PGE_312_Fall_09_Test_3_Solution_Key

# PGE_312_Fall_09_Test_3_Solution_Key - 1 PGE 312 - FALL 2009...

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1 PGE 312 - FALL 2009 Physical and Chemical Behavior of Petroleum Fluids I Test 3 Solution Key 1. ( 20 points ) Your company has discovered a dry gas reservoir with the following characteristics: Reservoir area = 2 miles x 1 mile Reservoir thickness = 250 ft Initial reservoir pressure = 5000 psia Reservoir temperature = 200 F Average porosity = 15% Average water saturation = 20% Initial gas Z-factor = 1.01 Calculate the initial gas in place in standard cubic feet. Note: 1 mile = 5280 feet 1 square mile = 640 acres Solution to Problem 1 (2*5280*1*5280)(250)(0.15)(1 0.20) res cu ft 1.01*(200 460) res cu ft 0.02827 5000 scf g gi gi Ah S HCPV G BB     = 9 3 1.673 10 3.769 10 x scf x = 4.438x10 11 scf

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2 2. ( 15 points ) Figure 1 shows the production performance of a dry gas reservoir with no water influx. Solution to Problem 2 a. Determine the initial gas in place in standard cubic feet (scf). From the graph, G = 1.120x10 9 scf b. Predict the gas recovery factor in % if the gas field will be abandoned at a P/Z of 2000 psia. At abandonment, G pa = 8.0x10 8 scf Gas Recovery Factor = (G pa /G)100 = (8.0x10 8 / 1.120x10 9 ) = 71.4% P/Z vs Gp Plot 0 1000 2000 3000 4000 5000 6000 7000 8000 0.0E+00 2.0E+08 4.0E+08 6.0E+08 8.0E+08 1.0E+09 1.2E+09 Cumulative Gas Production (scf) P/Z (psia) Gp G Figure 1. Dry gas reservoir performance.
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## This note was uploaded on 01/29/2011 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas.

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PGE_312_Fall_09_Test_3_Solution_Key - 1 PGE 312 - FALL 2009...

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