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For constant porosity and permeability, the equation is $h_T+h_X={\nabla }_{X,Y}\cdot (h{\nabla }_{X,Y}h)$. Under $\tau =T$, $\xi =X-T$, $\eta =Y$, the chain rule gives $h_T=h_{\tau }-h_{\xi }$ and $h_X=h_{\xi }$. Thus the moving-frame porous medium equation isIts conserved mass is $M=\iint hd\xi d\eta$.

There is a source inconsistency in the printed volume formula. Actual liquid volume is $V=P\iint hdxdy$. The stated coordinates have Jacobian determinantso the volume Jacobian for downslope stretched coordinates requiresThe printed factor ${\tan }^2\theta$ in the first expression is its reciprocal and cannot be proved with these coordinates. For example, at $\theta =\pi /6$ an integral equal to one corresponds to actual volume $3P$, whereas the printed formula gives $P/3$. The physical-volume normalization below uses the corrected Jacobian.

Let $r=\sqrt{{\xi }^2+{\eta }^2}$. For the source-type axisymmetric similarity solution, putMass conservation imposes $a=2b$, and balancing $h_{\tau }$ against $\nabla \cdot (h\nabla h)$ gives $a+1=2a+2b$. Hence $a=1/2$ and $b=1/4$. The profile equation isMultiplying by $s$, the left side is $-\frac{1}{4}(s^{2}f{)}^{\prime }$. Integration and regularity at the origin giveWhere $f>0$, divide by $sf$ to obtain $f^{\prime }=-s/4$, soSet the profile to zero beyond this front. The mass normalization isTherefore the two-dimensional Barenblatt profile for quadratic porous-medium diffusion iswhere $[q{]}_{+}=\max (q,0)$. Its radius and central thickness areThe profile is regular at the center, its flux $-h{h}_r$ vanishes at the dry edge, and the front speed $r_N^{\prime }=r_N/(4\tau )$ equals the limiting advective speed $-h_r$ there. Its support shrinks to the release point as $\tau \downarrow 0$ while its mass stays $M$, giving the concentrated initial release in the weak sense.

In physical coordinates its center travels atand $r=\tan \theta \sqrt{(x-x_c{)}^2+y^2}$ with $\tau =\rho gK\sin \theta \tan \theta t/(\mu P)$. If the printed volume expression were instead adopted merely as a definition of a formal parameter $\tilde{V}$, the same profile would use $M=\tilde{V}/(P{\tan }^2\theta )$; that parameter would equal $V{\tan }^4\theta$, not the actual released fluid volume.

Far downslope let the typical height be $H(X)$ and the cross-slope half-width be $W(X)$. Slenderness gives $W\ll X$ and slowly varying longitudinal thickness. In a steady porous gravity current, the longitudinal spreading term is smaller than advection by the ratio $H/X$, while longitudinal diffusion relative to transverse diffusion is of order $(W/X{)}^2$. Thus the leading balance isIntegrating across the finite current, with zero flux through its dry side edges, makes the leading downslope volume flux constant:Balancing the two differential terms givesEliminate $H\propto W^{-1/\beta }$ to find the far-downslope width of a constant-flux porous current:The typical height is correspondingly $H\propto X^{-1/(2\beta +1)}$. For every $\beta >0$, the width exponent is less than one half, so $W/X\to 0$ and $H/X\to 0$, consistent with the discarded longitudinal terms. The porosity exponent $\alpha$ drops out because it enters storage, which has zero time derivative in this steady problem.

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.

Use constant porosity $\varphi$, isotropic rock permeability of a porous medium $K$, common liquid dynamic viscosity $\mu$ and gravitational acceleration $g$. Treat the two liquids as separated by a sharp interface, with negligible capillary pressure, molecular mixing, ambient-current resistance and inertia. At distances beyond the initial source scale the current is shallow relative to its horizontal length, so its pressure is hydrostatic. The initial release is symmetric about $x=0$ and has actual liquid volume $V$ per unit length of source. Consequently its geometric thickness integral is $V/\varphi$, not $V$.

