Interference test – required production rate: 24-Pet-B5 Reservoir Mechanics, May 2016, Q2

Question 2: Interference test – required production rate (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
Interwell distance	$r$	330 ft
Permeability	$k$	2500 mD
Formation thickness	$h$	20 ft
Porosity	$\phi$	0.30
Oil viscosity	$\mu$	2 cP
Total compressibility	$c_t$	$5\times10^{-5}\ \text{psi}^{-1}$
Formation volume factor	$B_o$	1.2 bbl/STB
Observation time	$t$	5 hr
Target pressure drop	$\Delta p$	1 psi

Find. The active well's production rate $q$ needed to give $\Delta p=1$ psi at the observation well 5 hr after production starts.

Approach. Compute the observation well's dimensionless time $t_D$ at $r=330$ ft; since $t_D$ turns out far below 100, evaluate $p_D$ with the exact line-source ($E_i$) form (not the log approximation), then solve the radial-flow $\Delta p$ equation for $q$.

Dimensionless time at the observation well. $$t_D=\frac{0.0002637kt}{\phi\mu c_tr^2}=\frac{0.0002637(2500)(5)}{(0.3)(2)(5\times10^{-5})(330)^2}\approx 1.01$$ Since $t_D\approx 1.01\ll 100$, the log approximation is invalid and the exact line-source form must be used.
Dimensionless pressure (exact line source). $$p_D=0.5\left[-Ei\!\left(-\frac{1}{4t_D}\right)\right]=0.5\,E_1(0.248)\approx 0.526$$
Solve for the required rate. Rearranging $\Delta p=141.2q\mu B_op_D/(kh)$, $$q=\frac{\Delta p\,kh}{141.2\,\mu B_op_D}=\frac{(1)(2500)(20)}{141.2(2)(1.2)(0.526)}$$ $$\boxed{q\approx 281\ \text{STBD}}$$

Result	Value
$t_D$ at $r=330$ ft, $t=5$ hr	≈ 1.01
$p_D$ (exact line source)	≈ 0.526
Required production rate, $q$	≈ 281 STBD

Topic. Interference-test design via the exact line-source solution.

Key relations. $t_D=0.0002637kt/(\phi\mu c_tr^2)$; $p_D=0.5[-Ei(-1/4t_D)]$ (exact) $\to 0.5[\ln t_D+0.809]$ only for $t_D>100$; $\Delta p=141.2q\mu B_op_D/(kh)$.

Why this works. An observation well far from the active well sees a much smaller dimensionless time than the active wellbore itself at the same real time, because $t_D\propto 1/r^2$ — at typical interwell spacings and test durations $t_D$ is usually well under 100, so the convenient log approximation cannot be used and the full exponential-integral solution is required.

Common pitfall. Applying $p_D=0.5[\ln t_D+0.809]$ without checking $t_D>100$ first; here it would give $p_D\approx0.409$, a 22% error that propagates directly into a wrong required rate.

Source. Lee, Well Testing, Ch. 1 (line-source solution); Earlougher, Advances in Well Test Analysis, Ch. 5 (interference test design).

Question 2 of 6: Interference test – required production rate

Question 2: Interference test – required production rate (20 marks)