24-Pet-A3 Fundamental Reservoir Engineering · May 2014
Question 3 of 7: Core Flood Experiment — Absolute and Relative Permeability
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Pet-A3 — Fundamental Reservoir Engineering · National Exams, May 2014 · 3 hours, closed book, non-communicating calculator only · first five questions in the answer book are marked, all questions equal value, all parts of a multipart question equal weight. All seven questions are answered here as a complete study resource.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (material balance, transient well testing, radial flow); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (material balance, skin/productivity, relative permeability); McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PVT properties, Z-factor); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed. (core analysis, capillary pressure).
Given. Core: diameter $d=2$ cm, length $L=10$ cm, porosity $\phi=15\%$. 100%-water saturation: $\mu_w=1$ cp, $q=0.02$ cm$^3$/s, $\Delta p=1$ atm. Oil flood: $\mu_o=2$ cp, connate water saturation established at $S_{wc}=25\%$. Waterflood-of-oil at a later time: $\Delta p=1.2$ atm, $q_w=0.005$, $q_o=0.015$ cm$^3$/s.
Find. (a) absolute permeability $k$; (b) pore volume; (c) oil volume in place at end of the oil flood; (d) $k_{ro}$ and $k_{rw}$ at the stated waterflood condition.
Approach. Use Darcy's linear-flow law in the c.g.s.-Darcy unit set given on the formula sheet, $q=kA\Delta p/(\mu L)$, first for the single-phase (100% water) calibration to get $k$ and pore volume, then for the two-phase step where each flowing phase obeys the same law scaled by its own relative permeability.
Part (a) — absolute permeability. Cross-section $A=\pi d^2/4=\pi(2)^2/4=3.1416$ cm$^2$. Rearranging Darcy's law, $k=\dfrac{q\mu L}{A\Delta p}=\dfrac{0.02\times1\times10}{3.1416\times1}=0.06366$ Darcy. So $\boxed{k=63.66\text{ mD}}$.
Part (b) — pore volume. $V_p=\phi\, A\, L=0.15\times3.1416\times10$, so $\boxed{V_p=4.712\text{ cm}^3}$.
Part (c) — oil volume at end of oil flood. At $S_{wc}=25\%$ the oil occupies the balance of the pore volume: $V_o=V_p(1-S_{wc})=4.712\times0.75$, so $\boxed{V_o=3.534\text{ cm}^3}$.
Part (d) — relative permeabilities. With $k$ and $A$ already fixed, $k\,A=0.06366\times3.1416=0.2000$ Darcy$\cdot$cm$^2$ carries through unchanged (it is a rock/geometry property, not a flow-condition property). Applying Darcy's law to each phase separately at the new pressure drop, $k_{r,\text{phase}}=\dfrac{q_{\text{phase}}\,\mu_{\text{phase}}\,L}{k\,A\,\Delta p}$:
$k_{ro}=\dfrac{0.015\times2\times10}{0.2000\times1.2}=\dfrac{0.30}{0.24}=1.25$
$k_{rw}=\dfrac{0.005\times1\times10}{0.2000\times1.2}=\dfrac{0.05}{0.24}=0.208$
Check: applying the formula literally to the flow-rate labelling exactly as stated ("0.005 and 0.015 cc/sec ... for water and oil, respectively") gives $\boxed{k_{ro}=1.25}$, which exceeds unity and is physically impossible for a relative permeability (by definition $0\le k_r\le1$). This is a real internal inconsistency in the exam's stated numbers rather than an arithmetic slip on our part — the two rates and the pressure drop as printed simply do not form a physically consistent two-phase Darcy system. If the two flow-rate labels are instead read as swapped ($q_w=0.015$, $q_o=0.005$, a plausible ambiguity given the sentence structure), the same formula gives the physically sensible pair $k_{ro}=0.417$ and $k_{rw}=0.625$. Both computations are shown; the literal reading is boxed as the primary answer since it follows the question exactly as printed, with the swapped-label result reported as the internally-consistent alternative.
