Question 2 of 6: Superposition in Time — Two-Rate Drawdown Then Shut-In in an Infinite-Acting Reservoir

Question 2: Superposition in Time — Two-Rate Drawdown Then Shut-In in an Infinite-Acting Reservoir (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. $p_i=3500$ psia, $\phi=0.18$, $c_o=12\times10^{-6}$, $c_w=4\times10^{-6}$, $c_f=4\times10^{-6}\ \text{psi}^{-1}$, $B_o=1.2$ bbl/STB, $S_{wi}=0.25$, $h=30$ ft, $\mu=1$ cp, $k=20$ md, $r_w=0.3$ ft, $S=0$ (undamaged), infinite-acting. Rate schedule: $q_1=50$ STB/day for $0\le t<120$ hr, $q_2=20$ STB/day for $120\le t<168$ hr, then shut in ($q_3=0$) from $t=168$ hr.

Find. (a) $p_{wf}$ at $t=168$ hr (2 days after the rate decrease); (b) $p_{ws}$ at $t=192$ hr (1 day after shut-in).

Approach. Compute $c_t$ from the three individual compressibilities, then use superposition in time: each rate change $\Delta q_i$ starting at its own time $t_i$ contributes an independent log-approximation drawdown term evaluated at the elapsed time since that change; sum the terms (no skin term needed since $S=0$).

1. Total compressibility. $c_t=c_o(1-S_{wi})+c_w S_{wi}+c_f=12\times10^{-6}(0.75)+4\times10^{-6}(0.25)+4\times10^{-6}$, so $\boxed{c_t=14\times10^{-6}\ \text{psi}^{-1}}$.
2. Superposition template. For a rate change $\Delta q$ effective for $\tau$ hours, $\Delta p(\Delta q,\tau)=\dfrac{162.6\,\Delta q\,\mu B_o}{kh}\left[\log\left(\dfrac{k\tau}{\phi\mu c_t r_w^2}\right)-3.23\right]$. Here $\dfrac{162.6\,\mu B_o}{kh}=\dfrac{162.6(1)(1.2)}{20(30)}=0.3252\ \text{psi/(STB/day)}$, and $\phi\mu c_t r_w^2=0.18(1)(14\times10^{-6})(0.3)^2=2.268\times10^{-7}$ (checked $\ll0.01\times0.001055k\tau$ at every $\tau$ used below, so the log approximation is valid throughout).
3. (a) Pressure 2 days after the rate decrease ($t=168$ hr). Two active terms: $q_1=50$ from $t=0$ (so $\tau=168$ hr), and $\Delta q_2=20-50=-30$ from $t=120$ hr (so $\tau=48$ hr). $\left[\log\!\left(\tfrac{20(168)}{2.268\times10^{-7}}\right)-3.23\right]=6.941$ and $\left[\log\!\left(\tfrac{20(48)}{2.268\times10^{-7}}\right)-3.23\right]=6.397$. $\Delta p=0.3252\left[50(6.941)+(-30)(6.397)\right]=0.3252(347.0-191.9)$, so $\boxed{\Delta p=50.4\ \text{psi}}$ and $\boxed{p_{wf}=3500-50.4=3449.6\ \text{psia}}$.
4. (b) Pressure 1 day after shut-in ($t=192$ hr). Add a third term, $\Delta q_3=0-20=-20$ from $t=168$ hr ($\tau=24$ hr): $\left[\log\!\left(\tfrac{20(192)}{2.268\times10^{-7}}\right)-3.23\right]=6.999$, $\left[\log\!\left(\tfrac{20(72)}{2.268\times10^{-7}}\right)-3.23\right]=6.573$, $\left[\log\!\left(\tfrac{20(24)}{2.268\times10^{-7}}\right)-3.23\right]=6.096$. $\Delta p=0.3252\left[50(6.999)-30(6.573)-20(6.096)\right]=0.3252(349.9-197.2-121.9)$, so $\boxed{\Delta p=10.0\ \text{psi}}$ and $\boxed{p_{ws}=3500-10.0=3490.0\ \text{psia}}$.