# 6 49 ht hd hl h u 2 v from eq 6 50 hd 0186 o2

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Unformatted text preview: conditions at the top of the column. From the continuity equation, uo = m/AhV uo = (1,944/3,600)(30.9)/[(2.26)(0.0703)] = 105 ft/s and Co = 0.73 1052 0.0703 From Eq. (3), hd = 0186 . = 5.73 inches of liquid 0.732 47.2 Superficial velocity based on bubbling area = Ua = 105(2.26/22.6) = 10.4 ft/s From Eq. (6-53), Ks = U a V L - V 1/ 2 Exercise 7.49 (continued) (2) (3) (4) = 10.4 0.0703 47.2 - 0.0703 1/ 2 = 0.40 ft/s From Eq. (6-52), e = exp(-4.257Ks0.91) = exp[-4.257(0.40)0.91] = 0.16 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 = 945(30.9)/[60(8.33)(47.2/62.4)] = 77.2 gpm From Eq. (6-51), hl = e hw + Cl qL Lw e 2/3 77.2 = 0.16 2 + 0.362 ( 42.5 ) 0.16 2/3 = 0.61 in. liquid From Eq. (6-55), with maximum bubble size = DH = 3/16 inch = 0.00476 m and liquid density = 47.2 lb/ft3 or 756 kg/m3, h = 6 / g L DBmax = 6(20 / 1000) / (9.8)(756)(0.00476) = 0.0034 m = 0.13 in. liquid From Eq. (2), ht = hd + hl +...
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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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