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Unformatted text preview: Problem 64 Temperature = 528 R, Pressure = 1000psia, V= 3.2cu ft, Z = 0.89 , , 0.89 528 0.02827 0.02827 0.01328 1000 3.2 240.9 0.01328 g g T P T P g sc sc g ZT B B P V V B V cu ft V B = = = = = = = Therefore, the standard cubic feet in the cylinder is 240.9 Alternative solution: 10.2 0.6359 16.04 / m lbmass n lbmol M lbmass lbmol = = = From Avegadros law: V sc = n*379.4 = 241.26 scf Problem 68 A = 40 acres = 27878400 ft 2 , h = 21 ft , = 0.18 , S w = 0.33, Z = 0.951 from example 310 43560 (1 ) wi g Ah S G B  = Where, 3 0.0282 0.951 (194 460) 0.0282 0.0046 3810 g Z T ft B P scf + = = = 10 43560 (1 ) 27878400 21 0.18 (1 0.33) 1.53 10 0.0046 w g Ah S G scf B   = = = Problem 69 T pc = 350.17 R and P pc = 666.9 psia From example 310 194 460 1.87 350.17 5000 14.7 7.522 666.9 pr pc pr pc T T T P P P + = = = + = = = From fig. 37 Z = 1.05 3 0.0282 1.05 (194 460) 0.0282 0.00386 5000 14.7 g Z T res ft B P scf + = = = + Problem 612 D = 5.761 in, P = 1000 psia, q = 2.4 MMscfd, T = 100 F, Assumptions: Heptanes plus has the same properties of nHeptane Component mol. Percent Tc R yiTci Pc psia yiPci Hydrogen sulfide 672.4 1300 Carbon dioxide 1.67 547.9 9.2 1071...
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This note was uploaded on 09/01/2011 for the course PGE 312 taught by Professor Peters during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Peters

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