Question 3 of 7: Interference Test — Time to a Given Pressure Drop at an Observation Well

Question 3: Interference Test — Time to a Given Pressure Drop at an Observation Well (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. Distance between wells $r=1000$ ft; producing well rate $q=100$ STBD; target pressure drop at the observation well $\Delta p=2.5$ psi; reservoir/fluid data as tabulated above.

Find. The elapsed time $t$ for the observation well to record a 2.5 psi pressure drop.

Approach. The observation well records the transient pressure disturbance created by the producing well at radius $r=1000$ ft; solve the exam's own dimensionless-pressure equation $p_D=0.5[-\text{Ei}(-1/4t_D)]$ for $t_D$ at the required $p_D$ (checking whether the simpler log approximation is valid), then convert $t_D$ back to real time using this reservoir's $\eta=6.33k/(\phi\mu c_t)$.

1. Skin does not apply here. The skin factor $S=-2$ describes an extra (here, negative/stimulated) pressure loss confined to the near-wellbore zone of the well that is actually flowing. Because the observation well is 1000 ft away and carries zero rate, the skin term of the producing well has no effect on the pressure recorded there; it is a distractor for this part of the problem.
2. Required dimensionless pressure. $p(r,t)=p_i-\dfrac{0.141\,q\mu B_o}{kh}\,p_D\ \Rightarrow\ p_D=\dfrac{\Delta p\cdot kh}{0.141\,q\mu B_o}=\dfrac{2.5(0.4)(20)}{0.141(100)(1)(1.3)}$ (k in Darcy per the formula sheet, $k=400\ \text{mD}=0.4$ Darcy), so $\boxed{p_D=1.091}$.
3. Check the log approximation. If $p_D=0.5(\ln t_D+0.809)$ were used, solving gives $t_D\approx3.7$, which is far below the formula sheet's own validity cutoff of $t_D\gt100$; the log approximation cannot be used, and the exact line-source form $p_D=0.5[-\text{Ei}(-1/4t_D)]$ must be solved instead.
4. Solve for $t_D$ from the exact line-source solution. Numerically inverting $p_D=0.5[-\text{Ei}(-1/4t_D)]=1.091$ (bisection on the exponential-integral function) gives $\boxed{t_D=3.69}$ — consistent with the log-approximation estimate above, confirming $t_D\lt100$ and that the exact form was the right choice.
5. Convert to real time. $\eta=\dfrac{6.33k}{\phi\mu c_t}=\dfrac{6.33(0.4)}{0.15(1)(3\times10^{-6})}=5.627\times10^{6}\ \text{ft}^2/\text{day}$. Since $t_D=\eta t/r^2$, $t=\dfrac{t_D r^2}{\eta}=\dfrac{3.69(1000)^2}{5.627\times10^{6}}$, so $\boxed{t=0.656\ \text{days}\approx15.8\ \text{hours}}$.