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HW_7
Course: PGE 312, Spring 2009School: University of Texas
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312 PGE Physical and Chemical Behavior of Petroleum Fluids I Homework 7, Spring 2008 Problem 8-2 o API = 141.5 o - 131.5 = 141.5 - 131.5 = 30.2 0.875 Problem 8-4 o API = 141.5 o o - 131.5 = 35.2 o = 141.5 141.5 = = 0.8488 API + 131.5 35.2 + 131.5 Problem 8-7 Volume of Oil + dissolved gas leaving reservoir at Reservoir Conditions Volume of Oil in Stock Tank at Standard Conditions 227.0 res cc res bbl...
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312 PGE Physical and Chemical Behavior of Petroleum Fluids I
Homework 7, Spring 2008 Problem 8-2
o
API =
141.5
o
- 131.5 =
141.5 - 131.5 = 30.2 0.875
Problem 8-4
o
API =
141.5
o
o
- 131.5 = 35.2
o =
141.5 141.5 = = 0.8488 API + 131.5 35.2 + 131.5
Problem 8-7 Volume of Oil + dissolved gas leaving reservoir at Reservoir Conditions Volume of Oil in Stock Tank at Standard Conditions 227.0 res cc res bbl = = 1.356 STB 167.4 ST cc To calculate Bo Bo = To calculate Rs
Rs = Volume of Gas Produced at surface at Standard Conditions Volume of Oil entering Stock Tank at Standard Conditions 0.537 scf + 0.059 scf scf = = 566.1 -6 167.4 ST cc 6.2898 10 bbl / cc STB
Problem 8-9 To calculate Bo Volume of Oil + dissolved gas leaving reservoir at Reservoir Conditions Volume of Oil in Stock Tank at Standard Conditions 86.3 res bbl / d Bo = = 1.49 res bbl / STB 57.9 STB / d Bo =
To calculate Rs
Volume of Gas Produced at surface at Standard Conditions Volume of Oil entering Stock Tank at Standard Conditions (43150 + 7240) scf Rs = = 870.3scf / STB 57.9STB Rs = As reservoir pressure is decreased from initial pressure to bubble-point pressure, Bo is increased but Rs is constant. Therefore, applies Rs at the bubble point. Problem 8-10
Bo 1.60 1.50 Bo, res bbl/STB 1.40 1.30 1.20 1.10 1.00 0 1000 2000 3000 4000 Reservoir Pressure, psig 5000 6000
Rs 1000 900 800 700 Rs, scf/STB 600 500 400 300 200 100 0 0 1000 2000 3000 4000 Reservoir Pressure, psig 5000 6000
Problem 8-11
Bt = Bo + Bg ( Rsb - Rs ) Bt = 1.441 + 0.001090 (941 - 732) Bt = 1.669 res bbl / STB
Problem 8-12 P=3500 psig > Pb=3054 psig
Therefore, Bt = Bo = 1.527 res bbl/STB
Problem 8-16 For between 3500 and 4000 psig above the bubble point pressure Pb=3054 psig,
co = - 1 Bo Bo P T
From the graphs of B0 and Rs in Exercise 8-10,
Bo 1.517 - 1.527 = = -2 10-5 P 4000 - 3500 Bo = 1.522 res bbl / STB
co = -
1 -2 10-5 = 13.14 10-6 psi -1 1.522
Increases Increases Constant Decreases Increases
Problem 8-18 Formation volume factor of oil Total formation volume factor Solution gas-oil ratio Oil viscosity Coefficient of isothermal compressibility of oil Problem 8-19 Formation volume factor of oil Total formation volume factor Solution gas-oil ratio Oil viscosity Coefficient of isothermal compressibility of oil Formation volume factor of free gas Decreases Increases Decreases Increases Increases Increases
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University of Texas - PGE - 312
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Problem 2-21From the above Fig.2-37, lines are drawn through 300psia, 600psia and 900psia to draw the Isobars shown below. This is done by reading the compositions at the corresponding temperatures as they cross the Bubble Point and the Dew Point Curves.
University of Texas - PGE - 312
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University of Texas - PGE - 312
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