Extended flow test – reservoir oil in place: 24-Pet-B5 Reservoir Mechanics, December 2015, Q5

Question 5: Extended flow test – reservoir oil in place (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The 9-point $p_{wf}$-vs-$t$ table above, $S_{wi}=0.2$, and $q$, $h$, $B_o$, $\phi$, $c_t$, $\mu_o$ as listed. No initial pressure is needed for this method.

Find. Reservoir oil in place, $N$ (STB).

Approach. This is the classic reservoir-limit test: identify the late-time real-time (not log-time) straight-line pressure decline, whose slope is the exam's own formula-sheet relation $dp/dt=0.234qB_o/(c_tV_p)$, invert for pore volume $V_p$, then convert to oil in place with $S_{wi}$.

Solid red line: constant-slope fit through the last 7 points ($t=11$ to 41 hr, $R^2>0.9999$), showing the reservoir has already gone pseudo-steady-state by $t\approx11$ hr; the $t=1$ and 6 hr points (dotted, curved) are still transient and excluded.

Identify the PSS decline rate. Consecutive real-time slopes are constant at $-1.72$ psi/hr from $t=11$ hr onward ($t=1$-6 hr is still transient, at $-3.9$ psi/hr). Regression on $t\ge11$ hr gives $R^2=0.99999$ with $$\left|\frac{dp}{dt}\right|=1.724\text{ psi/hr}$$
Pore volume from the reservoir-limit relation. $$V_p=\frac{0.23395\,qB_o}{c_t\,|dp/dt|}=\frac{0.23395(100)(1.1)}{(1\times10^{-6})(1.724)}$$ $$\boxed{V_p\approx 14.92\text{ MMbbl}}$$
Equivalent drainage radius (cylindrical reservoir). $$r_e=\sqrt{\frac{5.615\,V_p}{\pi h\phi}}=\sqrt{\frac{5.615(14{,}924{,}731)}{\pi(50)(0.25)}}\approx1461\text{ ft}$$
Oil in place. With $S_{wi}=0.2$: $$N=\frac{V_p(1-S_{wi})}{B_o}=\frac{14{,}924{,}731(0.8)}{1.1}$$ $$\boxed{N\approx 10.85\text{ MMSTB}}$$

Quantity	Result
PSS decline rate, $\|dp/dt\|$	1.724 psi/hr
Reservoir pore volume, $V_p$	14.92 MMbbl
Implied drainage radius, $r_e$	1461 ft
Oil in place, $N$	10.85 MMSTB

Topic. Reservoir-limit (extended flow) test for volumetric oil in place.

Key relations. $dp/dt=0.234qB_o/(c_tV_p)$; $V_p=\pi r_e^2h\phi/5.615$; $N=V_p(1-S_{wi})/B_o$.

Why this works. Once pseudo-steady-state is reached the entire closed drainage volume declines together at a rate set purely by material balance (total fluid withdrawn vs total system compressibility) – measuring that one late-time real-time slope is exactly equivalent to running a full material-balance depletion study, without needing a separate pressure survey.

Common pitfall. Fitting the slope on a semilog ($\log t$) axis instead of a linear (real) time axis – the reservoir-limit-test straight line is linear in $t$, not $\log t$; the early ($t=1$-6 hr) points are still curved/transient on a linear-$t$ plot and must be excluded exactly as in Question 3.

Source. Earlougher, Advances in Well Test Analysis, ch.2 (reservoir limit test); Lee, Well Testing, ch.1 (material balance in PSS).

Question 5 of 7: Extended flow test – reservoir oil in place

Question 5: Extended flow test – reservoir oil in place (20 marks)