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Unformatted text preview: Eq. (6-112), DP = 6 =6 = 0.00150 m or 0.0049 ft a 92.3 Rectifying section of the column: Column diameter = DT = 3.8 ft 1 2 1 DP 2 1 0.0049 From Eq. (6-111), = 1+ = 1+ = 1037 . KW 3 1 - DT 3 1 - 0.977 3.8 Therefore, KW = 1/1.037 = 0.964 Superficial gas velocity at 70% of flooding from above = 7.84 ft/s Gas viscosity = 0.0091 cP and gas density = 0.171 lb/ft3. From Eq. (6-114), N ReV = Exercise 7.53 (continued) uV DP V (7.84)(0.0049)(0.171) KW = (0.964) = 45, 000 (1 - ) V (1 - 0.977 ) (0.0091)(0.000672) 64 1.8 64 1.8 + 0.08 = 0.421 + = 0.322 N ReV N ReV 45, 000 45, 0000.08 From Eq. (6-113), o = C p From Eq. (2), 92.3(0.3048) 7.84 2 (0171) . 1 Po = 0.322 = 162 lbf/ft3 or 0.0113 psi/ft . 3 0.977 2(32.2) 0.977 lT From Eq. (1), with hL = 0.0294 m3/m3 and liquid Froude number = 2.46 x 10-4 P 0.977 - 0.0294 = 0.0113 0.977 lT Stripping section of the column: Column diameter = DT = 3.6 ft From Eq. (6-102), 3/ 2 exp 1/ 2 13300 ( 2.46 10-4 ) = 0.0137 psi/ft 92.33/ 2 1 2 1 DP 2 1 0.0049 = 1+ = 1+ = 1039 . K...
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