hw6key - HOMEWORK # 6 SOLUTION KEY 6-1. Assuming ideal gas,...

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HOMEWORK # 6 SOLUTION KEY 6-1. Assuming ideal gas, PV = nRT , nRT V P = 100 10.732 (60 459.7) 14.65 xx + = scf = 38071.13 scf 6-8. Given, Area = 640 acres = 27878400 sq. ft. H = 21 ft Φ = 0.18 S W = 33% z = 0.951, from example 3-10. 0.0282 g zT B P = 0.0282 0.951 (194 459.7) 3810 + = cu ft/scf = 0.0046 cu ft/scf res g scf V B V = (1 ) w scf A HS V φ = = 15348.8 MMscf 6-10. Component Composition, mole fraction T c , R P c , psia y i T c , R y i P c , psia H 2 S 0.1 672.4 1300 67.24 130 CO 2 0.05 547.9 1071 27.395 53.55 N 2 0.021 227.5 493 4.7775 10.353 CH 4 0.703 343.3 666.4 241.3399 468.4792 C 2 H 6 0.062 549.9 706.5 34.0938 43.803 C 3 H 8 0.037 666.1 616 24.6457 22.792
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n-C 4 H 10 0.027 734.5 527.9 19.8315 14.2533 1 419.3234 743.2305 ε = 21 T’ pc = T pc ε = 419.3234-21 ºR = 398.32 ºR 22 ' ' (1 ) pc pc pc p c HS PT P Ty y ε = +− 743.23 398.32 419.3234 0.1(1 0.1)21 x = = 702.83 psia pr pc P P P = 5420 14 702.83 + = = 7.72 pr pc T T T = 257 459.7 398.32 + = = 1.8 z = 1.02, from fig.3-7. 0.00502 / g zT B resbbl cuft P = 0.00502 1.02 (257 459.7) 5420 14 g xx B + = + res bbl/scf = 0.000675 res bbl/scf or 0.0038 res ft 3 / scf 6-12. Given, I.D. = 6 inches = 0.5 ft V = 2.4 MMscfd T = 100 ºF = 559.7 ºR P = 1000 psia
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Component Mole percent T c , R P c , psia y i T c , R y i P c , psia M, lb/lbmol M i y i H 2 S Nil - - - - CO 2 1.67 547.9 1071 9.14993 17.8857 44 0.7348 N 2 0.32 227.5 493 0.728 1.5776 28 0.0896 CH 4 71.08 343.3 666.4 244.0176 473.6771 16.04 11.40123 C 2 H 6 15.52 549.9 706.5 85.34448 109.6488 30.07 4.666864 C 3 H 8 7.36 666.1 616 49.02496 45.3376 44.1 3.24576 i-C 4 H 10 0.92 734.5 527.9 6.7574 4.85668 58.1 0.53452 n-C 4 H 10 1.98 765.6 550.6 15.15888 10.90188 58.1 1.15038 i-C 5 H 12 0.33 829.1
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This note was uploaded on 09/14/2009 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.

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hw6key - HOMEWORK # 6 SOLUTION KEY 6-1. Assuming ideal gas,...

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