Question 2 of 7: Core-Plug Porosity, Absolute Permeability and End-Point Oil Relative Permeability

Question 2: Core-Plug Porosity, Absolute Permeability and End-Point Oil 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. A cylindrical core, $L=10$ cm, $d=3$ cm, initially 100% water-saturated with $V_p=22\ \text{cm}^3$ of water.

Find. (a) porosity $\phi$; (b) absolute permeability $k$; (c) connate water saturation $S_{wc}$; (d) oil relative permeability $k_{ro}$ at $S_{wc}$.

Approach. Get porosity from the pore volume over the measured bulk volume; get absolute permeability from the initial 100%-water-saturated linear Darcy flow (Darcy units: $k$ in Darcy, $q$ cm$^3$/s, $A$ cm$^2$, $\mu$ cp, $L$ cm, $\Delta p$ atm); get $S_{wc}$ from how much of the original water volume the oil flood could not displace; then, since no more water can be produced once $S_{wc}$ is reached, the stabilized oil-flood flow is effectively single-phase oil moving through the reduced pore space — its Darcy-law "effective" permeability divided by the absolute permeability gives $k_{ro}$ at $S_{wc}$.

1. Porosity. Bulk volume $V_b=\pi(d/2)^2L=\pi(1.5)^2(10)=70.69\ \text{cm}^3$. $\boxed{\phi=V_p/V_b=22/70.69=0.311\ (31.1\%)}$.
2. Absolute permeability. $A=\pi(1.5)^2=7.069\ \text{cm}^2$; $q_w=1/60=0.01667\ \text{cm}^3/\text{s}$; $\Delta p_w=20/14.696=1.361\ \text{atm}$ (Darcy-unit $q=\frac{kA}{\mu L}\Delta p$, $k$ in Darcy). $k=\dfrac{q_w\mu_wL}{A\,\Delta p_w}=\dfrac{(0.01667)(1)(10)}{(7.069)(1.361)}$, so $\boxed{k=0.01733\ \text{Darcy}=17.3\ \text{mD}}$.
3. Connate water saturation. Of the original 22 cm$^3$ of water, the oil flood displaced 15 cm$^3$ before stopping, leaving $22-15=7\ \text{cm}^3$ of water trapped and immobile. $\boxed{S_{wc}=7/22=0.318\ (31.8\%)}$.
4. Oil relative permeability at $S_{wc}$. At the stabilized oil-flood end point no more water moves, so the oil flows essentially alone through the plug; treating this as single-phase linear Darcy flow of oil gives its effective permeability, $k_o$: $\Delta p_o=30/14.696=2.042\ \text{atm}$, $q_o=0.1/60=0.001667\ \text{cm}^3/\text{s}$. $k_o=\dfrac{q_o\mu_oL}{A\,\Delta p_o}=\dfrac{(0.001667)(12)(10)}{(7.069)(2.042)}=0.01386\ \text{Darcy}$. $\boxed{k_{ro}=k_o/k=0.01386/0.01733=0.800}$.