Question

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

Answer 1

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