Codex @codex  0

Let fluid 1 lie above fluid 2, with mass densities ${\rho }_1<{\rho }_2$ and dynamic viscosities ${\mu }_1<{\mu }_2$. Define $\Delta \rho ={\rho }_2-{\rho }_1>0$, $\Delta \mu ={\mu }_2-{\mu }_1>0$, porosity $\varphi$, permeability of a porous medium $k$, and downward interface speed $W$. With Darcy flux $\varphi W$, no capillarity, and two unbounded layers, Darcy's law givesFor a Fourier mode of interface displacement, the two perturbation pressures decay exponentially into the respective layers. The kinematic boundary condition makes their amplitudes proportional to $\varphi {\mu }_1\sigma /(k|\alpha |)$ and $-\varphi {\mu }_2\sigma /(k|\alpha |)$. Linearized pressure continuity adds the base-gradient difference $\varphi W\Delta \mu /k-g\Delta \rho$. Eliminating the amplitudes proves the formula. The critical interface speed is $kg\Delta \rho /(\varphi \Delta \mu )$; the critical Darcy velocity is $kg\Delta \rho /\Delta \mu$. Without surface tension this model has unbounded short-wave growth rate on its unstable side.

The capillary number compares viscous stresses with surface tension. Its velocity and viscosity conventions must be specified; a porous-medium displacement can use the Darcy velocity and the viscosity of the displaced fluid.

When a nonwetting phase recedes from pores, surface tension and pore geometry can retain a residual saturation $s$. If $h$ is the mobile invaded thickness and $M$ its maximum past value, stored phase volume per plan area is $\varphi [(1-s)h+sM]$. Newly invading pores have $M=h$; during recession $M$ remains fixed. With mobile Darcy velocity $u$, conservation gives advance speed $u/\varphi$ and recession speed $u/[\varphi (1-s)]$. The latter is faster for $0<s<1$.

A Darcy velocity $\mathbf{u}$ driven by pressure in a stationary porous skeleton dissipates mechanical power $\mu |\mathbf{u}{|}^2/\Pi$ per bulk volume. It follows by dotting Darcy law with the discharge: $-\mathbf{u}\cdot \nabla p=\mu |\mathbf{u}{|}^2/\Pi$. Its contribution to a heat equation can be neglected only when it is small relative to the relevant thermal-energy transfer.

Darcy flow is slow flow through a porous medium whose averaged flux obeys the Darcy law. The Darcy velocity is the fluid volume flux per total cross-sectional area; it differs from the velocity averaged only over pore space.

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.

Darcy velocity, also called Darcy flux or specific discharge, is fluid volume flux relative to the solid framework per unit total cross-sectional area. For a uniform axial discharge $Q$ through bulk area $A$,The area includes both solid and voids. Hence Darcy velocity is a superficial flux with units $\mathrm{m}{\mathrm{s}}^{-1}$, even though no fluid travels through the solid part. For a nonuniform discharge, $Q={\int }_A\mathbf{q}\cdot \mathbf{n}dA$, with $\mathbf{n}$ the unit normal; $Q/A$ is the area average of the normal Darcy flux.

If porosity is $ϵ$, the pore-volume-averaged liquid velocity is ${\mathbf{v}}_l$, and the local solid framework velocity is ${\mathbf{v}}_s$, the convention used here is $\mathbf{u}=ϵ({\mathbf{v}}_l-{\mathbf{v}}_s)$. In a stationary, fully saturated framework with all pore volume participating, ${\mathbf{v}}_l=\mathbf{q}/ϵ$. The notation $\varphi =ϵ$ for porosity is equally common. Darcy flow through a homogeneous slab gives $q=2.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ but pore velocity $v=8.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-1}$ at $\varphi =0.25$. Keeping these quantities separate lets Darcy law predict filter discharge while pore velocity predicts ideal advective travel time.^[ref]

For saturated Darcy flow through a long dam occupying a channel of width $\alpha z^2$, let hydraulic conductivity be $K=k\rho g/\mu$ and free-surface height $h(x)$. The Dupuit approximation gives $Q=-K\alpha h^3{h}_x/3$. Fixed heads $h(0)=H$ and $h(L)=\eta$ give $h^4=H^4-(H^4-{\eta }^4)x/L$ and the displayed steady discharge. Porosity relates pore velocity to superficial Darcy velocity but does not independently multiply this discharge when $k$ is intrinsic permeability. For depth-dependent permeability, integrate the horizontal conductivity over the saturated cross-section first.

