4 ft 1 2 1 dp 2 1 0012 from eq 6 111 1 1 1044

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Unformatted text preview: 70) KW = (0.957) = 19, 900 (1 - ) V (1 - 0.947 ) (0.012)(0.000672) For Montz, (13)(0.00459)(0.0370) N ReV = (0.990) = 3,870 (1 - 0.930 ) (0.012)(0.000672) From Eq. (6-113), for NOR PAC, o = Cp 64 1.8 64 1.8 + 0.08 = 0.350 + = 0.287 N ReV N ReV 19,900 19,9000.08 64 18 . + = 0.279 3,870 3,8700.08 For Montz, o = 0.295 Exercise 7.52 (continued) Analysis: (e) Stripping section (continued) From Eq. (2), for NOR PAC, 2 Po a uV V 1 86.8(0.3048) 202 (0.0370) 1 = o 3 = 0.287 = 2.15 lbf/ft3 or 0.0149 psi/ft 3 lT 2 g c KW 0.947 2(32.2) 0.957 For Montz, Po 300(0.3048) 8.4 2 (0.0370) 1 = 0.279 = 311 lbf/ft3 or 0.0216 psi/ft . 3 lT 0.930 2(32.2) 0.990 From Eq. (1), for NOR PAC, with hL = 0.0269 m3/m3 and liquid Froude number = 4.28x10-4 P 0.947 - 0.0269 = 0.0149 lT 0.947 3/ 2 exp 1/ 2 13300 ( 4.28 10-4 ) = 0.0150 psi/ft 3003/ 2 From Eq. (1), for Montz, with hL = 0.0341 m3/m3 and liquid Froude number = 6.39x10-4 P 0.930 - 0.0341 = 0.0216 lT 0.930 Summary of Results: 1.5 exp 1/ 2 13300 6.39 10-4 ) = 0.021...
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