Below the interface the excess pressure relative to the ambient hydrostatic field is $p^{\prime }=\Delta \rho g(h-z)$, so Darcy law gives horizontal discharge $q_x=-K\Delta \rho g{h}_x/\mu$. The vertically integrated discharge is $Q_x=-K\Delta \rho gh{h}_x/\mu$. Take the specified downward leakage $v=\gamma h$ to mean Darcy flux per bulk horizontal area. Mass conservation then givesIf $v$ is instead defined as interstitial seepage speed, the actual leakage discharge is $\varphi v$ and the corresponding coefficient is $\lambda =\gamma$. Stating the velocity convention removes that otherwise ambiguous porosity factor. For a shale of thickness $d_s$ and permeability $K_s$, a simple hydrostatic leakage model has $\gamma =K_s\Delta \rho g/(\mu d_s)$, if the reservoir below it has the same ambient pressure reference and no capillary entry threshold.

Let $R(t)$ be the half-width of the support; the full end-to-end length will be $L(t)=2R(t)$. Appropriate conditions are symmetry $h_x(0,t)=0$, $h(\pm R,t)=0$, no flux beyond the support, and the moving-front condition $\dot{R}=-\beta h_x(R^{-},t)$. Initially $\varphi h(x,t)dx$ converges weakly to $V{\delta }_0$, expressing the concentrated line release. The equations describe the spreading porous gravity current, not the singular immediate release itself.

With $\lambda =0$, conserved volume and balance of $h_t$ against $\beta (h{h}_x{)}_x$ give width proportional to $t^{1/3}$ and height proportional to $t^{-1/3}$. Write $h=t^{-1/3}f(\eta )$, $\eta =x/t^{1/3}$. Substitution givesIntegrating from the symmetry axis gives $\beta f{f}^{\prime }=-\eta f/3$. Inside the support $f>0$, so $f^{\prime }=-\eta /(3\beta )$ and $f=A-{\eta }^2/(6\beta )$. Thus the compact Barenblatt solution isThe positive-part notation makes $h=0$ outside the current. Volume fixes its remaining constant:If “length” denotes the distance from the source to one front, it is $R$ rather than $L$; both conventions are given explicitly. Differentiating gives $\dot{R}=R/(3t)=-\beta h_x(R^{-},t)$, so the front condition holds. Using $V/2$ on each reflecting half-line gives the same normalization.

For positive leakage define actual retained volume $\mathcal{V}(t)=\varphi \int hdx$. Since the edge thickness and boundary flux vanish, integration of the equation yieldsPut $h=e^{-\lambda t}H(x,\tau )$, where $\tau =(1-e^{-\lambda t})/\lambda$. Because $\dot{\tau }=e^{-\lambda t}$, substitution cancels the leakage term and leaves $H_{\tau }=\beta (H{H}_x{)}_x$. This derives the exponential drainage transform for porous-medium diffusion. Its conserved geometric volume is still $V/\varphi$, soThe full length is again $L(t)=2R(t)$. In particular,At early times $\tau \sim t$, recovering impermeable-bed spreading. At late times, drainage reduces the current thickness and therefore its hydrostatic driving pressure; the effective spreading time saturates at $1/\lambda$. The current approaches a finite horizontal reach while its retained volume and thickness vanish exponentially. This does not mean a positive-volume pool remains at a finite final depth. For small positive $\lambda$ the length scale diverges as ${\lambda }^{-1/3}$, consistently with unbounded spreading when leakage is absent.

Let $z>0$ measure height from the lower edge and $y>0$ distance from one face of the vertical plate. Use the steady Boussinesq approximation, homogeneous isotropic permeability of a porous medium $K$, reference fluid mass density ${\rho }_0$, dynamic viscosity $\mu$, thermal expansion coefficient $\alpha$, effective thermal conductivity $k_e$ and liquid volumetric heat capacity $C_f={\rho }_0{c}_f$. Define $B={\rho }_{0}g\alpha K/\mu$ and $\kappa =k_e/C_f$, consistently with the thermal convention used above. Write $\theta =T-T_a$, where $T_a$ is the uniform ambient temperature.

