24-Pet-A3 Fundamental Reservoir Engineering · December 2018
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
17-Pet-A3 — Fundamental Reservoir Engineering · National Exams, December 2018 · 3 hours, closed book, approved Casio/Sharp 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: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, well testing, relative permeability, Darcy flow); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, transient well testing, gas PVT, decline-curve analysis); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.; McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PVT properties, Z-factor correlations).
Question 3: Interference Test — Pressure at an Observation Well (20 marks)
Given. Interwell distance $r=3000$ ft, $t=5$ days, $q_o=500$ STBD (Well #2), $k=50$ mD $=0.05$ D, $h=50$ ft, $\phi=0.15$, $\mu=1$ cP, $c_t=5\times10^{-6}\ \text{psi}^{-1}$, $B_o=1.2$ bbl/STB, $p_i=2000$ psia; Well #1's own skin $S=1$ is stated, but Well #1 is not flowing.
Find. The pressure at observation Well #1 after 5 days of production from Well #2.
Approach. The pressure drop felt at a distant, non-producing observation well is the line-source (Ei-function) transient solution evaluated at $r=3000$ ft; skin only adds an extra pressure drop at the sandface of the well that is actually flowing, so it must be excluded when evaluating the pressure seen at Well #1.
Fig. 1: plan view of the two-well interference test. Well #1's skin acts only at its own wellbore and does not enter the interference pressure it observes from Well #2.
Dimensionless time at the observation well. Using the formula sheet's $\eta=6.33k/(\phi\mu c)$ (k in Darcy) and $t_D=\eta t/r^2$: $\eta=6.33(0.05)/[0.15(1)(5\times10^{-6})]=422{,}000\ \text{ft}^2/\text{day}$, so $t_D=422{,}000(5)/3000^2$, giving $\boxed{t_D=0.2344}$.
Choice of $p_D$ form. Since $t_D=0.2344<100$, the log approximation $p_D=0.5(\ln t_D+0.809)$ is NOT valid; the exact line-source form must be used: $p_D=0.5\left[-\text{Ei}\!\left(-\dfrac{1}{4t_D}\right)\right]$. Argument $\dfrac{1}{4t_D}=\dfrac{1}{4(0.2344)}=1.0664$; evaluating the exponential integral, $-\text{Ei}(-1.0664)=0.1965$, so $\boxed{p_D=0.0983}$.
Pressure drop at Well #1 (no skin term). $p(r,t)=p_i-\dfrac{0.141q_o\mu B_o}{kh}(p_D+S)$, with $S=0$ here — Well #1's own skin factor is a distractor: Hawkins' skin term only represents the extra pressure drop across the damaged annulus of a well that is itself flowing, and Well #1 produces nothing. $\dfrac{0.141(500)(1)(1.2)}{0.05(50)}=33.84\ \text{psi/(dimensionless pressure)}$, so $\Delta p=33.84(0.0983)$, giving $\boxed{\Delta p=3.33\ \text{psi}}$.
Observed pressure. $p(r,t)=2000-3.33$, so $\boxed{p_{\text{Well\ \#1}}=1996.7\ \text{psia}}$.
Check: the given $r_e=4000$ ft (reservoir external radius) exceeds the ~3000 ft interwell spacing but is not itself used in this calculation — it only confirms the reservoir is large enough that the 5-day pulse ($r_{\text{investigated}}\sim\sqrt{\eta t}\approx1450$ ft from Well #2) has not yet reached the outer boundary, so the infinite-acting line-source solution remains valid.
