# 99 lbmolh with an r 159 l 15935099 5580 lbmolh

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Unformatted text preview: = 320 K, vapor density = V = 2.64 lb/ft3 and liquid density = L = 29.0 lb/ft3. FLV LM L V = VMV L 1/ 2 (5,580)(42.1) 2.64 = (5,931)(42.1) 29.0 0.5 = 0.284 Assume 24-inch tray spacing. From Fig. 6.24, CF = 0.25 From below Eq. (6-44), Ad/A = 0.1+ (0.284 - 0.1)/9 = 0.120 Assume surface tension, , = 10 dyne/cm. From below Eq. (6-42), FST = (10/20)0.2 = 0.87 Assume that the foaming factor, FF = 1.0 and take FHA = 1.0. From Eq. (6-42), C = 0.87(1.0)(1.0)(0.25) = 0.22 ft/s - V From Eq. (6-40), U f = C L V From Eq. (6-44), 4VM V DT = fU f (1 - Ad / A)V 1/ 2 1/ 2 = 0.22 29.0 - 2.64 2.64 0.5 = 0.70 ft / s 4(5931/ 3600)(42.1) = 0.85(0.70)(3.14)(1 - 0.12)(2.64) 0.5 = 8.0 ft Bottom tray of first column: Use the properties of pure propane. L in first column = L in second column + feed = 5,580 + 360 + 240 = 6,180 lbmol/h. V in first column = L in first column - B = 6,180 - (600 - 351) = 5,931 lbmol/h From Perry's Handbook, at 136oF = 596oR = 331 K, vapor density = V = 2.93 lb/ft3 and liquid density = L = 27.0 lb/ft3. Analysis: (continued) FLV Exercise 7.44 (continued) LM L V = VMV L 1/ 2 = (6,180)(44.1) 2.93 (5,931)(44.1)...
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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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