Steady-State Core Flood — Effective and Relative Permeability: 24-Pet-A3 Fundamental Reservoir Engineering, December 2018, Q4

Question 4: Steady-State Core Flood — Effective and Relative Permeability (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $L=10$ cm, $A=5\ \text{cm}^2$, $k_{abs}=0.5$ Darcy, $\mu_o=3.52$ cp, $\mu_w=1$ cp; steady-state co-flood rates vs. $S_w$ above (each column run at the SAME fixed differential pressure, a standard steady-state relative-permeability procedure).

Find. $k_{eff,w}$, $k_{eff,o}$, $k_{rw}$, $k_{ro}$ at $S_w=0.40$, and the inlet–outlet pressure difference at that saturation.

Approach. Use the two single-phase end points ($S_w=0$: 100% oil; $S_w=1$: 100% water) with Darcy's law and the GIVEN $k_{abs}$ to back out the common differential pressure $\Delta p$ used throughout the flood; then apply the same Darcy equation to each phase's rate at $S_w=0.40$ to get $k_{eff,w}$ and $k_{eff,o}$, and normalize by $k_{abs}$ for the relative permeabilities.

Differential pressure from the single-phase end points. Darcy's law in Darcy units, $q=\dfrac{kA\Delta p}{\mu L}$. At $S_w=0$ (100% oil, $q_o=0.0780$): $\Delta p=\dfrac{q_o\mu_oL}{k_{abs}A}=\dfrac{0.0780(3.52)(10)}{0.5(5)}$, giving $\Delta p=1.098$ atm. At $S_w=1$ (100% water, $q_w=0.2750$): $\Delta p=\dfrac{0.2750(1)(10)}{0.5(5)}=1.100$ atm. The two endpoints agree, confirming a single fixed $\Delta p$ was used throughout the test: $\boxed{\Delta p=1.10\ \text{atm}}$ (average of the two).
Effective permeabilities at $S_w=0.40$. Rates there: $q_w=0.0220$, $q_o=0.0260$ mL/s. $k_{eff,w}=\dfrac{q_w\mu_wL}{A\Delta p}=\dfrac{0.0220(1)(10)}{5(1.10)}$, so $\boxed{k_{eff,w}=0.0400\ \text{D}}$. $k_{eff,o}=\dfrac{q_o\mu_oL}{A\Delta p}=\dfrac{0.0260(3.52)(10)}{5(1.10)}$, so $\boxed{k_{eff,o}=0.1665\ \text{D}}$.
Relative permeabilities. $k_{rw}=k_{eff,w}/k_{abs}=0.0400/0.5$, so $\boxed{k_{rw}=0.080}$. $k_{ro}=k_{eff,o}/k_{abs}=0.1665/0.5$, so $\boxed{k_{ro}=0.333}$.
Inlet–outlet pressure difference at $S_w=0.40$. The differential pressure was fixed throughout the flood (Step 1), so $\boxed{\Delta p=1.10\ \text{atm}\ (\approx16.2\ \text{psi})}$ at $S_w=0.40$ as well.

Check: taking $\Delta p$ as constant across the whole flood is confirmed, not assumed — the two independent single-phase end points (Sw=0 and Sw=1) both back out $\Delta p\approx1.10$ atm from the given $k_{abs}=0.5$ D, so the two-phase rates at every intermediate saturation were clearly measured at that same fixed differential pressure.

Quantity	Value
Fixed differential pressure, $\Delta p$	1.10 atm
Water effective permeability, $k_{eff,w}$ (at $S_w=0.4$)	0.0400 D (40.0 mD)
Oil effective permeability, $k_{eff,o}$ (at $S_w=0.4$)	0.1665 D (166.5 mD)
Water relative permeability, $k_{rw}$	0.080
Oil relative permeability, $k_{ro}$	0.333

Topic: Steady-state relative-permeability measurement from a co-flood core test.

Key relations: Darcy's law (Darcy units) $q=kA\Delta p/(\mu L)$; $k_{r\alpha}=k_{eff,\alpha}/k_{abs}$ for phase $\alpha$.

Why this works: in the steady-state method, oil and water are co-injected at a fixed ratio until both saturation and both phase rates stop changing with time; because the differential pressure is imposed externally and held fixed, each phase independently obeys Darcy's law with its own effective permeability, so the absolute permeability (measured with a single, known phase) calibrates $\Delta p$, and every other column of the table gives a fresh $k_{eff}$ pair at its own $S_w$.

Common pitfall: summing $k_{rw}+k_{ro}$ and expecting 1 — the two curves are independent and their sum is always ≤1, reflecting that two immiscible phases sharing the pore space interfere with each other's flow paths.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 4 (relative permeability); Craft & Hawkins, Applied Petroleum Reservoir Engineering, 3rd ed., Ch. 1.

Question 4 of 7: Steady-State Core Flood — Effective and Relative Permeability

Question 4: Steady-State Core Flood — Effective and Relative Permeability (20 marks)