Question:

A depleted and shut-in oil reservoir originally contained \(25 \times 10^6\) STB of oil with a formation volume factor of 1.35 res bbl/STB and a connate water saturation of 0.25. Cumulative oil production to date has been \(2.5 \times 10^6\) STB of oil. The oil formation volume factor is now 1.25 res bbl/STB. Assuming no water influx, the gas saturation in the reservoir is \(\underline{\hspace{2cm}}\) % (round off to one decimal place).

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Updated On: Jan 9, 2026
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Correct Answer: 11 - 14

Solution and Explanation

Step 1: Calculate the Remaining Oil in Reservoir (STB)
Original Oil in Place (OOIP) = \(25 \times 10^6\) STB
Cumulative Oil Produced = \(2.5 \times 10^6\) STB
Remaining Oil = OOIP - Cumulative Oil Produced = \((25 - 2.5) \times 10^6 = 22.5 \times 10^6\) STB

Step 2: Convert Remaining Oil to Reservoir Barrels
Using current oil formation volume factor (OFVF):
Remaining Oil (reservoir bbl) = Remaining Oil (STB) × Current OFVF = \(22.5 \times 10^6 \times 1.25 = 28.125 \times 10^6\) res bbl

Step 3: Calculate Pore Volume
Original Pore Volume = Initial Oil (res bbl) = OOIP × Initial OFVF = \(25 \times 10^6 \times 1.35 = 33.75 \times 10^6\) res bbl

Step 4: Calculate Water Volume in Reservoir
Connate Water Saturation = 0.25
Water Volume = Original Pore Volume × Connate Water Saturation = \(33.75 \times 10^6 \times 0.25 = 8.4375 \times 10^6\) res bbl

Step 5: Calculate Remaining Hydrocarbon Pore Volume
Remaining Hydrocarbon Pore Volume = Original Pore Volume - Water Volume = \(33.75 \times 10^6 - 8.4375 \times 10^6 = 25.3125 \times 10^6\) res bbl

Step 6: Calculate Gas Volume in the Reservoir
Gas Volume = Remaining Hydrocarbon Pore Volume - Remaining Oil Volume = \(25.3125 \times 10^6 - 28.125 \times 10^6\) = \(3.1875 \times 10^6\) res bbl

Step 7: Calculate Gas Saturation
Gas Saturation = (Gas Volume / Remaining Hydrocarbon Pore Volume) × 100%
Gas Saturation = \(\frac{3.1875 \times 10^6}{25.3125 \times 10^6} \times 100\% \approx 12.6\%\)
Upon rounding off to one decimal place, Gas Saturation = 12.6%

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