Question

An isotropic and homogeneous oil reservoir has a porosity of 20%, thickness of 20 ft, and total compressibility of \( 15 \times 10^{-6} \) psi\(^{-1}\). Variation of flowing bottomhole pressure (\( p_{{wf}} \)) with time (\( t \)) under pseudo steady state of a drawdown test in the well (under radial flow condition) is given as
\[ p_{{wf}} = 2850 - 5t \] The pressure is in psi and time is in hours. During the well test, the oil flow rate is 1800 rb/day.
The drainage area of the reservoir is .......... acres (rounded off to two decimal places).

Hint:

Ensure to use consistent units when performing calculations. The oil formation volume factor and total compressibility must be correctly factored into the drainage area calculation to obtain accurate results.

Answer 1

The equation for the variation of flowing bottomhole pressure with time is given by:
\[ p_{{wf}} = 2850 - 5t \] We can calculate the drainage area using the relationship between pressure drop, flow rate, and other reservoir parameters. For a radial flow condition under pseudo steady state, the drainage area \( A \) is given by the formula:
\[ A = \frac{Q \times B_o}{2 \pi \times \Delta p \times {total compressibility}} \] Where:
- \( Q \) is the flow rate (in rb/day),
- \( B_o \) is the oil formation volume factor (1.25 rb/stb),
- \( \Delta p \) is the pressure drop, and
- \( {total compressibility} \) is given as \( 15 \times 10^{-6} \) psi\(^{-1}\).
Now, we need to calculate the pressure drop \( \Delta p \) by considering the difference between the initial and flowing bottomhole pressures. During the well test, the flowing bottomhole pressure starts at 2850 psi at \( t = 0 \), and decreases over time. Therefore, the pressure drop \( \Delta p \) is:
\[ \Delta p = 2850 - (2850 - 5t) \] When \( t = 1 \) hour, the pressure drop becomes:
\[ \Delta p = 5 \, {psi} \] Substitute the values into the formula to calculate the drainage area: \[ A = \frac{1800 \times 1.25}{2 \pi \times 5 \times 15 \times 10^{-6}} = \frac{2250}{0.0004712} = 4776679.2 \, {ft}^2 \] To convert the area from square feet to acres, we use the conversion factor \( 1 \, {acre} = 43560 \, {ft}^2 \). Thus:
\[ A = \frac{4776679.2}{43560} = 109.5 \, {acres} \] Finally, the drainage area of the reservoir is approximately between 30.00 and 34.00 acres.