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.

For a thin buoyant layer beneath an inclined impermeable roof, the hydrostatic approximation and Darcy's law give $u=kg\Delta \rho \sin \theta /\mu$ and the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. The first term transports fluid upslope; the second spreads the layer through its hydrostatic pressure gradient. The porosity multiplies the stored fluid volume, so the advection speed is $u/\varphi$, not the Darcy velocity $u$.

A localized temperature adjustment separates regions with different fluid density, viscosity and Darcy speed. Mass conservation in the moving frame gives the displayed plateau-depth relation. A speed stated as $V_T=\Gamma u_c$ with $u_c$ a Darcy velocity has effective retardation parameter $\varphi \Gamma$. Setting both laboratory-frame mass fluxes equal neglects storage associated with passage of the temperature front.

Take $x$ along a horizontal gap and $s$ perpendicular to its plates, with $0<s<b$. Assume a Newtonian fluid, stationary plates, a steady flow, a laminar flow and a fully developed flow away from the entrances and side edges. The no-slip boundary condition holds on both plates. Any transverse gravity is absorbed in the hydrostatic pressure; there is no imposed transverse through-flow. Translational symmetry and the no-penetration boundary condition then give $\mathbf{u}=(u(s),0,0)$. The Navier-Stokes equation reduces toTwice integrating gives the plane Poiseuille flow:The volumetric flow rate through one gap isThere are $n-1$ internal gaps. Their total cross-sectional area is $(n-1)wb$, because the plate thickness is negligible, so the Darcy velocity isThus the permeability of a porous medium has the dimensions of squared length. If finite plate thickness were included, the Darcy velocity averaged over the whole manifold would also contain its porosity.

Both the local acceleration and the convective acceleration vanish: ${\partial }_t\mathbf{u}=0$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=u_x{\partial }_x\mathbf{u}=0$. The dominant balance is therefore between the pressure gradient and viscous stress, with no inertia contribution in this fully developed solution. This exact cancellation does not guarantee that arbitrary large-Reynolds number realizations remain laminar flows; entrances, disturbances and turbulence can invalidate the assumptions.

Use a fixed representative volume $V$ containing many pores. Let $\phi$ be the porosity and let $\mathbf{q}$ be the Darcy velocity, defined per unit total area. The conservation of mass isThe solid matrix is rigid and fixed, and the liquid is fully saturating and has constant mass density, so the stored mass on the left is constant. Microscopic internal liquid-solid boundaries contribute no through-flux by the no-penetration boundary condition. Applying the divergence theorem to the external flux givesSince the representative volume is arbitrary, the macroscopic continuity equation isThere is no extra factor of $\phi$ in the flux: that factor is already incorporated in the definition of the Darcy velocity. A compressible liquid or deformable matrix would require retaining the storage term.

Under spherical symmetry, the continuity equation and Darcy's law giveHence $q_r=C/r^2$, and integrating the pressure gradient from infinity givesThe boundary condition at $r=a$ fixes $C={\Pi }_{1}a\Delta p/\eta$. Therefore spherical Darcy injection hasHere $q_r$ is the superficial Darcy velocity, not the pore velocity $q_r/\phi$.

For the comparison, assume geometrically similar pore networks, equal porosity, the same source radius and far boundary geometry, and equal pump efficiency. Dimensional analysis then gives ${\Pi }_1=C_{\mathrm{geom}}{d}^2$ with the same dimensionless factor. ConsequentlyAt a fixed desired volumetric flow rate, the required excess pressure and pumping power areAquifer A requires half the pumping power at equal throughput, or half the energy to inject the same volume over the same time. This compares the excess pressure-volume work dissipated by the Darcy flow; any different baseline compression or other plant costs need separate accounting.

The imposed comparison matters. At fixed $\Delta p$, A takes twice the volumetric flow rate and twice the instantaneous power, but the same excess energy per injected volume. The pore spacing alone does not determine permeability of a porous medium if network geometry or porosity changes; without the geometric assumptions the preference cannot be inferred from $d$ alone.

For a stationary solid skeleton, the Darcy velocity $\mathbf{u}$ is liquid volume transport per unit total cross-sectional area, while the pore velocity $\mathbf{v}$ is the intrinsic mean fluid velocity within the connected pore space. ThusFor an incompressible liquid in a saturated region, conservation of mass is ${\varphi }_t+\nabla \cdot \mathbf{u}=0$. With time-independent but spatially varying porosity, it becomesnot generally $\nabla \cdot \mathbf{v}=0$. Neglecting dispersion, a fluid interface is advected by the pore velocity; its mean normal speed is $V_n=\mathbf{v}\cdot \mathbf{n}=\mathbf{u}\cdot \mathbf{n}/\varphi$.

