= Solution
\b[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 $H/L$. Define
$$
\xi=\frac{x}{L},\qquad K(\xi)=k_2+(k_1-k_2)\xi.
$$
<Darcy's law> and the fixed two-dimensional <volume flux> give
$$
Q=-\frac{H K(\xi)}{\mu}p_x,\qquad
q_i(x)=\frac{Qk_i}{HK(\xi)}.
$$
Here $q_i$ is <Darcy velocity>; the parcel speed is $q_i/\phi$. Integrating the <pressure gradient> over the length and defining $Q=Hk_{\rm eff}\Delta p/(\mu L)$ yields
$$
\boxed{k_{\rm eff}=\frac{k_1-k_2}{\log(k_1/k_2)}.}
$$
The equal-<porous permeability> limit is $k_{\rm eff}=k_1=k_2$. 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.
\b[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 is
$$
\psi=\begin{cases}q_1(x)y,&y\leq H\xi,\\Q-q_2(x)(H-y),&y\geq H\xi.\end{cases}
$$
Let $r=y_0/H$ 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 at
$$
\boxed{\xi_c(r)=\frac{rk_2}{k_1(1-r)+k_2r}.}
$$
Before 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 $\tau_V=\phi HL/Q$, the pore-volume throughput time. Integrating $dt=\phi\,dx/q_i$ on the two portions of the path gives
$$
\boxed{\begin{aligned}
\tau(y_0)&=\tau_V\left[\frac1{k_2}\int_0^{\xi_c}K(\xi)d\xi
+\frac1{k_1}\int_{\xi_c}^{1}K(\xi)d\xi\right]\\
&=\tau_V\left[\frac{k_1+k_2}{2k_1}
+\frac d{k_1}\xi_c+\frac{d^2}{2k_1k_2}\xi_c^2\right].
\end{aligned}}
$$
Its derivative with respect to $\xi_c$ is $\tau_V dK/(k_1k_2)$, 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. Hence
$$
\boxed{\Delta\tau_{\max}=\frac{\phi HL}{2Q}
\frac{|k_1^2-k_2^2|}{k_1k_2}.}
$$
The printed expression assumes $k_1\geq k_2$. The absolute value is needed for a nonnegative maximum difference without that ordering. For equal <porous permeabilities> every parcel has time $\tau_V$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-78-layered-streamlines.png]
{title=Streamlines crossing an inclined permeability interface in a slender layer with lower-wedge permeability ten times the upper-wedge permeability}
\b[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.
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