Gas Well Near a Sealing Fault — Image-Well Superposition: 24-Pet-B2 Oil and Gas Evaluation and Economics, December 2015, Q6

Question 6: Gas Well Near a Sealing Fault — Image-Well Superposition (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. $L=500$ ft to the fault; $p_i=2000$ psia; $T=580^{\circ}$R; $h=39$ ft; $\mu=0.0158$ cP; $\phi=0.15$; $k=20$ mD; $r_w=0.4$ ft; $c_i=0.00053$ psi$^{-1}$; $q_{sc}=5$ MMSCFD $=5000$ MSCFD; $t=1$ day. $\psi(p)$ given graphically.

Find. $p_{wf}$ after 1 day of production, accounting for the sealing fault.

Approach. Model the no-flow fault boundary with the method of images: a fictitious image well of the same rate and sign at twice the well-to-fault distance ($r=2L$) superposes its own pressure drop onto the real well’s. Since $t_D$ at the real wellbore radius is enormous (log approximation valid) but $t_D$ at the much larger image distance is small (log approximation invalid there), each term uses whichever $p_D$ form its own $t_D$ calls for — the given $p_D=0.5[-Ei(-1/4t_D)]$ chart is exactly this exact-form solution.

Fig. 3 — Method-of-images geometry: a same-rate, same-sign image well at $r=2L=1000$ ft reproduces the sealing fault's no-flow boundary condition at the real wellbore.

Dimensionless times. $t_D=6.33\times10^{-3}kt/(\phi\mu c_ir^2)$ ($t$ in days, $k$ in mD). At the real wellbore ($r=r_w=0.4$ ft): $\boxed{t_{D,real}=6.30\times10^5}$ (using $t_D>100$, log approximation). At the image well ($r=2L=1000$ ft): $\boxed{t_{D,image}=0.1008}$ (using the exact line-source form, since $t_D<100$).
Dimensionless pressures. Real well: $p_D=\tfrac12(\ln t_{D,real}+0.809)=\tfrac12(\ln(6.30\times10^5)+0.809)$: $\boxed{p_{D,real}=7.081}$. Image well: $p_D=\tfrac12[-Ei(-1/4t_{D,image})]=\tfrac12[-Ei(-2.480)]$, evaluated via the standard exponential-integral rational approximation: $\boxed{p_{D,image}=0.0128}$ — the image well's contribution is small but not negligible at only 500 ft from the fault.
Superposition. $p_{D,total}=p_{D,real}+p_{D,image}=7.081+0.0128=\boxed{p_{D,total}=7.094}$.
Pseudopressure drop. $\Delta\psi=\dfrac{1.422\,q_{sc}T}{kh}\,p_{D,total}=\dfrac{1.422(5000)(580)}{20(39)}(7.094)$: $\boxed{\Delta\psi=3.75\times10^7\ \text{psia}^2/\text{cp}}$.
Invert the $\psi(p)$ chart. From the chart, $\psi(p_i=2000)=3.400\times10^8$ psia$^2$/cp. $\psi(r_w,1\text{ day})=3.400\times10^8-3.75\times10^7=\boxed{\psi_{wf}=3.025\times10^8\ \text{psia}^2/\text{cp}}$. Interpolating the same chart backward for $p$: $\boxed{p_{wf}=1864\ \text{psia}}$.

Fig. 4 — Real gas pseudopressure function $\psi(p)$ (given chart, digitized), with $\psi(p_i)$ and the well pressure after 1 day marked.

Quantity	Value
$t_D$ at real wellbore	$6.30\times10^5$
$t_D$ at image well ($r=2L$)	0.1008
$p_D$, real + image (superposed)	7.094
Pseudopressure drop, $\Delta\psi$	$3.75\times10^7$ psia$^2$/cp
Well pressure after 1 day, $p_{wf}$	1864 psia

Check: the $\psi(p)$ chart was digitized at nine points across its 0–2500 psia range and interpolated piecewise-linearly; a smoother interpolant would shift $p_{wf}$ by at most a few psia. The image-well distance uses the standard method-of-images convention for a single sealing (no-flow) linear fault: an image of the same sign and rate at twice the well-to-boundary distance.

Topic: Bounded-reservoir transient flow via the method of images (single sealing fault).

Key relations: $\psi(r,t)=\psi_i-\dfrac{1.422q_{sc}T}{kh}\sum p_D$; $t_D=6.33\times10^{-3}kt/(\phi\mu c_ir^2)$; $p_D=\tfrac12(\ln t_D+0.809)$ for $t_D>100$, else $p_D=\tfrac12[-Ei(-1/4t_D)]$.

Why this works: A single linear no-flow boundary can be reproduced exactly, for the region on the well's side of the fault, by superposing the real well's own pressure response with that of a fictitious well of identical rate and sign mirrored across the fault — the two solutions sum to satisfy the zero-flow-across-the-fault condition automatically.

Common pitfall: Using the log-time approximation for the image well's $p_D$ without first checking its own $t_D$ — at a much larger radius than $r_w$, $t_D$ can easily fall below the $t_D>100$ validity threshold even when the real well's own $t_D$ is enormous, silently corrupting the superposed answer if applied blindly.

Source: Lee & Wattenbarger, Gas Reservoir Engineering, Ch. 2 & 4 (bounded reservoirs, method of images); formula sheet (this exam).

Question 6 of 7: Gas Well Near a Sealing Fault — Image-Well Superposition

Question 6: Gas Well Near a Sealing Fault — Image-Well Superposition (20 marks)