# 8 and above nor pac montz 23 fp ft ft 14 33 2 3 a m m

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Unformatted text preview: ntz, column diameter = DT = 5.3 ft 1 2 1 DP 2 1 0.00459 From Eq. (6-111), = 1+ = 1+ = 1008 . KW 3 1 - DT 3 1 - 0.930 5.3 Therefore, KW = 1/1.008 = 0.992 Superficial gas velocity at 70% of flooding from above = 13 ft/s for NOR PAC. Superficial gas velocity at 70% of flooding from above = 8.4 ft/s for Montz Gas viscosity = 0.011 cP and gas density = 0.0688 lb/ft3. From Eq. (6-114), for NOR PAC, u D (13)(0.012)(0.0688) N ReV = V P V KW = (0.966) = 26,500 (1 - ) V (1 - 0.947 ) (0.011)(0.000672) For Montz, N ReV = (8.4)(0.00459)(0.0688) (0.992) = 5, 090 (1 - 0.930 ) (0.011)(0.000672) From Eq. (6-113), for NOR PAC, o = Cp 64 1.8 64 1.8 + 0.08 = 0.350 + = 0.280 N ReV N ReV 26, 500 26,5000.08 64 18 . + = 0.272 5,090 5,0900.08 For Montz, o = 0.295 From Eq. (2), for NOR PAC, 2 Po a uV V 1 86.8(0.3048) 132 (0.0688) 1 = o 3 = 0.280 = 1.63 lbf/ft3 or 0.0113 psi/ft 3 lT 2 g c KW 0.947 2(32.2) 0.966 For Montz, Po 300(0.3048) 8.4 2 (0.0688) 1 = 0.272 = 2.35 lbf/ft3 or 0.0163 psi/ft 3 lT 0.930 2(32.2) 0.992 From Eq....
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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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