Question 4 of 6: Core-Flood Permeability, Relative Permeability and Buckley-Leverett Frontal Advance

Question 4: Core-Flood Permeability, Relative Permeability and Buckley-Leverett Frontal Advance (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 $L=1$ ft, $A=0.1\ \text{ft}^2$, $\phi=0.15$, $B_o=B_w=1.0$ bbl/STB. Single-phase oil flood: $q_o=1.127$ STB/day, $\mu_o=2.5$ cp, $\Delta P=100$ psi. Two-phase point at $S_w=0.5$: $q_o=0.4$, $q_w=0.2$ STB/day (same $\Delta P=100$ psi), $\mu_w=1$ cp. Fig. 3 gives $f_w(S_w)$ and $-df_w/dS_w$ from the full relative-permeability curve built off this core. $S_{wi}=0.15$, injection $q_t=0.1$ STB/day, $t=10$ min.

Find. (a) absolute permeability $k$; (b) $k_{ro}$, $k_{rw}$ at $S_w=0.5$; (c) front saturation $S_{wf}$ and the distance travelled after 10 min; (d) qualitative effect of higher oil viscosity on $S_{wf}$ and swept-zone oil recovery.

Approach. Linear Darcy flow gives absolute and phase-effective permeabilities directly (parts a, b); the Buckley-Leverett frontal-advance equation with a Welge tangent construction from $(S_{wi},0)$ gives the front (part c); a mobility-ratio argument on the fractional-flow curve answers part d without further computation.

1. (a) Absolute permeability. $q_o=\dfrac{0.001127\,kA\Delta P}{\mu_o B_o L}\ \Rightarrow\ k=\dfrac{q_o\mu_o B_o L}{0.001127\,A\,\Delta P}=\dfrac{1.127(2.5)(1.0)(1)}{0.001127(0.1)(100)}$, so $\boxed{k=250\ \text{md}}$.
2. (b) Relative permeabilities at $S_w=0.5$. Effective permeabilities from the same Darcy form: $k_{\text{eff},o}=\dfrac{q_o\mu_o B_o L}{0.001127 A\Delta P}=\dfrac{0.4(2.5)(1)(1)}{0.001127(0.1)(100)}=88.7\ \text{md}$; $k_{\text{eff},w}=\dfrac{q_w\mu_w B_w L}{0.001127 A\Delta P}=\dfrac{0.2(1)(1)(1)}{0.001127(0.1)(100)}=17.7\ \text{md}$. Dividing by $k=250$ md: $\boxed{k_{ro}=88.7/250=0.355}$, $\boxed{k_{rw}=17.7/250=0.0710}$.
3. (c) Front saturation. Because the $f_w(S_w)$ curve itself starts at $(S_{wi},0)=(0.15,0)$, the Welge tangent line from $(S_{wi},0)$ touches the curve exactly at the point where $df_w/dS_w$ is largest — the peak of the $-df_w/dS_w$ curve in Fig. 3, read at $\boxed{S_{wf}=0.50}$, where $(df_w/dS_w)_{S_{wf}}\approx3.2$.
4. Distance travelled by the front. $x(S_{wf},t)=\dfrac{5.615\,q_t B_w t}{\phi A}\left(\dfrac{df_w}{dS_w}\right)_{S_{wf}}$, with $t=10\ \text{min}=6.944\times10^{-3}$ day: $x=\dfrac{5.615(0.1)(1)(6.944\times10^{-3})}{0.15(0.1)}(3.2)$, so $\boxed{x=0.83\ \text{ft}}$ — about 83% of the 1-ft core, i.e. close to breakthrough after 10 minutes at this rate.

(d) Effect of higher oil viscosity on the front and swept-zone recovery. Raising $\mu_o$ (with everything else fixed) raises the water/oil mobility ratio $M=(k_{rw}/\mu_w)/(k_{ro}/\mu_o)$, making water relatively MORE mobile than oil. On the fractional-flow curve this steepens $f_w$ at low-to-moderate $S_w$, so the Welge tangent line from $(S_{wi},0)$ now touches the curve at a LOWER saturation: $\boxed{S_{wf}\ \text{decreases}}$. Because the average water saturation behind the front is tied to the same tangent construction ($1/(df_w/dS_w)_{S_{wf}}=(\bar S_w-S_{wi})$), a lower $S_{wf}$ with a shallower governing slope means less of the pore volume swept ahead of breakthrough has actually been displaced to water: $\boxed{\text{oil recovery behind the front decreases}}$. Physically, a more viscous oil lets the low-viscosity water "finger" through it and reach the producer earlier while leaving more oil bypassed — the classic adverse-mobility-ratio penalty that motivates thermal or chemical EOR for viscous crudes rather than plain waterflooding.