Question

The initial water saturation in an oil reservoir with a free gas cap is 30%. The initial gas saturation is 40%. At the end of water flooding, all the free gases are dissolved due to the elevated pressure and the oil formation volume factor reaches a value of 1.20 rb/stb. The final water saturation at the end of water flooding is 50%. If the two-phase formation volume factor at the initiation of the water flood is 2.3 rb/stb, the pore-to-pore displacement efficiency under the current reservoir condition is ____________ % (rounded off to one decimal place).

Hint:

Always convert oil saturations into reservoir barrels using formation volume factors before calculating displacement efficiency.

Answer 1

Correct Answer: 68.7

The initial oil saturation is computed as:
\[ S_{oi} = 1 - S_{wi} - S_{gi} \]
\[ S_{oi} = 1 - 0.30 - 0.40 = 0.30 \]
At the end of water flooding, all gas is dissolved, so:
\[ S_{of} = 1 - S_{wf} = 1 - 0.50 = 0.50 \]
The oil volumes must be corrected using formation volume factors:
Initial oil volume in reservoir barrels:
\[ V_{oi} = \frac{S_{oi}}{B_{oi}} = \frac{0.30}{2.3} \]
Final oil volume in reservoir barrels:
\[ V_{of} = \frac{S_{of}}{B_{of}} = \frac{0.50}{1.20} \]
The pore-to-pore displacement efficiency is:
\[ E_D = \frac{V_{of} - V_{oi}}{V_{of}} \]
\[ E_D = \frac{\left( \frac{0.50}{1.20} \right) - \left( \frac{0.30}{2.3} \right)}{\left( \frac{0.50}{1.20} \right)} \]
Evaluate numerator and denominator:
\[ \frac{0.50}{1.20} = 0.4167,\quad \frac{0.30}{2.3} \approx 0.1304 \]
\[ E_D = \frac{0.4167 - 0.1304}{0.4167} = \frac{0.2863}{0.4167} = 0.687 \]
Thus, in percent:
\[ E_D = 68.7% \]
Final Answer: 68.7%