Bottom-Hole Pressure Near a Sealing Fault (Method of Images): 24-Pet-A3 Fundamental Reservoir Engineering, May 2016, Q3

Question 3: Bottom-Hole Pressure Near a Sealing Fault (Method of Images) (20 marks)

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Find. The bottom-hole (flowing wellbore) pressure of the well after 1 day of production at 250 STBD, accounting for the nearby sealing fault.

Fig. 1 — A sealing fault is modelled by the method of images: an imaginary well of the same rate is placed a mirror distance $L$ beyond the fault, so its added pressure drop reproduces the fault's no-flow condition.

Approach. A single sealing (no-flow) fault is modelled by superposition: place an image well of the same rate at twice the well-to-fault distance ($2L=500$ ft) and add its pressure-drop contribution to the real well's own (infinite-acting) contribution. Compute $t_D$ at $r=r_w$ (real well) and at $r=2L$ (image well); since the exam's own formula sheet states the log approximation $p_D=0.5(\ln t_D+0.809)$ is valid "only if $t_D>100$", check each $t_D$ and use the exact line-source form $p_D=0.5[-Ei(-1/4t_D)]$ wherever it is not.

Diffusivity constant. With $k=100\ \text{mD}=0.1$ Darcy (the formula sheet's transient equations use $k$ in Darcy, consistent with its steady-flow equation), $\eta=\dfrac{6.33k}{\phi\mu c_t}=\dfrac{6.33(0.1)}{(0.20)(2)(6\times10^{-6})}=263{,}750\ \text{ft}^2/\text{day}$.
Real well contribution. At $r=r_w=0.33$ ft, $t=1$ day: $t_D=\eta t/r_w^2=263{,}750/0.33^2=2.42\times10^6$. Since $t_D>100$, use the log form: $\boxed{p_D=0.5(\ln t_D+0.809)=0.5(14.70+0.809)=7.755}$.
Image-well contribution. At $r=2L=500$ ft: $t_D=\eta t/(500)^2=263{,}750/250{,}000=1.055$. Since $t_D<100$, the log form is invalid; use the exact line-source form: $\boxed{p_D=0.5[-Ei(-1/(4\times1.055))]=0.5[-Ei(-0.2370)]=0.543}$.
Bottom-hole pressure. $\dfrac{0.141\,q\mu B_o}{kh}=\dfrac{0.141(250)(2)(1.25)}{(0.1)(100)}=8.81$ psi. Superposing both wells (no skin given), $\Delta p=8.81\,(p_{D,\text{real}}+p_{D,\text{image}})=8.81(7.755+0.543)=8.81(8.298)=73.1$ psi. $\boxed{p_{wf}=p_i-\Delta p=3000-73.1=2926.9\ \text{psia}}$.

Quantity	Value
$t_D$, real well ($r=r_w$)	$2.42\times10^6$
$p_D$, real well	7.755
$t_D$, image well ($r=2L=500$ ft)	1.055
$p_D$, image well	0.543
Total pressure drop, $\Delta p$	73.1 psi
Bottom-hole pressure after 1 day	2926.9 psia

Topic: Method of images for a single sealing (no-flow) fault boundary in transient radial flow.

Key relations: $\eta=6.33k/(\phi\mu c_t)$; $t_D=\eta t/r^2$; exact $p_D=0.5[-Ei(-1/4t_D)]$ (any $t_D$); log approximation $p_D=0.5(\ln t_D+0.809)$ (valid only for $t_D>100$); image well at $r=2L$ with the same (not opposite) sign of rate.

Why this works: A no-flow boundary can be reproduced exactly, for a well in an otherwise infinite reservoir, by superposing an identical "image" well reflected across the boundary at twice the true distance and flowing at the same rate and sign; the extra pressure drop this image well contributes at the real well is mathematically identical to the effect of the physical fault reflecting the pressure disturbance back.

Common pitfall: Using the log-approximation $p_D$ formula for the image-well term without checking its $t_D>100$ validity condition — here $t_D=1.055$ for the image well, so the log form would understate $p_D$ and understate the total pressure drop.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 7 (transient flow, boundary effects, method of images); Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 8.

Question 3 of 7: Bottom-Hole Pressure Near a Sealing Fault (Method of Images)

Question 3: Bottom-Hole Pressure Near a Sealing Fault (Method of Images) (20 marks)