Question

The fluid flow through an under-saturated oil reservoir is driven by solution gas drive mechanism. The reservoir parameters are as given below. 
\(\text{Compressibility of water, } c_w = 1 \times 10^{-6} \, \text{psi}^{-1}, \\ \text{Compressibility of formation, } c_f = 1 \times 10^{-5} \, \text{psi}^{-1},\)
\( \text{Connate water saturation, } S_{wc} = 0.2, \\ \text{Initial reservoir pressure, } p_i = 4000 \, \text{psi}, \\ \text{Reservoir pressure at bubble-point, } p_b = 3000 \, \text{psi},\)
\( \text{Oil formation volume factor, } B_{oi} = 1.24 \, \text{rb/STB}, \\ \text{Formation volume factor at bubble point pressure, } B_{ob} = 1.26 \, \text{rb/STB.}\)
The percentage of oil recovered as a fraction of the Original Oil in Place (OOIP) is \(\underline{\hspace{2cm}}\)% (round off to one decimal place).

Hint:

For gas drive calculations, use the formation volume factors at the initial and bubble-point pressures to estimate the recovery factor.

Answer 1

Correct Answer: 2.6

The recovery factor for solution gas drive is given by the following formula: \[ \text{Recovery Factor} = \frac{B_{oi} - B_{ob}}{B_{oi}} \cdot 100. \] Substitute the given values: \[ \text{Recovery Factor} = \frac{1.24 - 1.26}{1.24} \cdot 100 = \frac{-0.02}{1.24} \cdot 100 = -1.61%. \] Thus, the percentage of oil recovered is approximately \( 1.6% \).