9 f 2 2 095 or 95 unacceptable d u f 1 ad a v 6

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Unformatted text preview: Use the entrainment flooding correlation of Fig. 6.24, with densities and molecular weights from a simulation program, where the abscissa is, FLV LM L V = VM G L 1/ 2 = 3,351(181) 0.0408 . 1,789(18.8) 59.5 1/ 2 = 0.0473 From Fig. 6.24, for 24-inch tray spacing, CF = 0.37 ft/s. Because FLV < 1, Ad /A = 0.1. FHA = 1.0, FF = 1.0, and since = 20 dynes/cm, FST =(58/20)0.2 = 1.24 From Eq. (6-24), C = FSTFFFHACF = (1.24)(1)(1)(0.37) = 0.46 ft/s From Eq. (6-40), U f = C L - V / V 1/ 2 = 0.46 59.5 - 0.04 / 0.0408 1/ 2 = 17.6 ft / s From a rearrangement of Eq. (6-44), solving for the fraction of flooding, 4VMV 4(1,789 / 3,600)(18.8) f = 2 = 2 = 0.51 or 51% D U f 1 - Ad / A V 6 (17.6)(314)(1 - 01)(0.0408) . . Analysis: (continued) (b) Vapor pressure drop per tray is given by Eq. (6-49), ht = hd + hl + h u 2 V From Eq. (6-50), hd = 0186 o2 . Co L Column area = A = D2/4 = 3.14(6)2/4 = 28.3 ft2 Take downcomer area = Ad = 0.1(28.3) = 2.83 ft2 Bubbling area = Aa = A - 2Ad = 28.3 - 2(2.83) = 22.6 ft2 Hole area = Ah = 0.1(22.6) = 2.26 ft2 Consider first, the...
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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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