Question 7 of 7: Gas Well Buildup Test — Permeability, Skin, and End of Wellbore Storage

Question 7: Gas Well Buildup Test — Permeability, Skin, and End of Wellbore Storage (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. $p_{wf}=1379$ psia (initial flowing pressure); $h=45$ ft; $r_w=0.512$ ft; $\phi=0.105$; $q=9500$ MSCFD; $c_g=8\times10^{-4}$, $c_w=3\times10^{-6}$, $c_f=6\times10^{-6}$ psia$^{-1}$; $S_{wc}=S_{wi}=0.28$; $\bar\mu_g=0.02$ cp; $\psi_{wf}(\Delta t=0)=1.73\times10^8$ psi$^2$/cp; $T=659.67^{\circ}$R (assumed, see the check note). Horner semilog plot ($\psi_{ws}$ vs. $\log[(t_p+\Delta t)/\Delta t]$) and log-log diagnostic plot ($\Delta\psi$ vs. $\Delta t$) given.

Find. Formation permeability $k$, skin factor $S'$, and the approximate end of wellbore storage.

Approach. Fit the Horner semilog middle-time-region (MTR) straight line to get the slope $m$ and hence $k$; read $\psi$ at $\Delta t=1$ hr directly off the log-log plot for the skin calculation; and use the log-log plot's departure from (and eventual flattening relative to) the early unit-slope trend to bound the end of wellbore storage.

1. Total system compressibility. $c_t=c_g(1-S_{wi})+c_wS_{wi}+c_f=8\times10^{-4}(0.72)+3\times10^{-6}(0.28)+6\times10^{-6}$: $\boxed{c_t=5.828\times10^{-4}\ \text{psi}^{-1}}$.
2. Semilog slope (MTR). Fitting the Horner plot's declining middle-time-region trend (roughly Horner ratio 200–3000, between the early wellbore-storage-distorted plateau at large Horner ratio and the late-time flattening as the well approaches average reservoir pressure) by least squares against $\log_{10}[(t_p+\Delta t)/\Delta t]$: $\boxed{m=1.281\times10^8\ \text{psia}^2/\text{cp per cycle}}$.
3. Permeability. $k=1637qT/(mh)=1637(9500)(659.67)/(1.281\times10^8\times45)$: $\boxed{k=1.78\ \text{mD}}$.
4. Pseudopressure at $\Delta t=1$ hr. Read directly off the log-log plot's plateau near $\Delta t=1$ hr: $\Delta\psi(1\text{hr})\approx3.55\times10^8$, so $\psi_{1hr}=\psi_{wf}(0)+\Delta\psi(1\text{hr})=1.73\times10^8+3.55\times10^8=\boxed{\psi_{1hr}=5.28\times10^8\ \text{psia}^2/\text{cp}}$.
5. Skin factor. $S'=1.151\left[\dfrac{\psi_{1hr}-\psi_{wf}(0)}{m}-\log_{10}\!\left(\dfrac{k}{\phi\mu c_tr_w^2}\right)+3.23\right]$. With $k/(\phi\mu c_tr_w^2)=1.78/(0.105\times0.02\times5.828\times10^{-4}\times0.512^2)=5.51\times10^5$, $\log_{10}(\cdot)=5.741$, and $(\psi_{1hr}-\psi_{wf}(0))/m=3.55\times10^8/1.281\times10^8=2.771$: $S'=1.151(2.771-5.741+3.23)$: $\boxed{S'=-0.86}$ — essentially an undamaged well, mildly negative (no meaningful stimulation or damage).
6. End of wellbore storage. The log-log $\Delta\psi$-vs-$\Delta t$ plot follows a near-unit-slope trend out to about $\Delta t\approx0.03$–0.05 hr, then curves away toward the flattened infinite-acting-radial-flow trend; applying the standard rule of thumb (WBS ends roughly 1 to 1.5 log cycles after departure from unit slope): $\boxed{\Delta t_{WBS,end}\approx0.3\ \text{hr}}$ (an estimate, not an exact boundary).