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13 November, 19:37

An oil reservoir has a average porosity of 20%, an area of 100 acres, and a av - erage thickness of 10 feet. The connate water saturation is 25% and the resid ual oil saturation is 35%. No gas is present.

a. Calculate the recovery factor for the oil reservoir.

b. Calculate the volume of oil which may be recovered per acre-ft of reser voir Calculate the bulk volume of the reservoir in units of acre-ft.

c. Calculate the total volume of oil in cubic feet which may be recovered from the entire reservoir

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  1. 13 November, 23:37
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    Answer:a) recovery factor = 53.3%, bi) volume of oil per acre-ft = 11090.64bbl/acre-ft

    bii) bulk volume of the reservoir in acre-ft = 1000 acre-ft

    c) 62214.74 cubic feet

    Explanation: a) recovery factor is the percentage amount of oil that can recovered from a reservoir, it is the oil produced divided by oil initially in place

    Recovery factor = 1 - (Soi/1-Swi)

    = 1 - (0.35/1-0.25)

    = 0.533 * 100%

    = 53.3%

    bi) volume of oil which may be recovered per acre-ft = 7758. porosity. (1 - Swi-Soi)

    = 7758 x 0.2 x (1-0.25-0.35)

    = 620.64 bbl/acre-ft

    bii) bulk volume of the reservoir in acre-ft = Area x thickness

    = 100 acres x 10 ft

    = 1000 acre-ft

    c) total volume of oil in cubic feet

    since we have gotten volume as 1000 acre-ft we simply multiply it by the volume of oil gotten in answer bi)

    = 1000 acre-ft x 620.64 bbl / acre-ft

    = 620640bbl

    So we convert from barrel (bbl) to cubic feet, and 1 barrel is equal to 5.609 cubic feet, so to convert the answer from barrel to cubic feet we multiply the answer 620640 bbl by 5.609

    = 620640 bbl x 5.609

    = 3481580.711 cubic feet
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