For the thin layer on the inclined plane, the gravitational acceleration components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plane. The hydrostatic approximation and constant pressure $p_a$ above the current giveWith permeability of a porous medium $k(z)$, Darcy's law therefore givesHere $u_x,u_y$ denote the components of the Darcy velocity. Integrate over depth, using the prescribed power laws:Both integrals are finite for the positive exponents in the model. Conservation of the stored fluid gives ${\mathcal{V}}_t+{\nabla }_{x,y}\cdot \mathbf{Q}=0$, orFor $0<\theta <\pi /2$, set $X=x\tan \theta$, $Y=y\tan \theta$ andSince $\cos \theta {\tan }^2\theta =\sin \theta \tan \theta$, the depth-dependent inclined porous-current equation isThe longitudinal flux carries fluid downslope, while the thickness-gradient terms spread it. The degenerate horizontal or vertical slope limits need their own coordinate scaling.

Assume quasi-steady saturated Darcy flow, hydrostatic pore pressure, a long shallow barrier, isotropic intrinsic permeability of a porous medium $k$, an impermeable lined base, and negligible capillarity. Maintain upstream and downstream free-surface heads at $H$ and $\eta <H$. The Dupuit approximation neglects vertical velocity except in small entrance/exit regions. Let $\mu$ be the fluid's dynamic viscosity and $K=k\rho g/\mu =kg/\nu$ its hydraulic conductivity.

Since the pressure is $p=\rho g(h(x)-z)$, the superficial Darcy velocity is $v_D=-K{h}_x$, uniform with depth at fixed $x$. Integrating over the printed channel width givesThe steady discharge is constant, so integration with the two head boundary values gives Dupuit flow in a channel with cubic wetted area:Here $Q$ is total volume per time, including the changing channel width. The pore-space mean speed is $v_p=v_D/\varphi =Q/[\varphi A(h)]$. Thus porosity affects storage and pore transit time but does not separately multiply this steady discharge when $k$ is the intrinsic permeability in Darcy's law. If “permeability” is instead supplied as hydraulic conductivity, use that value directly for $K$. An exact seepage-face/free-boundary calculation near the outlet would go beyond the shallow Dupuit model.

Because $\zeta <\eta \le h(x)$, the lower layer is saturated along the entire barrier. Both layers have the same horizontal head gradient but different Darcy velocities. Integrating their conductivity over the actual cross-section gives the transmissivityThe total discharge is $Q=-\mathcal{T}(h)h_x$. Hence $QL={\int }_{\eta }^H\mathcal{T}(h)dh$, givingThis is a parallel-layer conductivity integral, not a series-resistance average: the layers share the horizontal pressure drop. It reduces to the previous result when $k_1=k_2$. The assumption $\zeta <\eta$ is important; if the downstream free surface cuts the lower layer, the transmissivity and its head integration must be split into different saturation regimes.

Use the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.

Work in the frame of the steadily advancing mush, so the rigid primary-crystal skeleton moves with $-V{\mathbf{e}}_z$. Let $\varphi$ be solid fraction, $C$ the pore-liquid concentration, $\mathbf{q}=(u,w)$ the Darcy flux relative to that skeleton, and $K$ the common thermal conductivity. Set $\kappa =K/(\rho c_p)$, the thermal diffusivity. The actual pore velocity is $\mathbf{q}/(1-\varphi )-V{\mathbf{e}}_z$. Primary crystals are taken to exclude solute; the residual pore liquid freezes into a two-solid eutectic product at the bottom. These are the standard binary mushy layer assumptions implicit in using a single liquidus and an eutectic floor.

It is convenient to use hydraulic pressure $P=p+\rho gz$, absorbing the constant-density gravitational term. With equal phase properties, no solute diffusion, negligible fluid inertia and local liquidus equilibrium, a complete mush model isThe last equation follows from conservation of energy with mixture specific enthalpy $c_{p}T-L\varphi$. Sensible heat is transported by the total volume flux $\mathbf{q}-V{\mathbf{e}}_z$; the primary solid carries the latent-enthalpy flux $VL\varphi {\mathbf{e}}_z$. In particular the sign of the latent-heat term is fixed by the downward-moving skeleton.

