Question

Oil is produced from an undersaturated oil reservoir that has no initial gas cap. The reservoir rock compressibility and connate water saturation effects are considered negligible. There is no water influx into the reservoir and no water production. The initial oil formation volume factor was 1.24 RB/STB. After producing the reservoir for a short period, while the reservoir pressure is still above the bubble point pressure, the oil formation volume factor changed to 1.25 RB/STB. The calculated oil recovery factor (in fraction, rounded to three decimal places) is ________.

Hint:

For an undersaturated reservoir above bubble point with negligible rock and water compressibility and no water influx, recovery factor equals 1 minus Boi over Bo.

Answer 1

Correct Answer: 0.008

Step 1: Identify the drive mechanism for this undersaturated reservoir:
Since the pressure stays above the bubble point pressure throughout, no free gas comes out of solution and the oil remains a single, undersaturated liquid phase. The rock compressibility and water saturation effects are negligible, and there is no water influx or water production. This means the only mechanism available to push oil out of the reservoir and into the wellbore is the expansion of the oil itself as the reservoir pressure declines.
Step 2: Write the reservoir volume balance:
Let N be the original oil in place in STB and Np be the cumulative oil produced in STB, both at surface conditions. At the initial condition, this oil occupies a reservoir volume of N times Boi. Since nothing else in the reservoir changes volume, no rock expansion, no water influx, no water production, the total hydrocarbon pore volume that was originally filled by the oil stays fixed. After producing Np STB, the remaining oil, which is N minus Np STB, has expanded because pressure has dropped, and it must now occupy that same fixed reservoir volume by itself, evaluated at the new, larger Bo. This gives the balance: \[ (N - N_p) \, B_o = N \, B_{oi} \]
Step 3: Rearrange to obtain the recovery factor formula:
Expanding the left side gives \[ N B_o - N_p B_o = N B_{oi} \] which rearranges to \[ N_p B_o = N B_o - N B_{oi} = N (B_o - B_{oi}) \] Dividing both sides by N times Bo gives the recovery factor, RF, which is the fraction of original oil in place that has been produced: \[ RF = \frac{N_p}{N} = \frac{B_o - B_{oi}}{B_o} = 1 - \frac{B_{oi}}{B_o} \]
Step 4: Substitute the given formation volume factors and compute:
Here Boi is 1.24 RB/STB and Bo is 1.25 RB/STB. First divide: \[ \frac{B_{oi}}{B_o} = \frac{1.24}{1.25} = 0.992 \] Then subtract from 1: \[ RF = 1 - 0.992 = 0.008 \]
Final Answer:
\[ \boxed{0.008} \]