24-Pet-A3 Fundamental Reservoir Engineering · December 2014
Question 3 of 7: Interference Test — Pressure at an Observation Well
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Pet-A3 — Fundamental Reservoir Engineering · National Exams, December 2014 · 3 hours, closed book, Casio/Sharp approved calculator only · five (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked), all questions equal value, all parts of a multipart question equal weight.
Reference texts: Ahmed, T., Reservoir Engineering Handbook, 5th ed. (Darcy's law and fluid potential, transient well testing, p/Z and oil material balance, capillary pressure/relative permeability); Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (reservoir drive mechanisms, material balance fundamentals); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed. (Standing–Katz Z-factor correlation); McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (capillary pressure and relative permeability laboratory data).
Question 3: Interference Test — Pressure at an Observation Well (20 marks)
Check: the source gives a "skin factor of well #1" of 1, but well #1 is the non-producing observation well. Skin (Hawkins' formula, $\Delta p_{skin}=141.2\,q\mu B_o S/(kh)$) is an additional pressure drop generated only across the flowing well's own damaged near-wellbore zone, proportional to that well's own flow rate; since well #1 carries zero rate, its skin contributes zero additional pressure drop at its own gauge. This value is treated as a distractor and is not applied — only well #2's rate and the formation properties govern the pressure well #1 sees.
Find. The pressure at the observation well (well #1), 3000 ft from the producer, after 5 days of production from well #2 at 500 STBD.
Fig. 2 — Well #2 produces; well #1, 3000 ft away, only records the transient pressure response — its own skin is inert.
Approach. Compute the dimensionless time $t_D$ at the 3000 ft observation distance; because $t_D<100$ the formula sheet's log approximation does not apply (it is valid only for $t_D>100$), so use the exact exponential-integral (line-source) solution shown on the same sheet's $p_D$–$t_D$ chart, then convert to a pressure drop with no skin term.
Diffusivity and dimensionless time. With $k=50\ \text{mD}=0.05$ Darcy, $\eta=\dfrac{6.33k}{\phi\mu c_t}=\dfrac{6.33(0.05)}{(0.15)(1)(5\times10^{-6})}=422{,}000\ \text{ft}^2/\text{day}$. At $r=3000$ ft, $t=5$ days: $t_D=\dfrac{\eta t}{r^2}=\dfrac{422{,}000\times5}{3000^2}=\boxed{0.234}$.
Dimensionless pressure (line-source solution). Since $t_D=0.234<100$, the log form $p_D=\tfrac12(\ln t_D+0.809)$ is invalid; use the exact form on the formula sheet, $p_D=\tfrac12\left[-Ei\!\left(-\dfrac{1}{4t_D}\right)\right]=\tfrac12\left[-Ei(-1.067)\right]=\boxed{p_D=0.0983}$.
Pressure at well #1. $\dfrac{0.141\,q\mu B_o}{kh}=\dfrac{0.141(500)(1)(1.2)}{(0.05)(50)}=33.84$ psi. With no skin term (Step-callout above), $\Delta p = 33.84\times0.0983=3.33$ psi, so $\boxed{p(r,t)=2000-3.33=1996.7\ \text{psia}}$.
