# 2 016 2 0362 425 016 23 061 in liquid from

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Unformatted text preview: h =5.73 + 0.61 + 0.13 = 6.47 inches liquid = 0.175 psi/tray. This is a very high pressure drop/tray. Consider next, the conditions at the bottom tray of the column. From the continuity equation, uo = m/AhV uo = (1,789/3,600)(18.1)/[(2.26)(0.0408)] = 97.9 ft/s and Co = 0.73 97.9 2 0.0408 From Eq. (3), hd = 0186 . = 2.28 inches of liquid 0.732 59.5 From Eq. (6-53), Ks = U a V L - V 1/ 2 (4) Superficial velocity based on bubbling area = Ua = 97.9(2.26/22.6) = 9.8 ft/s = 9.8 0.0408 59.5 - 0.0408 1/ 2 = 0.255 ft/s From Eq. (6-52), e = exp(-4.257Ks0.91) = exp[-4.257(0.255)0.91] = 0.293 Analysis: (b) (continued) Exercise 7.49 (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 = 3,351(18.1)/[60(8.33)(59.5/62.4)] =127 gpm From Eq. (6-51), hl = e hw + Cl qL Lw e 2/3 127 = 0.293 2 + 0.362 ( 42.5 ) 0.293 2/3 = 1.08 in. liq From Eq. (6-55), with maximum bubble size = DH = 3/16 inch = 0.00476 m and...
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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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