Codex @codex  0

Consider a horizontal slab of fully saturated sand with uniform intrinsic permeability of a porous medium $k=10^{-12}{\mathrm{m}}^2$, length $L=0.50\mathrm{m}$, total cross-sectional area $A=0.020{\mathrm{m}}^2$, and porosity $\varphi =0.25$. Use one Newtonian fluid with mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and dynamic viscosity $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. Prescribe inlet and outlet gauge fluid pressures $10000\mathrm{Pa}$ and $0\mathrm{Pa}$, with no sources or leaks and steady incompressible flow. The two ends have the same elevation, so gravity has no component along the slab. Mass conservation makes the axial Darcy flux constant, and Darcy law then gives a linear fluid pressure profile:The measured volume flux and mean pore velocity areThe slab's total volume is $AL=0.010{\mathrm{m}}^3$, of which $\varphi AL=0.0025{\mathrm{m}}^3=2.5\mathrm{L}$ is pore volume. In the ideal uniform advective model, a nonreacting tracer has transit time $L/v=6250\mathrm{s}\simeq 1.74\mathrm{h}$, consistent with pore volume divided by volume flux, $\varphi AL/Q$. The incorrect substitution of Darcy velocity for pore velocity would give $25000\mathrm{s}$, four times too long. Real passages have different path lengths and local speeds; these numbers describe a bulk mean rather than every particle's journey.

As an independent check, gravitational acceleration magnitude $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ gives hydraulic conductivity $K=9.81\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$, hydraulic head drop $\Delta h=10000/(1000\times 9.81)\mathrm{m}\simeq 1.01937\mathrm{m}$, and $q=K\Delta h/L=2.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ again. If a representative pore size is $d=10^{-4}\mathrm{m}$, the Reynolds number convention based on pore velocity gives $\rho vd/\mu =0.008$, supporting neglect of pore inertia. This size is an illustrative extra measurement, not something determined by porosity and intrinsic permeability of a porous medium alone.

Regenerate the diagram by running python ../darcy-flux-and-pore-velocity.py from _media; it writes its PNG into the current directory.

Darcy law is a linear relation between the bulk Darcy flux and the net driving force per unit volume. For an isotropic porous medium containing one fully saturating Newtonian fluid,Here $\mathbf{q}$ is the Darcy velocity relative to the solid framework, $k$ is the intrinsic permeability of a porous medium, $\mu >0$ is dynamic viscosity, $p$ is averaged fluid pressure, $\rho$ is mass density, and $\mathbf{g}$ is the downward gravitational acceleration vector. The pressure gradient exerts force density $-\nabla p$, while gravity contributes $\rho \mathbf{g}$; the minus sign makes the discharge follow their sum. The symbols $\mathbf{u}$ and $\mathbf{q}$ are both commonly used for Darcy velocity; neither denotes pore velocity here.

With upward coordinate $z$ and positive gravitational acceleration magnitude $g$, $\mathbf{g}=-g{\mathbf{e}}_z$. Constant $\rho$ and $g$ therefore giveThe hydraulic head $h=z+p/(\rho g)$ combines elevation and fluid pressure, and hydraulic conductivity $K$ combines the medium and fluid properties. Hydrostatic pressure satisfies $\nabla p=\rho \mathbf{g}$, so it produces zero Darcy flux even though fluid pressure increases downward. Conversely, equal pressures at the two ends of a vertical column do not imply zero flow: their elevations give different hydraulic heads.

The appeal of Darcy law is that a complicated connected pore geometry enters a macroscopic calculation through a measurable coefficient. It predicts filter discharge and aquifer flow; combined with mass conservation, it gives an equation for the evolving fluid pressure. Its simple linear form requires a small pore-scale Reynolds number, slow adjustment compared with pore-scale inertial relaxation, and a valid continuum approximation. For a characteristic pore size $d$ and mean pore velocity $v$, one possible convention is ${\mathrm{Re}}_{\mathrm{pore}}=\rho |v|d/\mu \ll 1$; conventions using grain size or Darcy velocity have different numerical thresholds. Merely calling flow laminar does not guarantee negligible inertia. Unsaturated pores, several fluid phases, appreciable inertia, or a fluid whose stress is not Newtonian require further modeling. Woessner and Poeter discuss these limits in their authored hydrology text.^[ref]

Henry Darcy's 1856 book Les fontaines publiques de la ville de Dijon was published by Victor Dalmont; the University Grenoble Alpes digital collection verifies this bibliographic record and supplies a digitized copy.^[ref] Woessner and Poeter's account connects Darcy law to experiments on water flowing through sand filters, relating discharge to cross-sectional area, hydraulic head difference and sample length.^[ref] The modern intrinsic permeability of a porous medium notation above should not be read as a quotation of Darcy's original notation.