Interpret $F$ as uniform wall heat-flux density: a plate segment of height $dz$ inputs heat $Fdz$ per unit horizontal span. The cumulative heat input up to height $z$ is therefore $Fz$. If $F$ instead denotes the sum of flux densities into two identically heated faces, replace it by $F/2$ in each face formula before summing. A specification only of total power per span, without its distribution in height, would not determine this uniformly heated-wall problem. We calculate one heated face; identical heating on both faces doubles the integrated fluxes.

Away from the leading edge, suppose the boundary layer is slender, $\delta (z)\ll z$, and porous inertia and longitudinal heat conduction are negligible. Lateral pressure adjustment leaves the vertical Darcy velocity proportional to buoyancy:At the plate, $v=0$ and $-k_e{\theta }_y=F$; in the far field $w,\theta \to 0$. Darcy law permits tangential discharge at an impermeable plate, so imposing a further no-slip condition $w=0$ would be inappropriate in this leading porous model.

Balance $\theta \sim F\delta /k_e$, $w\sim B\theta$ and $w\theta /z\sim \kappa \theta /{\delta }^2$ to obtainThus the warmed layer thickens and accelerates upward. Define a streamfunction by $w={\psi }_y$, $v=-{\psi }_z$ and setThe last two expressions agree because ${\delta }^3=\kappa k_{e}z/(BF)$. Differentiation gives $v=-(\kappa /3\delta )(2f-\eta f^{\prime })$. Substituting in the thermal equation leaves the single nonlinear equation for the uniformly heated vertical plate in a porous medium:The first condition is impermeability; the second fixes the wall heat flux; the third fixes ambient temperature and vertical velocity. Seek the positive-temperature branch with $f^{\prime }\ge 0$ and finite $f(\infty )$. Decay also gives $f^{\prime \prime }(\infty )=0$. The far-field transverse inflow need not vanish: $v_{\infty }=-(2\kappa /3\delta )f(\infty )$ supplies the increasing discharge.

An approximate solution can be obtained without a numerical shooting calculation. Integrate the differential equation over $0<\eta <\infty$. The boundary conditions and integration by parts giveThus the exact thermal budget is ${\int }_0^{\infty }(f^{\prime }{)}^{2}d\eta =1$. Assume an exponential thermal shape, $f=C(1-e^{-\eta /d})$. The wall derivative requires $C/d^2=1$, and the integral condition requires $C^2/(2d)=1$. Hence $d=2^{1/3}$, $C=2^{2/3}$ and the integral exponential profile for porous wall convection isIt satisfies all boundary conditions, has positive decaying temperature, and matches the exact integrated heat budget. It is an integral approximation, not an exact pointwise solution of the differential equation; its mass-flux coefficient should be read accordingly.

Define the volume discharge, kinematic Darcy momentum flux and buoyancy flux per horizontal span byWith these standard discharge conventions the mass flux is $\dot{m}={\rho }_{0}Q$. Their similarity expressions areThe integral identity evaluates two of them exactly:Equivalently $C_f\int w\theta dy=Fz$: all the heat supplied below height $z$ is carried upward through that section. Although these fluxes are exact within the boundary-layer equations, porous drag means that the momentum flux itself is not conserved with height.

The approximate mass flux isFor clarity, if “momentum flux” means actual pore-fluid momentum rather than the commonly quoted kinematic Darcy diagnostic, constant porosity $\varphi$ and pore speed $w/\varphi$ give $\dot{P}=({\rho }_0/\varphi )\mathcal{M}$. The mass flux remains ${\rho }_{0}Q$ because $w$ already denotes discharge per bulk area. Stating both prevents an implicit porosity error. Validity also requires $\alpha \theta \ll 1$ and a sufficiently slender, low-inertia porous boundary layer; the approximation is not the immediate leading-edge conduction field.

