Reservoir limit test – pore volume and initial pressure: 24-Pet-B5 Reservoir Mechanics, December 2014, Q5

Question 5: Reservoir limit test – pore volume and initial pressure (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
Oil production rate	$q$	1000 STB/D
Oil formation volume factor	$B_o$	1.2 bbl/STB
Total compressibility	$c_t$	$1\times10^{-5}\ \text{psi}^{-1}$
Wellbore radius	$r_w$	0.3 ft
Oil viscosity	$\mu$	1 cp
Formation thickness	$h$	25 ft
External radius of reservoir	$r_e$	2185 ft
Reservoir permeability	$k$	50 mD
Skin factor	$S$	2

Find. Reservoir pore volume $V_p$ and initial reservoir pressure $p_i$.

Approach. Fit the late-time (pseudosteady-state) portion of the $p_{wf}$-vs-$t$ data with a straight line; its slope gives $V_p$ directly from the PSS decline-rate formula. The same PSS pressure-profile formula, evaluated using that slope, then isolates $p_i$ without needing porosity separately.

Fig. 4 – Flowing bottomhole pressure vs. time. The dashed line is the least-squares fit through the late-time ($t\ge20$ hr), linear pseudosteady-state portion of the data; the earlier points still reflect the transient (infinite-acting) period and are excluded from the fit.

Fit the PSS straight line. The ten data points from $t=20$ to $88$ hr lie on a clean straight line (the earlier points, $t\le15$ hr, are still transient and are excluded). Least squares gives $$\frac{dp_w}{dt}\approx-0.416\ \text{psi/hr},\qquad p(t)\approx 494.6-0.416\,t\ \ (\text{psia, fit intercept at }t=0)$$
Pore volume from the decline rate. Rearranging the formula sheet's PSS decline relation for $V_p$, $$V_p=\frac{-0.234\,qB_o}{c_t\left(dp_w/dt\right)}=\frac{-0.234(1000)(1.2)}{(1\times10^{-5})(-0.416)}$$ $$\boxed{V_p\approx 6.74\times10^{7}\ \text{ft}^3}\ \ (\approx1.20\times10^{6}\ \text{bbl reservoir volume})$$
Skin/geometry drawdown, independent of $\phi$. The formula sheet's exact PSS pressure profile at the wellbore is $p(r_w,t)=p_i-\dfrac{0.0744qB_ot}{\phi c_th r_e^2}+\dfrac{q\mu B_o}{0.00708kh}\!\left[\ln(r_e/r_w)-\tfrac34+S\right]$. Because the material-balance coefficient $0.0744\approx0.234/\pi$ (both constants on the same formula sheet), the first correction term collapses to exactly $-t\,(dp_w/dt)$ once $V_p$ is known – so $\phi$ never has to be evaluated separately. The skin/flow-convergence term alone is $$\Delta p_{skin}=\frac{q\mu B_o}{0.00708kh}\left[\ln\!\left(\frac{r_e}{r_w}\right)-\frac34+S\right]=\frac{(1000)(1)(1.2)}{0.00708(50)(25)}\Big[\ln(2185/0.3)-0.75+2\Big]\approx1375\ \text{psi}$$
Initial pressure. A physically consistent PSS profile must sit below $p_i$ by both the depletion term and the skin/convergence loss (never above by one and below by the other, as a literal plus-sign reading of the printed formula would require – check: treated as a source sign inconsistency). Matching the fitted line's own $t=0$ intercept to $p_i$ minus that fixed skin/geometry loss, $$p_i=(\text{fit intercept})+\Delta p_{skin}=494.6+1375$$ $$\boxed{p_i\approx 1874\ \text{psia}}$$

Result	Value
PSS decline rate, $dp_w/dt$	≈ −0.416 psi/hr
Reservoir pore volume, $V_p$	≈ 6.74×10⁵ ft³ (≈1.20×10⁶ bbl)
Skin + flow-convergence loss	≈ 1375 psi
Initial reservoir pressure, $p_i$	≈ 1874 psia

Check: the exam's own formula sheet prints the wellbore PSS pressure profile with the skin/geometry bracket ADDED to $p_i$ minus the depletion term; taken literally this gives an unphysical negative $p_i$ ($\approx-877$ psia). A physically consistent profile subtracts both loss terms from $p_i$ (used above); flagged here rather than silently reconciled, since it changes the sign convention of a formula printed on the exam's own formula sheet.

Topic. The reservoir-limit test (Q1-j in this same paper).

Key relations. $dp_w/dt=-0.234qB_o/(c_tV_p)$; $p(r_w,t)=p_i-(\text{depletion term})-(\text{skin/geometry term})$; $V_p=\pi\phi h r_e^2$ (so the depletion term reduces to $t\times|dp_w/dt|$ once $V_p$ is known, independent of $\phi$).

Why this works. Once the outer no-flow boundary is fully felt (pseudosteady state), the ENTIRE reservoir depletes at one uniform rate set purely by how much oil is being removed relative to how much pore volume and compressibility is available to absorb it – that single number, read straight off the slope of a late-time $p$-vs-$t$ plot, is a full material-balance measurement of drainage volume with no separate volumetric (area × net pay × porosity) survey required.

Common pitfall. Fitting the straight line through the EARLY data (here $t<20$ hr) – those points are still in the transient, infinite-acting period (curved, not linear, on $p$ vs. $t$) and including them biases the fitted slope, understating $V_p$.

Source. Earlougher, Advances in Well Test Analysis, ch.5 (reservoir limit test); Lee, Well Testing, ch.4 (pseudosteady-state material balance).

Question 5 of 7: Reservoir limit test – pore volume and initial pressure

Question 5: Reservoir limit test – pore volume and initial pressure (20 marks)