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PGE 312_Spring10_Solution_HW6

PGE 312_Spring10_Solution_HW6 - PGE 312 Physical and...

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PGE 312 Physical and Chemical Behavior of Fluids Home Work 6 Solutions Problem 6-4 Given: Example 3-6 Component Methane Standard Pressure, P sc = 14.7psia Temperature inside the container, T sc = 60 o F = 520 o R Mass of methane inside the container, m g = 10.2lb Solution: Using the ideal gas equation at standard conditions, we have P sc V sc = n g RT sc ----- (1) Molecular weight of methane, M= 16 ?? ?????? No. of moles of methane gas n g = ? ? ? = 10.2 16 = 0.6375 lb mole Substituting all the quantities in equation (1) we get V sc = 0.6375 10.732 520 14.7 = 242.02 scf Problem 6-9 Given: Example: 3-10 Pressure, P= 5014.7psia Temperature, T = 194 o F = 654 o R Solution: Formation volume factor B g = 0.02827 ?? 𝑃 ??? ?? . ?? ??? ---- (1)
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To find Z-factor at 5000psia and 654 o R: Component Mole fraction y j Critical Temperature, o R T cj y j T cj Critical Pressure, psia P cj y j P cj methane 0.9712 343.33 333.44 666.4 647.21 ethane 0.0242 549.92 13.31 706.5 17.10 propane 0.0031 666.06 2.06 616.0 1.91 i-butane 0.0005 734.46 0.37 527.9 0.26 n-butane 0.0002 765.62 0.15 550.6 0.11 hexane 0.0002 913.60 0.18 436.9 0.09 heptanes plus 0.0006 1082.00 0.65 372.0 0.22 1.0000 T pc 350.17 o R P pc 666.90 psia Calculate pseudoreduced properties, T pr = ? ? ?? = 654.0 350.2 = 1.87 P pr = 𝑃 𝑃 ?? = 5014.7 666.9 = 7.51 Using Fig. 3-7, we get Z = 1.02 Substituting back in equation (1) we get B g = 0.02827* 1.02 654 5014.7 = 3.818 x 10 -3 ??? 𝒄? . ?? ?𝒄? Problem 6-17 Given: Pressure, P = 900psia Temperature, T = 200 o F Pressure, psia Molar Volume, cu ft/lb mole 600 10.6 700 9.7 800 8.9 900 8.4 1000 8.1 1100 7.9 1200 7.8
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Solution: Plotting the values we get, The coefficient of isothermal compressibility is given by equation (6-4) c g = 1 ?
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