Steady-State Core Flood — Effective and Relative Permeabilities: 24-Pet-A3 Fundamental Reservoir Engineering, December 2015, Q2

Question 2: Steady-State Core Flood — Effective and Relative Permeabilities (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.

Core length, $L$	10 cm
Cross-sectional area, $A$	5 cm$^2$
Absolute permeability, $k$	0.5 Darcy
Oil viscosity, $\mu_o$	3.52 cP
Water viscosity, $\mu_w$	1 cP
$q_w,\,q_o$ at $S_w=0.4$	0.0220, 0.0260 mL/s

Find. The effective permeabilities $k_o$, $k_w$ and relative permeabilities $k_{ro}$, $k_{rw}$ at $S_w=40\%$, and the pressure difference across the core at that saturation.

Fig. 1 — Relative permeability curves computed from every row of the core-flood table using the constant test pressure differential; the dashed line marks the asked-for $S_w=0.4$.

Approach. A steady-state relative-permeability rig holds the total pressure differential across the core constant while it varies the oil/water injection split; recover that constant $\Delta p$ from Darcy's linear-flow law at the two single-phase end points ($S_w=0$ and $S_w=1$, where the effective permeability equals the absolute permeability), then apply the same $\Delta p$ to the two-phase rates at $S_w=0.4$.

Recover the constant test $\Delta p$ from the single-phase end points. In Darcy (cgs) units, $q=\dfrac{kA\,\Delta p}{\mu L}\ \Rightarrow\ \Delta p=\dfrac{q\mu L}{kA}$. At $S_w=0$ (100% oil, so $k_o=k_{abs}=0.5$ D): $\Delta p=\dfrac{(0.0780)(3.52)(10)}{(0.5)(5)}=1.098$ atm. At $S_w=1$ (100% water, $k_w=k_{abs}=0.5$ D): $\Delta p=\dfrac{(0.2750)(1)(10)}{(0.5)(5)}=1.100$ atm. The two end points agree to within rounding, confirming $\boxed{\Delta p \approx 1.10\ \text{atm} \;(16.2\ \text{psi})}$ was held constant throughout the test — this is also the answer to the pressure-difference part of the question, since $\Delta p$ does not change with saturation in this rig design.
Effective permeabilities at $S_w=0.4$. Applying the same Darcy relation to each phase's rate at $S_w=0.4$ ($q_w=0.0220$, $q_o=0.0260$ mL/s): $k_w=\dfrac{q_w\mu_w L}{A\,\Delta p}=\dfrac{(0.0220)(1)(10)}{(5)(1.10)}=0.0400$ D $=\boxed{40.0\ \text{mD}}$; $k_o=\dfrac{q_o\mu_o L}{A\,\Delta p}=\dfrac{(0.0260)(3.52)(10)}{(5)(1.10)}=0.1665$ D $=\boxed{166.5\ \text{mD}}$.
Relative permeabilities. $k_{rw}=k_w/k_{abs}=0.0400/0.5=\boxed{0.080}$; $k_{ro}=k_o/k_{abs}=0.1665/0.5=\boxed{0.333}$.

Quantity	Value
Test pressure differential, $\Delta p$	1.10 atm (16.2 psi)
Effective water permeability, $k_w$	40.0 mD
Effective oil permeability, $k_o$	166.5 mD
Relative water permeability, $k_{rw}$	0.080
Relative oil permeability, $k_{ro}$	0.333

Concept & Theory

Topic: Steady-state relative permeability measurement.

Key relations: Darcy's law per phase, $q_p=\dfrac{k_p A\,\Delta p}{\mu_p L}$, with $k_{rp}=k_p/k_{abs}$; the sum $k_{ro}+k_{rw}<1$ everywhere two phases flow simultaneously (each phase blocks part of the pore network for the other).

Why this works: Because the rig fixes total $\Delta p$ and only redistributes the oil/water injection ratio to reach each steady-state saturation, the single-phase end points ($S_w=0$, $S_w=1$) directly calibrate that constant differential without needing a separate pressure-transducer reading — a useful cross-check on lab data quality, exactly as the two end points agreed here to within 0.2%.

Common pitfall: Forgetting that $k_{ro}$ and $k_{rw}$ are each normalized by the single-phase absolute permeability $k_{abs}$, not by the effective permeability of the other phase, and mixing cgs Darcy units (cm, cP, atm) with field units mid-calculation.

Source: Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 5 (relative permeability); McCain, The Properties of Petroleum Fluids, 3rd ed. (steady-state core-flood method).

Question 2 of 7: Steady-State Core Flood — Effective and Relative Permeabilities

Question 2: Steady-State Core Flood — Effective and Relative Permeabilities (20 marks)