24-Pet-A3 Fundamental Reservoir Engineering · December 2014
Question 2 of 7: Interwell Flow Direction and Velocity
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 2: Interwell Flow Direction and Velocity (20 marks)
Find. (a) The direction of flow between the two wells; (b) the Darcy (superficial) fluid velocity along the line joining them, in ft/day.
Fig. 1 — The two wells sit at different subsea depths, so a raw pressure comparison is misleading; flow direction requires comparing fluid potential at a common datum.
Approach. Correct each well's pressure to a common datum to get its fluid potential $\Phi$ (flow runs from high $\Phi$ to low $\Phi$), then apply Darcy's linear-flow velocity equation along the line joining the wells, using the true (inclined) separation distance.
Fluid potential and flow direction. Referencing both wells to an arbitrary common datum (sea level, $D=0$) with the given fluid gradient, $\Phi = p - (0.35)D$: $\Phi_A = 4000-0.35(9000)=850$ psia; $\Phi_B = 4020-0.35(9200)=800$ psia. Since $\Phi_A>\Phi_B$, $\boxed{\text{flow is from Well A to Well B}}$, a potential drop $\Delta\Phi=\Phi_A-\Phi_B=50$ psi (the raw pressure difference alone, $4020-4000=20$ psi, would have wrongly suggested B$\to$A).
True separation distance. The wells are offset both horizontally (3000 ft) and vertically ($9200-9000=200$ ft), so the flow path length is $L=\sqrt{3000^2+200^2}=3006.7$ ft (the 200 ft vertical offset changes $L$ by under 0.3%).
Darcy velocity. With $k$ already in Darcy (per the formula sheet's unit convention), the linear-flow velocity is $v=1.127\dfrac{k}{\mu}\dfrac{\Delta\Phi}{L}$ in bbl/day per ft$^2$: $v=1.127\times\dfrac{0.2}{0.6}\times\dfrac{50}{3006.7}=1.127\times0.3333\times0.01663=0.006247$ bbl/day-ft$^2$. Converting to ft/day (1 bbl $=5.615$ ft$^3$): $\boxed{v=0.006247\times5.615=0.0351\ \text{ft/day}}$, directed from A to B.
