Unformatted text preview: ne from wastewater with air in a packed column.
Given: Column operation at 2 atm (1,520 torr) and 25oC. Wastewater enters at 600 gpm
containing 10 ppm by weight of benzene. Suggested air rate is 1,000 scfm (60oF and 1 atm).
Exit water to contain just 0.005 ppm benzene. Vapor pressure of benzene = 95.2 torr at 25oC.
Solubility of benzene in water = 0.180 g/100 g water at 25oC. Packing is 2inch polypropylene
Flexirings. Mass transfer coefficients are: kLa = 0.067 s1 and kGa = 0.80 s1 (both driving
forces are in concentration units).
Assumptions: No stripping of water. No absorption of air.
Find: (a)
rate.
(b)
(c)
(d)
(e) Minimum air stripping rate. Is it less than suggested? If not, use 1.4 times minimum
Stripping factor
NOG
KGa in units of s1 and mol/m3skPa, and which phase controls mass transfer
Volume of packing in m3 Analysis: Water flow rate = 600(8.33)(60)/18.02 = 16,640 lbmol/h
For benzene, xin = (10/1,000,000)(18.02/78) = 2.31 x 106
xout = (0.005/1,000,000)(18.02/78) = 1.16 x 109
Fraction of benzene stripped = 1  (2.31 x 106/1.16 x 109) = 0.9995
(a) System is very dilute with respect to benzene, therefore use Kremser equation to determine
the minimum gas rate. From Eq. (612),
Vmin = L (fraction stripped)/ K
(1)
For the Kvalue, use Eq. (4) in Table 2.3, a modified Raoult's law, K = γLPs/P
(2)
To obtain the liquidphase activity coefficient, use the given benzene solubility data. For liquidliquid equilibrium with respect to benzene, which is distributed between a waterrich phase (2)
and a benzenerich phase (1), using Eq. (230), γ ( 2 ) = x (1) / x ( 2 ) γ (1) . But if the benzenerich phase is nearly pure benzene, γ(1) = 1 and x(1) = 1. Therefore, γ(2) = 1/x(2)
018(18.02)
.
From the given benzene solubility in water, x =
= 4.2 × 10 −4
100(78)
1
Therefore from Eq. (3), γ L =
= 2,380
4.2 × 10 − 4 (3) From Eq. (2), K = (2,380)(95.2)/1,520 = 149
From Eq. (1), Vmin = 16,640(0.9995)/150 = 111.6 lbmol/h
At 60oF and 1 atm, there are 379 scf/lbmol. Therefore, Vmin = 111(379)/60 = 705 scfm.
This is less than the 1,000 scfm suggested by the expert. Therefore, use the value suggested by
the expert, rather than revise it. Exercise 6.26 (continued) Analysis: (continued) (b) Operating V = 111.6(1,000/705) = 158.3 lbmol/h
From Eq. (551), S = KV/L = 149(158.3)/16,640 = 1.417 (a good value)
(c) For NOG, use Eq. (693), with A = 1/S = 1/1.417 = 0.7057
Benzene stripped = Linxin(fraction stripped) = 16,640(2.31 x 106)(0.9995) = 0.0384 lbmol/h
Therefore, yout = 0.0384/158.3 = 0.0002427. Also, xin = 2.31 x 106 and yin = 0.0. Therefore, ln
N OG = 0.7057 − 1
0.7057 0.0 − 149(2.31× 10−6 )
1
+
−6
0.0002427 − 149(2.31× 10 )
0.7057 = 14.2
(0.7057 − 1) / 0.7057
(d) Note that both kL and kG are given for concentration driving forces. Therefore, we write the
rate of mass transfer of benzene as:
r = kG acG ( yb − yi ) = k L acL ( xi − xb ) = K G acG ( yb − y* ) (4) where cG and cL are total gas and liquid concentrations, respectively, and y* is the vapor mole
fraction in equilibrium with the bulk li...
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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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