Applied Petroleum Reservoir Engineering (Second Edition) by B. C. Craft, M. Hawkins, Ronald E. Terry

By B. C. Craft, M. Hawkins, Ronald E. Terry

The most up-tp-date, utilized publication out there for petroleum engineers, geologists and others operating within the improvement and construction of oil and gasoline fields and people taken with the stream of floor water, this revised variation displays the advances made in reservoir engineering calculation strategies. a number of actual global examples make clear the cloth, offering the reservoir engineer with the sensible info to make utilized calculations.

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Extra info for Applied Petroleum Reservoir Engineering (Second Edition)

Example text

7 pounds per mole. 3 Real Gas Law All that has just been said applies to a perfect or ideal gas. Actually there are no perfect gases; however, many gases near atmospheric temperature and pressure approach ideal behavior. All molecules of real gases have two tendencies (1) to fly apart from each other because of their constant kinetic motion, and (2) to come together because of electrical attractive forces between the molecules. Because the molecules are quite far apart, the attractive , and the gas behaves close to ideal.

4. , at standard conditions, Pw and 7w). They are generally expressed in either cubic feet or barrels of reservoir volume per standard cubic foot of gas. 00) at reservoir pressure p and temperature T by Eq. 17) The constants in Eqs. 7 psia and 60°F only, and different constants must be calculated for other standards. 00533 cu ft of space in the reservoir at 3250 psia and 213°F. Because oil is usually expressed in barrels and gas in cubic feet, when calculations are made on combination reservoirs containing both gas and oil, either the oh volume must be expressed in cubic feet or the gas volume in barrels.

4. To find the compressibility of a gas from thc gas deviation factor curve. Given: The gas deviation (actor curve for a gas at 150°F, Fig. 6. SoluTIoN At 1000 psia, the slope dzldp is shown graphically in Fig. 83: as — 127 I I 1(Y6) 6=llS3xlO6psi the slope dz/dp is zero, so the At 2500 psia Cg = Fig. 6. Gas compressibility from Example 1 4) compressibility is simply: = 400 X 10 the gas deviation factor versus pressure plot (Sec introduction to Reservoir Engineenng 26 At 4500 psia the slope dz/dp is positive and as shown in Fig.

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