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

# 005 99994 tce 650 120 0005 99996 tca 275 145 0200

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Unformatted text preview: Exercise 6.10 (continued) Analysis: (continued) (a) Plot of results: (b) Plot of results: Exercise 6.11 Subject: Stripping of VOCs from water by air in a trayed tower to produce drinking water. Given: 1,500 gpm of groundwater containing ppm amounts of DCA, TCE, and TCA given below. Stripping at 1 atm and 25oC to reduce the ppm amounts to the very low levels below. K-values of VOCs given below. Assumptions: No stripping of water and no absorption of air. System is dilute with respect to the VOCs. Find: Minimum air flow rate in scfm (60oF and 1 atm). Number of equilibrium stages for 2 times the minimum air flow. Composition in ppm for each VOC in the resulting drinking water. Analysis: Flow rate of water = L = (1,500 gpm)(8.33 lb/gal)/18.02 lb/lbmol = 693 lbmol/min % stripping of a VOC = (inlet ppm - outlet ppm)/inlet ppm) VOC K-value Inlet ppm Outlet ppm % stripping DCA 60 85 0.005 99.994 TCE 650 120 0.005 99.996 TCA 275 145 0.200 99.862 Because of its high % stripping and a K-value that is much lower than for the other two VOCs, the stripper is likely to be controlled by DCA. So base the calculations on DCA and then check to see that the % stripping for TCE and TCA exceed the above requirements. Because of the dilute conditions, use Kremser's method. 693 L From Eq. (6-12), Vmin = (fraction stripped) = (0.99994) = 11.55 lbmol./min K 60 or minimum air flow rate = 11.55 (379 scf/lbmol) = 4,377 scfm at 60oF and 1 atm For 2 times the minimum value, V = 2(11.55) = 23.1 lbmol/min. K V 60(23.1) The stripping factor, given by Eq. (6-16), is for DCA, S DCA = DCA = = 2.0 L 693 S N +1 − S 2 N +1 − 2 From Eq. (6-14), Fraction stripped = 0.99994 = N +1 = (1) S − 1 2 N +1 − 1 Solving Eq. (1), N = 13 stages. Now, for V = 11.55 lbmol/min and N = 13 stages, compute the fraction stripped for TCE and TCA from Eq. (6-14) with the stripping factor from Eq. (6-16). The results are: VOC S Fraction stripped DCA 2.0 0.99994 TCE 21.6 1.00000 TCA 9.17 1.00000 The drinking water contains 0.005 ppm of DCA and essentially zero ppm of TCE and TCA. Exercise 6.12 Subject: Stripping of a solution of SO2, butadienes, and butadiene sulfone with nitrogen. Given: 120 lbmol/h of liquid containing in lbmol/h,10.0 SO2, 8.0 1,3-butadiene (B3), 2.0 1,2butadiene (B2), and 100.0 butadiene sulfone (BS). Liquid effluent to contain < 0.05 mol% SO2 and < 0.5 mol% (B3 + B2). Stripping agent is pure nitrogen. Stripping at 70oC and 30 psia. Kvalues are 6.95 for SO2, 3.01 for B2, 4.53 for B3, and 0.016 BS. Assumptions: Negligible stripping of BS because of its low K-value. No absorption of N2. Find: Flow rate of nitrogen. Number of equilibrium stages. Analysis: Assuming no stripping of BS, the exiting liquid will contain 0.05 mol% SO2, 0.50 mol% (B3 + B2), and 99.45 mol% BS, or in lbmol/h, 100.0 BS, 0.0503 SO2, and 0.503 (B3 + B2). Therefore, % stripping of SO2 = (10.0 - 0.0503)/10.0 = 0.99497 and % stripping of (B3 + B2) = (8.0 + 2.0 - 0.503)/(8.0 + 2.0) = 0.9497. The minimum stripping ag...
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