# Ch. 6 c - PGE 312 Physical and Chemical Behavior of...

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PGE 312 – Physical and Chemical Behavior of Petroleum Fluids I Homework 6 Problem 6-4 P = 14.65 psia T = 60 o F = 520 o R lb moles of methane is n = m/M = 10.2 lb / 16.04 lb/lb mole = 0.64 lb mole Volume occupied by 0.64 lb moles of methane at standard conditions is: V = (10.732 psia cuft)*(0.64 lb mole)*(520 o R)/(14.65 psia) = 242.1 scf Problem 6-9 Given P pc = 666.9 psia and T pc =350.17 o R (from example 3-10) For a pressure of 5014.7 psia and 654 o R (194+460), the pseudo reduced properties are as follows: T pr = T/T pc = 654 o R/350.17 o R = 1.87 P pr = P/P pc = 5014.7 psia/666.9 psia = 7.52 From Figure 3-7, Z = 1.02 with the above values. The gas formation factor is calculated as follows: P ZT B g 02827 . 0 = B g = 0.02827*1.02*654 o R/5014.7 psia = 0.00378 res ft 3 /scf 6-17 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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The slope at 900 psia is (8.75-8.25)/(950-860) = -0.0056 = ( gG± ) The coefficient of Isothermal Compressibility is ³ ´ = − µ 1 ¶· ¸( ¹¶· ¹º ) ³ ´ = − µ 1 8.4 ¸ (−0.0056) ³ ´ = 0.000661º»¼½ ¾¿ = 661 x 10 -6 psia -1 If the gas was ideal, then c À = 1 p c À = 1 900 psia = 0.00111 psia ¾¿ Problem 6-20 From Fig. 3-11, gas with specific gravity of 0.862 at 3025 psia and 175+460 = 635 o R has the following pseudo critical properties: P pc = 640 psia T pc = 415 o R The pseudo critical properties are: 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 600 700 800 900 100011001200 Vm cu. ft/lb mole Pressure, psia Molar Volume
P pr = P/P pc = 3025/640 = 4.73 T pr = T/T pc = 635/415 =

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## This note was uploaded on 11/27/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas.

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Ch. 6 c - PGE 312 Physical and Chemical Behavior of...

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