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phase homework chapter 6

# phase homework chapter 6 - Phase hw chapter 6 6-4 The...

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Phase hw chapter 6 6-4 . The conditions in the cylinder from Example 3-6 are; P = 1000 Psia T = 528 o R V= 3.20 cu ft Z = 0.890 P,T g scf scf 0.02827 0.890*528 0.02827 1000 0.01328 V But, B V 3.2 V 240.9 cu ft 0.01328 Therefore, the standard cubic feet in the cylinder is 240.9 g g g ZT B P B B = = = = = = Alternative Solution scf 241.26 379.4 * n V law, s Avogadro' From lbmole 6359 . 0 ole lbmass/lbm 04 . 16 lbmass 2 . 10 sc = = = = = M m n 6.8. The gas in place is given by;

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o 10 43,560 (1 ) where 0.02827 From example 3-10 for the give gas at 3,810 Psia and 190 F Z=0.951 0.02827*0.951*654 0.004613 res cu ft/scf 3810 43,560*640*21*0.18*(1-0.33) G = 1.53 10 0.004613 w g g g Ah S G B ZT B P B s φ - = = = = = × cf 6.9. From example 3-10, P PC = 666.9 Psia and T TC = 350.17 o R Pressure = 5014.7 Psia, T = 654 o R 0.02827 654 1.87 350.2 5014.7 7.52 666.9 1.02 0.02827*1.02*654 0.0038 res cu ft/scf 5014.7 g pr pr g ZT B P T P Z B = = = = = = = = 6.12 We can solve this problem using the formation volume factor approach actual actual actual standard standard standard actual standard V V Q t V V Q t Also, 0.02827 actual cu ft/scf so, Q 0.02827*Q * cu ft Z factor can be determined for 1,000 Psia as follows g g B ZT B P ZT P = = = = = Component Mole Critical Critical
Fractions Pressure Temperature y i P ci , psia y i * P ci T c , o F T c , o R y i * T ci Hydrogen Sulfide 0.0000 1300.0 0.0 212.5 672.5 0.0 Carbon dioxide 0.0167 1071.0 17.9 87.9 547.9 9.2 Nitrogen 0.0032 493.1 1.6 -232.5 227.5 0.7 Methane 0.7108 666.4 473.7 -116.7 343.3 244.0 Ethane 0.1552 706.5 109.6 89.9 549.9 85.3 Propane 0.0736 616.0 45.3 206.1 666.1 49.0 i-Butane 0.0092 527.9 4.9 274.5 734.5 6.8

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phase homework chapter 6 - Phase hw chapter 6 6-4 The...

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