= Layered porous permeability
{title2=$k_\perp=(\sum_i f_i/k_i)^{-1},\quad k_\parallel=\sum_i f_i k_i$}
Consider planar horizontal layers with positive scalar intrinsic <intrinsic permeabilities> $k_i$, thicknesses $b_i$, and fractions $f_i=b_i/\sum_jb_j$. Assume each layer is uniform and <isotropic>, fully saturated by the same <Newtonian fluid> of constant <mass density> and <dynamic viscosity>, with perfect hydraulic contact and steady <incompressible flow>. Use boundaries that impose pure flow normal to the layers or a common gradient of <hydraulic head> along them; additional interface resistance and finite-edge effects are excluded.
For vertical flow normal to the layers, every layer carries the same normal <Darcy flux> by <mass conservation>. The <hydraulic head> drops add. Writing $B=\sum_i b_i$ and taking positive $\Delta h_i$ as the drop along the flow direction gives
$$
q=\frac{\rho g}{\mu}k_i\frac{\Delta h_i}{b_i},\qquad \Delta h=\sum_i\Delta h_i=\frac{\mu q}{\rho g}\sum_i\frac{b_i}{k_i},\qquad k_\perp=\frac{B}{\sum_i b_i/k_i}=\left(\sum_i\frac{f_i}{k_i}\right)^{-1}.
$$
This is the thickness-weighted <harmonic mean>. It is <hydraulic head>, rather than <fluid pressure> alone, that accounts correctly for <gravity> in the vertical calculation. A thin layer of very small <permeability of a porous medium> can control the whole discharge because all flow must pass through it.
For horizontal flow parallel to the layers, each layer experiences the same gradient of <hydraulic head>. For width $W$, each area is $A_i=Wb_i$, and the individual <volume fluxes> add:
$$
Q=\sum_i Q_i=-\frac{\rho g}{\mu}\frac{dh}{dx}\sum_i k_i A_i,\qquad \frac QA=-\frac{\rho g}{\mu}\frac{dh}{dx}\sum_i f_i k_i,\qquad k_\parallel=\sum_i f_i k_i.
$$
This is the thickness-weighted <arithmetic mean>. A highly permeable layer can carry most of the horizontal flow because it supplies a continuous route from inlet to outlet. Woessner and Poeter derive both cases for <hydraulic conductivity>; identical <fluid> properties allow the common factor $\rho g/\mu$ to cancel.https://books.gw-project.org/hydrogeologic-properties-of-earth-materials-and-principles-of-groundwater-flow/chapter/equation-derivation-for-equivalent-k-and-a-4-layer-application/{ref}
For two equally thick layers, take $k_1=k_0$ and $k_2=9k_0$, where $k_0=10^{-12}\,\mathrm{m^2}$. The values 1 and 9 are therefore permeabilities in units of $k_0$, rather than dimensionless intrinsic <intrinsic permeabilities>. Then
$$
k_\perp=\frac{2}{1/k_0+1/(9k_0)}=1.8k_0,\qquad k_\parallel=\frac{k_0+9k_0}{2}=5k_0.
$$
Across the layers, the less permeable half takes $90\%$ of the total <hydraulic head> drop and the more permeable half only $10\%$. Along the layers, the more permeable half carries $90\%$ of the total <volume flux>. At the same magnitude of imposed gradient of <hydraulic head>, horizontal <Darcy flux> is $5/1.8=25/9\simeq2.78$ times the vertical <Darcy flux>.
The arrangement thus creates physical <anisotropy> from locally <isotropic> ingredients. In coordinates with $x,y$ along the layers and $z$ normal to them, the effective <permeability of a porous medium> <tensor> is $\boldsymbol{k}_{\rm eff}=\operatorname{diag}(5,5,1.8)k_0$. A driving gradient of <hydraulic head> with both horizontal and vertical components generally produces a discharge in a different direction. This distinction matters when deciding whether a geological bed conducts <aquifer> flow along itself or acts as a barrier to flow across it.
\Image[/layered-porous-permeability.png]
{title=The same two layers give a harmonic mean across them and an arithmetic mean along them}
Regenerate the diagram by running `python ../layered-porous-permeability.py` from `_media`; it writes its PNG into the current directory.
Back to article page
