Question

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

Hint:

In pseudo-steady state, pressure decline rate is directly proportional to flow rate and inversely proportional to pore volume of the reservoir.

Answer 1

Step 1: Pseudo-steady state pressure decline.
For constant rate \(q\), \[ \frac{dp}{dt} = - \frac{q B}{\phi c_t h A} \] where \(A\) = drainage area.

Step 2: Substitute known values.
\[ \frac{dp}{dt} = -5 \, psi/hr \] Convert to per day: \[ -5 \times 24 = -120 \, psi/day \]

Step 3: Solve for area.
\[ 120 = \frac{1800 \times 1}{0.2 \times 15 \times 10^{-6} \times 20 \times A} \] Denominator = \(0.2 \times 15 \times 10^{-6} \times 20 = 6 \times 10^{-5}\). \[ 120 = \frac{1800}{6 \times 10^{-5} A} \] \[ A = \frac{1800}{120 \times 6 \times 10^{-5}} = \frac{1800}{0.0072} = 2.5 \times 10^5 \, ft^2 \] Convert to acres: \[ \frac{2.5 \times 10^5}{43560} = 5.74 \, acres \]

Final Answer: \[ \boxed{5.74 \, acres} \]