Question 3 of 6: Pressure-Drawdown Test — Permeability, Skin and a Volumetric OOIP Estimate

Question 3: Pressure-Drawdown Test — Permeability, Skin and a Volumetric OOIP Estimate (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. $p_i=4412$ psia, $q=250$ STB/day, $\mu=0.8$ cp, $B_o=1.136$ bbl/STB, $r_w=0.198$ ft, $h=69$ ft, $\phi=3.9\%$, $c_t=17\times10^{-6}\ \text{psi}^{-1}$, $S_{wi}=0.25$; $p_{wf}$ vs. $t$ recorded on the semilog (Fig. 1) and linear (Fig. 2) charts reproduced below.

Find. (a) permeability $k$; (b) skin factor $S$; (c) a volumetric estimate of initial oil in place $N$.

Approach. Digitize the semilog straight-line portion of Fig. 1 to get the slope $m$ and the $t=1$ hr intercept $p_{1hr}$; get $k$ from $m$ and $S$ from $p_{1hr}$ vs. $p_i$; then use the radius of investigation reached by the end of the recorded test (from this same exam's own formula-sheet transient solution) as a current/lower-bound drainage radius for a volumetric $N$.

1. Semilog slope and $p_{1hr}$. Fitting the straight-line portion of Fig. 1 ($t\gtrsim5$ hr) gives $\boxed{m=-75.8\ \text{psi/cycle}}$ and an extrapolated intercept $p_{1hr}=3666\ \text{psia}$ at $t=1$ hr.
2. Permeability. $|m|=\dfrac{162.6\,q\mu B_o}{kh}\ \Rightarrow\ k=\dfrac{162.6\,q\mu B_o}{h|m|}=\dfrac{162.6(250)(0.8)(1.136)}{69(75.8)}$, so $\boxed{k=7.06\ \text{md}}$.
3. Skin factor. $S=1.151\left[\dfrac{p_{1hr}-p_i}{m}-\log\!\left(\dfrac{k}{\phi\mu c_t r_w^2}\right)+3.23\right]$. $\dfrac{3666-4412}{-75.8}=9.83$; $\phi\mu c_t r_w^2=0.039(0.8)(17\times10^{-6})(0.198)^2=2.08\times10^{-8}$, so $\log(7.06/2.08\times10^{-8})=8.53$. $S=1.151[9.83-8.53+3.23]$, giving $\boxed{S=5.2}$ (a moderately damaged well).
4. Initial oil in place (radius of investigation). This exam's own formula sheet writes the transient drawdown as $p(r,t)=p_i+\dfrac{70.6q\mu B_o}{kh}\text{Ei}\!\left(-\dfrac{\phi\mu c_t r^2}{0.001055kt}\right)$; the pressure disturbance has meaningfully reached radius $r$ once the Ei-argument's magnitude is 1, i.e. $r_{\text{inv}}=\sqrt{0.001055\,k\,t/(\phi\mu c_t)}$. Using the last recorded test time, $t\approx322$ hr (Fig. 1's rightmost point, with no departure yet visible from the straight line): $r_{\text{inv}}=\sqrt{0.001055(7.06)(322)/(2.08\times10^{-8}\times69/69\ \ldots)}$; numerically, $r_{\text{inv}}=\sqrt{2.396/5.30\times10^{-7}}$, so $\boxed{r_{\text{inv}}=2125\ \text{ft}}$. Drainage area $A=\pi r_{\text{inv}}^2/43560=\pi(2125)^2/43560$, so $A=325.7$ acres, and $N=\dfrac{7758\,A\,h\,\phi(1-S_{wi})}{B_o}=\dfrac{7758(325.7)(69)(0.039)(0.75)}{1.136}$, giving $\boxed{N=4.49\times10^{6}\ \text{STB (4.49 MMSTB)}}$.