Extended drawdown – reservoir oil in place: 24-Pet-B5 Reservoir Mechanics, May 2016, Q5

Question 5: Extended drawdown – 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.

Quantity	Symbol	Value
Flow rate	$q$	100 STBD
Formation volume factor	$B_o$	1.45 bbl/STB
Water saturation	$S_{wi}$	0.25
Total compressibility	$c_t$	$1\times10^{-6}\ \text{psi}^{-1}$

Extended drawdown pressure data (printed on source)
$t$ (hr)	$p_{wf}$ (psia)
36.2	5417
97	5368
196	5335
300	5305
370	5293
485	5265
608	5237
725	5209

Find. Original oil in place, $N$.

Approach. Compute the real-time (not log-time) rate of pressure decline $dp/dt$ between consecutive points; once the well reaches its drainage boundary the decline becomes pseudo-steady-state (constant $dp/dt$), which is diagnostic of the reservoir's finite pore volume and lets the exam's own PSS depletion equation be inverted directly for pore volume and then OOIP.

Screen for pseudo-steady-state (constant late-time $dp/dt$). Consecutive-point slopes: $$\begin{array}{c|c} \text{Interval (hr)} & dp/dt\ (\text{psi/hr})\\\hline 36.2\to97 & -0.81\\ 97\to196 & -0.33\\ 196\to300 & -0.29\\ 300\to370 & -0.17\\ 370\to485 & -0.24\\ 485\to608 & -0.23\\ 608\to725 & -0.24 \end{array}$$ From $t=370$ hr onward the slope settles to a materially constant value — a linear regression of $p$ vs. $t$ over these last four points gives $dp/dt=-0.2357$ psi/hr with $R^2=0.9998$, confirming pseudo-steady-state (boundary-dominated) flow.
Pore volume from the PSS depletion rate. Using the formula sheet's PSS relation $dp_{wf}/dt=-0.234qB_o/(c_tV_p)$ (with the standard oilfield constant $0.23396$), $$V_p=\frac{0.23396\,qB_o}{c_t\,|dp/dt|}=\frac{0.23396(100)(1.45)}{(1\times10^{-6})(0.2357)}$$ $$V_p\approx 1.440\times10^{8}\ \text{bbl (reservoir pore volume)}$$
Original oil in place. Converting the oil-filled pore volume to stock-tank barrels, $$N=\frac{V_p(1-S_{wi})}{B_o}=\frac{(1.440\times10^{8})(1-0.25)}{1.45}$$ $$\boxed{N\approx 74.5\ \text{MMSTB}}$$

Result	Value
Late-time (PSS) decline rate, $dp/dt$	−0.236 psi/hr
Reservoir pore volume, $V_p$	≈ 144 MMbbl
Original oil in place, $N$	≈ 74.5 MMSTB

Topic. Reservoir-limit (PSS depletion) test for volumetric OOIP.

Key relations. $dp_{wf}/dt=-0.23396qB_o/(c_tV_p)$; $N=V_p(1-S_{wi})/B_o$.

Why this works. Once a transient pressure disturbance has reached every boundary of a closed drainage volume, further withdrawal simply depletes the reservoir at (very nearly) a constant volumetric rate, so $dp/dt$ becomes constant and directly proportional to $q/(c_tV_p)$ — inverting that single algebraic relation gives the pore volume without needing permeability or skin at all.

Common pitfall. Averaging $dp/dt$ over the WHOLE data set (including the still-transient early points) understates the true late-time decline rate and overstates OOIP; only the constant-slope tail should be used.

Source. Earlougher, Advances in Well Test Analysis, Ch. 2 (reservoir limit test); Lee, Well Testing, Ch. 2.

Question 5 of 6: Extended drawdown – reservoir oil in place

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