Sealing fault – pressure after 5 days (method of images): 24-Pet-B5 Reservoir Mechanics, May 2015, Q2

Question 2: Sealing fault – pressure after 5 days (method of images) (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
Oil rate	$q$	500 STBD
Wellbore radius	$r_w$	0.25 ft
Total compressibility	$c_t$	$5\times10^{-5}\ \text{psi}^{-1}$
Oil viscosity	$\mu$	2 cP
Porosity	$\phi$	0.3
Permeability	$k$	250 mD
Formation thickness	$h$	20 ft
Oil formation volume factor	$B_o$	1.2 bbl/STB
Initial pressure	$p_i$	3000 psi
Distance, well to fault	$L$	25 ft
Elapsed time	$t$	5 days = 120 hr

Find. Flowing wellbore pressure $p_{wf}$ at $t=5$ days.

Approach. A sealing fault is solved by the method of images – replace the barrier with an identical "image" well of the same rate at twice the well-to-fault distance, $2L$, then superpose the real well's own pressure drop and the image well's pressure drop at the actual wellbore. Each term's dimensionless time is evaluated separately and compared against the $t_D\gt 100$ threshold to decide whether the log approximation or the exact $Ei$-function form applies.

Fig. 1 – Plan view: the real well, the sealing fault, and its image well at twice the well-to-fault distance (method of images).

Dimensionless time for the real well (at $r_w$). Using the formula-sheet group $t_D=0.0002637kt/(\phi\mu c_tr_w^2)$ at $t=120$ hr, $$t_{D,rw}=\frac{0.0002637(250)(120)}{(0.3)(2)(5\times10^{-5})(0.25)^2}\approx 4.22\times10^{6}$$ which is far above 100, so the real well's own contribution uses the log approximation $p_D=0.5[\ln t_D+0.809]$: $$p_{D,real}=0.5[\ln(4.22\times10^{6})+0.809]\approx 8.03$$
Dimensionless time for the image well (at $r=2L=50$ ft). Repeating with $r=2L$ in place of $r_w$, $$t_{D,2L}=\frac{0.0002637(250)(120)}{(0.3)(2)(5\times10^{-5})(50)^2}\approx 105$$ This sits right at the $t_D=100$ threshold, so the image well's contribution is evaluated with the exact line-source form (safe on either side of the threshold at this value): $$p_{D,image}=0.5\left[-Ei\!\left(-\frac{1}{4t_D}\right)\right]\approx 2.73$$
Superpose the two contributions. With both wells producing at the same rate $q$ (the image well mimics the no-flow fault by mirroring the real well's own withdrawal), the total pressure drop at the real wellbore is the sum of both $p_D$ terms: $$\Delta p=\frac{141.2\,q\mu B_o}{kh}\left(p_{D,real}+p_{D,image}\right)=\frac{141.2(500)(2)(1.2)}{(250)(20)}(8.03+2.73)$$ $$\Delta p\approx 364.8\ \text{psi}$$
Wellbore pressure. $$p_{wf}=p_i-\Delta p=3000-364.8$$ $$\boxed{p_{wf}(5\text{ days})\approx 2635\ \text{psia}}$$

Result	Value
$t_{D}$ at $r_w$	≈ 4.22×10⁶
$t_D$ at $2L$	≈ 105
Total pressure drop, $\Delta p$	≈ 365 psi
Wellbore pressure at 5 days, $p_{wf}$	≈ 2635 psia

Topic. Method of images for a single linear (sealing-fault) boundary.

Key relations. $t_D=0.0002637kt/(\phi\mu c_tr_w^2)$; $p_D=0.5[\ln t_D+0.809]$ ($t_D\gt 100$) or $p_D=0.5[-Ei(-1/4t_D)]$ (exact); $\Delta p=141.2q\mu B_o(\sum p_D)/(kh)$.

Why this works. A sealing (no-flow) fault is mathematically equivalent to reflecting the drainage domain across the fault plane and adding a mirror-image well of the same sign and rate – the two wells' combined flow field automatically satisfies zero flux across the fault, exactly the no-flow outer boundary condition derived in this paper's own diffusivity-equation question.

Common pitfall. Using the SAME (log-approximation) formula for both wells regardless of their own $t_D$ – the image well sits twice as far away, so its own dimensionless time can still be below the $t_D=100$ validity threshold even when the real well's is enormous; each term must be checked and evaluated on its own merits, exactly as done in Step 2 here.

Source. Earlougher, Advances in Well Test Analysis, ch.2 (method of images, bounded reservoirs); Lee, Well Testing, ch.2.

Question 2 of 7: Sealing fault – pressure after 5 days (method of images)

Question 2: Sealing fault – pressure after 5 days (method of images) (20 marks)