Separation Process Principles- 2n - Seader & Henley - Solutions Manual

3 a modified raoults law k lpsp 2 to obtain the

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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 2-inch polypropylene Flexirings. Mass transfer coefficients are: kLa = 0.067 s-1 and kGa = 0.80 s-1 (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 s-1 and mol/m3-s-kPa, 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 10-6 xout = (0.005/1,000,000)(18.02/78) = 1.16 x 10-9 Fraction of benzene stripped = 1 - (2.31 x 10-6/1.16 x 10-9) = 0.9995 (a) System is very dilute with respect to benzene, therefore use Kremser equation to determine the minimum gas rate. From Eq. (6-12), Vmin = L (fraction stripped)/ K (1) For the K-value, use Eq. (4) in Table 2.3, a modified Raoult's law, K = γLPs/P (2) To obtain the liquid-phase activity coefficient, use the given benzene solubility data. For liquidliquid equilibrium with respect to benzene, which is distributed between a water-rich phase (2) and a benzene-rich phase (1), using Eq. (2-30), γ ( 2 ) = x (1) / x ( 2 ) γ (1) . But if the benzene-rich 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. (5-51), S = KV/L = 149(158.3)/16,640 = 1.417 (a good value) (c) For NOG, use Eq. (6-93), with A = 1/S = 1/1.417 = 0.7057 Benzene stripped = Linxin(fraction stripped) = 16,640(2.31 x 10-6)(0.9995) = 0.0384 lbmol/h Therefore, yout = 0.0384/158.3 = 0.0002427. Also, xin = 2.31 x 10-6 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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