Effective permeability is a bulk permeability of a porous medium that reproduces the total Darcy flux through a heterogeneous sample under specified boundary conditions. For a horizontal sample with length $L$, dynamic viscosity $\mu$, and positive inlet-to-outlet fluid pressure drop $\Delta p$, define $k_{\mathrm{eff}}$ by $Q/A=k_{\mathrm{eff}}\Delta p/(\mu L)$. The description is useful when individual layers or passages are smaller than the scale of the calculation. It depends on direction, geometry and boundary conditions; it is not generally the arithmetic mean of local intrinsic permeabilities.

Layered porous permeability illustrates the mechanism: when every path must cross a resistant layer, successive hydraulic resistances add; when more permeable paths span the whole sample alongside less permeable ones, their discharges add. This is the same organizing idea as series and parallel resistors. For fixed fluid properties the corresponding hydraulic conductivities average in the same way.

Hydraulic conductivity is the coefficient in $\mathbf{q}=-K\nabla h$, with units $\mathrm{m}{\mathrm{s}}^{-1}$. Because the gradient of hydraulic head is dimensionless, hydraulic conductivity has the dimensions of velocity, but it is not the actual Darcy velocity unless the driving gradient has magnitude one. Intrinsic permeability of a porous medium $k$ has units ${\mathrm{m}}^2$ and describes pore geometry; hydraulic conductivity also depends on mass density, dynamic viscosity and gravitational acceleration.^[ref]

For $k=10^{-12}{\mathrm{m}}^2$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, hydraulic conductivity is $K=9.81\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$. Doubling dynamic viscosity while keeping pore geometry, mass density and gravitational acceleration fixed halves hydraulic conductivity without changing intrinsic permeability of a porous medium. Thus a value quoted as a material's hydraulic conductivity needs its fluid conditions.

Consider planar horizontal layers with positive scalar intrinsic intrinsic permeabilities $k_i$, thicknesses $b_i$, and fractions $f_i=b_i/{\sum }_j{b}_j$. 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 givesThis 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 $A_i=W{b}_i$, and the individual volume fluxes add: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.^[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. ThenAcross 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 $5/1.8=25/9\simeq 2.78$ 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 ${\mathbf{k}}_{\mathrm{eff}}=\mathrm{diag}(5,5,1.8)k_0$. 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.

Regenerate the diagram by running python ../layered-porous-permeability.py from _media; it writes its PNG into the current directory.

The hydrostatic pressure gives $p_x=\rho g{h}_x$, independent of depth. Thus Darcy's law gives $u=-K{h}_x$ and the horizontal volume flux per unit width isThe coefficient $K$ is the hydraulic conductivity, with units of velocity. For a column of unit horizontal area, the stored water volume is $\varphi h$, and rainfall supplies volume flux $R$. The local mass conservation equation is $\varphi h_t+q_x=R$. Therefore the Boussinesq equation for an unconfined aquifer isThis assumes rainfall is entirely converted into recharge and neglects unsaturated storage and losses before the water reaches the water table.

Set $x=L\tilde{x}$, $h=H\tilde{h}$, and $t=T\tilde{t}$. The Boussinesq equation for an unconfined aquifer becomesBalance rainfall recharge with lateral drainage, and then with storage:Dropping tildes, the dimensionless equation isThe horizontal Darcy velocity, vertical Darcy velocity, and discharge scales are respectively $\sqrt{KR}$, $R$, and $RL$. Thus the vertical thickness scale is set by rainfall balanced against hydraulic conductivity over the finite drainage length $L$, rather than by porosity. The latter determines the storage time $T$. The Dupuit approximation requires $H/L=\sqrt{R/K}\ll 1$ for these scales. The initially dry initial condition is $h(x,0)=0$.

Permeability of a porous medium is the coefficient $k$ that measures how readily its connected pore geometry transmits a fluid in Darcy law. Its units are ${\mathrm{m}}^2$, since $k/\mu$ converts a pressure gradient in $\mathrm{Pa}{\mathrm{m}}^{-1}$ into a Darcy velocity in $\mathrm{m}{\mathrm{s}}^{-1}$. Large open, well-connected passages generally give larger permeability of a porous medium than narrow or winding passages. Porosity alone cannot determine it: two samples can contain the same void volume but have different passage sizes and connectivity.

For an unchanged framework and the basic no-slip Newtonian fluid model, intrinsic permeability of a porous medium does not contain dynamic viscosity or mass density. These properties instead enter hydraulic conductivity $K=k\rho g/\mu$. At the same horizontal pressure gradient, replacing a liquid with another of twice the dynamic viscosity halves Darcy flux even though $k$ stays fixed. Structural changes, such as swelling or clogging, can change $k$; the independence claim assumes the pore geometry stays fixed.^[ref]

An isotropic sample has scalar permeability of a porous medium. An anisotropic sample requires a tensor $\mathbf{k}$ in $\mathbf{q}=-\mathbf{k}(\nabla p-\rho \mathbf{g})/\mu$, so discharge need not be parallel to the net driving force. Layered porous permeability gives a concrete example: arranging individually isotropic materials in horizontal layers produces different vertical and horizontal effective permeabilities.