Question 4 of 7: Absolute and Effective Permeability from a Core-Flood Test

Question 4: Absolute and Effective Permeability from a Core-Flood Test (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.

Table 1 (relative permeability vs. $S_w$): $k_{rw}$ first reaches 0 at $S_w=0.20$ (irreducible water saturation, $k_{ro}=0.90$ there); $k_{ro}$ first reaches 0 at $S_w=0.85$ (residual oil saturation, $k_{rw}=0.20$ there).

[Figure not reproduced: Fig. Q4: measured flow rate vs. pressure drop, single-phase water flood (source Figure 2) — a straight line through the origin, slope 5 cc/min per atm. See the official exam paper.]

Find. (1) Absolute permeability $k$; (2) oil volume at $S_{wi}$; (3) $\Delta p$ for oil flow at $S_{wi}$; (4) $\Delta p$ for brine flow at $S_{or}$.

Approach. Read the slope of the linear water-flood test to get absolute permeability via Darcy's law in Darcy units; combine porosity, bulk volume and the saturation end-points from Table 1 for the pore-volume questions; then re-apply the two-phase Darcy equation with the relative permeability at each saturation end-point for the pressure-drop questions.

1. Absolute permeability from Figure 2. Core area $A=\pi(1.25)^2=4.909$ cm². The line's slope is $10/2.0=5$ cc/min/atm $=0.08333$ cc/s/atm. From Darcy's law $q=\dfrac{kA}{\mu}\dfrac{\Delta p}{L}$: $k=\dfrac{(\text{slope})\,\mu L}{A}=\dfrac{0.08333(1.05)(15)}{4.909}$, so $\boxed{k=0.2674\ \text{Darcy}=267.4\ \text{md}}$.
2. Oil volume at irreducible water saturation. $S_{wi}=0.20$ (where $k_{rw}$ first reaches zero). Bulk volume $V_b=A L=4.909(15)=73.63$ cm³; pore volume $V_p=\phi V_b=0.20(73.63)=14.73$ cm³. So $\boxed{V_o=(1-S_{wi})V_p=0.80(14.73)=11.78\ \text{cm}^3}$.
3. Pressure drop, oil flow at $S_{wi}$. At $S_{wi}=0.20$, $k_{ro}=0.90$: $\Delta p=\dfrac{q\,\mu_o L}{k\,k_{ro}\,A}=\dfrac{0.02(2.0)(15)}{0.2674(0.90)(4.909)}$, so $\boxed{\Delta p=0.508\ \text{atm}}$.
4. Pressure drop, brine flow at $S_{or}$. $S_{or}=0.85$ (where $k_{ro}$ first reaches zero), $k_{rw}=0.20$ there: $\Delta p=\dfrac{q\,\mu_w L}{k\,k_{rw}\,A}=\dfrac{0.02(1.05)(15)}{0.2674(0.20)(4.909)}$, so $\boxed{\Delta p=1.200\ \text{atm}}$.