pressure versus SO2
concentration in the liquid, similar to Fig. 6.31. First, the operating line is drawn as a straight
line connecting the column end points (0.0303, 0.0) and (0.0690, 0.00584), as (p , c). Then the
equilibrium curve is drawn, using Eq. (3). Then the point (pb , cb) is marked on the operating
line. A straight line is extended from this point, with a slope of (kL/kp) = 6.67, to the point of
intersection on the equilibrium line, giving the same result as above Exercise 6.26
Subject: Stripping of benze...
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 Spring '11
 Levicky
 The Land

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