From below eq 6 42 fst 302002 1084 assume that

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Unformatted text preview: 84(1.0)(1.0)(0.37) = 0.40 ft/s From Eq. (6-40), - V Uf =C L V 1/ 2 52 - 0.252 = 0.40 0.252 0.5 = 5.73 ft / s From Eq. (6-44), 4VM V DT = fU f (1 - Ad / A)V 1/ 2 4(336.5 / 3600)(78.1) = 0.85(5.73)(3.14)(1 - 0.1)(0.252) 0.5 = 2.9 ft Exercise 7.44 Subject: Diameters and heights for separation of propylene from propane in two columns. Given: Feed and product conditions and numbers of trays for sieve-tray columns in Fig. 7.47. Assumptions: Constant molar overflow. 85% of flooding. Find: Column diameters, tray efficiencies, numbers of actual trays, and column heights. Analysis: Column diameters: Assume that the diameter of the second column is controlled at the top tray, and than the diameter of the first column is controlled by the bottom tray. Top tray of second column: Use the properties of pure propylene. Distillate rate = 3.5 + 360 - 12.51 = 350.99 lbmol/h With an R = 15.9, L = 15.9(350.99) = 5,580 lbmol/h. V = L + D = 351 + 5,580 = 5,931 lbmol/h. From Perry's Handbook, at 116oF = 576oR...
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This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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