# 317 exp 352 0362 from eq 6 51 f 040 946 hl

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Unformatted text preview: 3.08 cm2 From Eq. (6-64), t L = (3.08)(23,900)/5,970 = 12.3 s Analysis: (d) (continued) The continuity equation is, mV = U a Aa V , so U a , ft/s = Exercise 7.41 (continued) mV 630(18.5)(2.205) / 3600 = = 6.8 ft/s = 2.07 m/s Aa V 28.3(0.91)(0.657 /16.02) From Eq. (6-65), tG = (0.617)(3.08) = 0.024 s (0.383)(6.8)(2.54)(12) From below Eq. (6-67), 0.5 F=Ua V = 2.07(0.657)0.5 = 1.68 (kg/m)0.5/s From Eq. (6-67), k L a = 78.8(8 10 -5 ) 0.5 (168 + 0.425) = 1.48 s-1 . From Eq. (6-66), k G a = 1,030(0.30)1/ 2 0.40 - 0.842(0.40) 2 3.081/ 2 = 85.3 s-1 From Eq. (6-63), N L = k L at L = 148(12.3) = 18.2 . From Eq. (6-62), N G = k G at G = 85.3(0.024) = 2.05 From the vapor-liquid equilibrium data at the bottom of the column, Kmethanol = 0.156/0.0246 = 6.34 Absorption factor = KV/L = (6.34)(630)/1,192 = 3.35 From Eq. (6-61), 1 1 ( KV / L) 1 3.35 = + = + = 0.488 + 0.184 = 0.672 N OG N G NL 2.05 18.2 NOG = 1/0.672 = 1.488 Thus, the gas-phase resistance is more important than the liquid-phase resistance. From Eq. (6...
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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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