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Effective permeability of a porous medium. Work to leading order in the slender-layer ratio $H/L$. Take the growing lower wedge to have porous permeability $k_1$, with interface $y=Hx/L$, and the upper wedge to have porous permeability $k_2$. The leading pressure is independent of $y$; vertical flow is smaller than horizontal flow by $H/L$. DefineDarcy's law and the fixed two-dimensional volume flux giveHere $q_i$ is Darcy velocity; the parcel speed is $q_i/\varphi$. Integrating the pressure gradient over the length and defining $Q=H{k}_{\mathrm{eff}}\Delta p/(\mu L)$ yieldsThe equal-porous permeability limit is $k_{\mathrm{eff}}=k_1=k_2$. This logarithmic mean comes from parallel layers at each cross-section followed by series addition of their local hydraulic resistances. It is a slender-layer result; a full two-dimensional transmission problem has small end and interface corrections.

Parcel paths and travel times. Pressure equalization does not mean that parcels stay at fixed $y$: mass conservation requires a small vertical flow. Define a streamfunction by $\psi (x,0)=0$ and $\psi (x,H)=Q$. Its leading expression isLet $r=y_0/H$ label a parcel released at the inlet. There $\psi =Qr$. At the inclined interface $\psi /Q=k_1\xi /K(\xi )$, so the parcel crosses from the upper wedge into the lower one atBefore crossing, $y/H=1-(1-r)K/k_2$; afterwards, $y/H=rK/k_1$. These expressions show why integrating the speed along a horizontal line would give the wrong parcel time.

Put $d=k_1-k_2$ and ${\tau }_V=\varphi HL/Q$, the pore-volume throughput time. Integrating $dt=\varphi dx/q_i$ on the two portions of the path givesIts derivative with respect to ${\xi }_c$ is ${\tau }_{V}dK/(k_1{k}_2)$, so it is monotone. The limiting streamline times at the lower and upper boundaries are ${\tau }_V(k_1+k_2)/(2k_1)$ and ${\tau }_V(k_1+k_2)/(2k_2)$ respectively. HenceThe printed expression assumes $k_1\ge k_2$. The absolute value is needed for a nonnegative maximum difference without that ordering. For equal porous permeabilities every parcel has time ${\tau }_V$.

Oil recovery. In the ideal passive-displacement model with $k_1/k_2=10$, the earliest and latest travel times are $0.55{\tau }_V$ and $5.5{\tau }_V$, a spread of $4.95{\tau }_V$. Preferential paths through the high-porous permeability wedge therefore give early water breakthrough while oil on slower paths remains unswept. Continued injection sends much water through paths already swept; complete displacement requires several pore volumes. The flux-weighted residence time in a porous layer remains ${\tau }_V$. Real waterflooding also depends on phase mobilities, relative permeabilities, capillary pressure and mixing, so these numbers illustrate heterogeneity rather than a quantitative two-phase recovery prediction.