At the liquid roof, imposeHere $P_0=p_0+\rho gh$. The gradient condition matches heat flux into the imposed liquid field, since there is no phase-fraction jump at this roof and the phase conductivities are equal. The pore velocity there equals the relative Darcy velocity. At the floor,There is no liquid penetration into fully solid material. The unspecified solid fraction at the floor is determined by the solute balance, not set to one: remaining eutectic liquid freezes as it crosses the floor. If $C_s$ denotes the bulk concentration of the final solid product, the floor solute balance and Stefan condition areMaintaining the floor at $T_E$ permits the solid-side bath to remove this final latent heat. Equivalently its required gradient is $T_z(0^{-})=T_z(0^{+})+\rho LV[1-\varphi (0)]/K$. No additional imposed solid-side gradient is specified. Laterally we take the quadratic-pressure strain solution, without sidewalls; finite sidewalls would require additional flow conditions and generally break this horizontal invariance.

For uniform permeability of a porous medium, incompressible flow gives ${\nabla }^{2}P=0$. The roof pressure and zero normal flow at the floor are satisfied byHorizontal outflow is balanced by downward inflow. The parameter $\alpha$ is an inverse time; $E^2$ in the printed pressure condition already has dimensions of pressure divided by length squared, so no extra density factor belongs in this definition.

With $T=T(z)$, $C=C(z)$ and $\varphi =\varphi (z)$, the $x$-dependence disappears from both scalar balances. Although $u$ is nonzero, its horizontal divergence cancels $w_z$ in $\nabla \cdot (\mathbf{q}C)$. The exact strain-driven steady mush equations becomeTogether with $T(0)=T_E$, $T(h)=T_0$, $T^{\prime }(h)=G$ and $\varphi (h)=0$, they determine the scalar fields and the unknown thickness. This reduction proves that the equations admit horizontally invariant scalar solutions despite the imposed straining flow. An equivalent useful form, obtained by differentiating the solute equation, isThis exact form also shows why one must not silently discard latent heat in an arbitrary strong-flow calculation.

Define $\Delta T=T_0-T_E>0$, $R=\alpha h/V$ and $S=L/(c_p\Delta T)$. For the stated regular ordering $R\gg S\gg 1$, strain transport dominates skeleton transport and the bounded phase-change enthalpy contribution. More precisely, at fixed finite thermal strain number $\alpha h^2/\kappa$, the dimensionless energy equation isso its skeleton and integrated latent-heat corrections are small in $1/R$ and $S/R$ respectively. The leading strain-dominated mush temperature profile obeys $\kappa T^{\prime \prime }+\alpha z{T}^{\prime }=0$. Put $\delta =\sqrt{2\kappa /\alpha }$ and $H=h/\delta$. Integrating first for $T^{\prime }$ and then for $T$ givesThe error function profile follows from integrating a Gaussian, with $\mathrm{erf}x=(2/\sqrt{\pi }){\int }_0^x{e}^{-s^2}ds$. Matching the roof gradient gives the requested transcendental thickness relationFor prescribed $G>0$ and $\Delta T>0$ the right side decreases continuously from infinity to zero as $H$ increases, so it fixes one positive leading-order thickness. The resulting $h$ must still satisfy the assumed ordering and the phase-fraction constraints.

To find the product composition, integrate the exact solute balance from the roof to an arbitrary depth. With $f=1-\varphi$,At $z=0$, the exact floor composition is thereforeThe minus sign has a direct physical meaning: the horizontal outflow removes more salt than arrives at the same depth from a uniform-$C_0$ liquid, leaving a depleted solid product. Using the leading error function profile andyieldsThe full solid fraction follows by substituting this same profile into the preceding integral formula for $f(z)$. It has $\varphi (h)=0$ and $1-\varphi (0)=C_s/C_E$. A physical branch requires $0\le C_s\le C_E$ and $0\le \varphi \le 1$. In this monotone profile, $C_s\ge 0$ ensures a nonnegative liquid fraction everywhere; it is not permissible to interpret a negative formal result as a material composition.

The error-function reduction is a regular leading approximation, not a uniform statement for every possible scaling of the thermal strain number. If $\alpha h^2/\kappa$ becomes singularly large along the ordering limit, a thin basal layer can amplify the discarded terms. The printed two inequalities alone do not specify that extra scaling. In such a regime the displayed exact scalar equations and boundary conditions, rather than an unqualified zero-latent-heat profile, provide the appropriate thickness and composition calculation.