362 425 023 23 074 in liquid from eq 6 55 with

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Unformatted text preview: Eq. (2), ht = hd + hl + h = 3.39 + 0.74 + 0.13 = 4.26 inches liquid = 0.115 psi/tray. Consider next, the conditions at the bottom tray of the column. From the continuity equation, uo = m/AhV (4) uo = (1,376/3,600)(18.1)/[(2.26)(0.0408)] = 75.0 ft/s and Co = 0.73 75.02 0.0408 From Eq. (3), hd = 0186 . = 135 inches of liquid . 0.732 59.5 Superficial velocity based on bubbling area = Ua = 75.0(2.26/22.6) = 7.5 ft/s From Eq. (6-53), Ks = U a V L - V 1/ 2 = 7.5 0.0408 59.5 - 0.0408 1/ 2 = 0.196 ft/s From Eq. (6-52), e = exp(-4.257Ks0.91) = exp[-4.257(0.196)0.91] = 0.381 Exercise 7.48 (continued) Analysis: (b) (continued) From Eq. (6-54), Cl = 0.362 + 0.317 exp(-3.5hw ) = 0.362 + 0.317 exp[-3.5(2.0)] = 0.362 Weir length = Lw = 42.5 in. Volumetric liquid rate = qL = 2,578(18.1)/[60(8.33)(59.5/62.4)] = 97.9 gpm From Eq. (6-51), hl = e hw + Cl qL Lw e 2/3 97.9 = 0.381 2 + 0.362 ( 42.5 ) 0.381 2/3 = 1.22 in. liq From Eq. (6-55), with maximum bubble size = DH = 3/16 inch = 0.00476 m and liquid density = 59.5 lb/ft3 or 954 kg/m3...
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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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