Assume the dense current lies on the bottom, $\Delta \rho >0$, and its height is small compared with the confined aquifer depth $w$. This permits the hydrostatic approximation and neglect of leading-order resistance from the return flow of the original fluid. The dense-layer pressure anomaly on the basal seal is $\Delta \rho gh$. The along-layer Darcy flux, integrated over the current thickness, and the downward leakage flux are respectivelyVolume conservation requires $\varphi h_t+{\partial }_x{q}_x=-q_{\ell }$. Thus the porous gravity current with hydrostatic leakage through a basal seal obeysHere permeability of a porous medium has units of area, and $\Omega$ has units of inverse time. The original PDF instead prints the sink coefficient $\lambda k\Delta \rho g/(\mu b)$. With $\lambda$ defined there as a permeability, that coefficient has units of area per time and cannot multiply $h$ to give $h_t$. It also omits the division of leakage by porosity. This is a genuine source error, not a Darcy law convention. The remaining parts solve the stated linear-drainage equation for a constant $\Omega >0$; the physically consistent coefficient is the one above. Formally treating the printed coefficient as an independently prescribed inverse-time constant leaves those mathematical solutions unchanged.

Let $k_1>k_2$ and let both original and injected fluids initially have dynamic viscosity $\mu$. Assume sharp displacement fronts, constant porosity and negligible exchange between layers. The imposed pressure difference gives Darcy velocities $q_i^0=k_i\Delta P/(\mu L)$ and interstitial velocities $u_i^0=q_i^0/\varphi$. The two arrival times are thereforeIf $A_i$ is the flow cross-section of layer $i$, its volume discharge is $Q_i^0=A_i{q}_i^0$. Equal viscosities make the hydraulic resistance independent of front position, so the total fluid discharge $Q_1^0+Q_2^0$ stays constant. The injected-fluid component at the outflow well isIt jumps from zero to the high-permeability discharge at the first breakthrough, then to the total discharge at the second. Before a layer breaks through, its outflow is original reservoir fluid. Cross-sections are needed to state an absolute volume discharge or its partition; the arrival times do not require them.

Take $Q$ as the fluid volume flux per unit source length into one side of the aquifer. For a symmetric release with total rate $Q$, use $Q/2$ on each side. Introduce a constant porosity $\varphi$ to distinguish Darcy velocity from the advecting interstitial velocity; if the given flux is already normalized to pore volume, this factor is absorbed in its definition. With a pressure gradient $G=-p_x>0$, Darcy law givesThus, with $\xi =z/h$, the pore velocity and its depth average areThe tracer concentration initially obeys the two-dimensional advection-diffusion equationReflecting transverse boundaries and constant effective molecular diffusion are assumed. No independent radioactive decay rate is supplied; it can be added separately if it is significant over the experiment.

For times long compared with $h^2/D$, transverse diffusion exchanges tracer between fast and slow streamlines. Write $c=C(x,t)+\chi (z)C_x+\cdots$, where $C=\overline{c}$ and $\overline{\chi }=0$. At the leading transverse-correction order,Integration yieldsIts mean and boundary derivatives are zero. Averaging the concentration equation gives $C_t+U{C}_x=[D-\overline{(u-U)\chi }]C_{xx}$. Integration by parts in the cell equation shows that the enhancement is positive:Consequently Taylor dispersion in a parabolic porous-layer velocity profile givesInjection fixes the inlet concentration or solute-flux boundary condition. This depth-averaged equation requires transverse equilibration and longitudinal concentration variations slow enough for the cell expansion. The supplied PDF hint has both a different polynomial and a false numerical value: its integral evaluates exactly to $-83/2520$, not $0.3525$. The cell problem above supplies the coefficient independently.

Choose $x$ upslope and $z$ normally downwards from the impermeable roof, so the buoyant layer occupies $0<z<h(x,t)$. Use the hydrostatic approximation and neglect the ambient return-flow pressure correction of relative order $h/H$. Pressure continuity at the interface gives the pressure inside the buoyant layer asThe along-slope Darcy velocity is thereforeDefine $u=kg\Delta \rho \sin \theta /\mu$. Integrating over depth gives the volume flux per unit width $F=uh-u\cot \theta h{h}_x$. Local fluid-volume conservation, including porosity, is $\varphi h_t+F_x=0$. Hence the inclined porous gravity current satisfiesThe first term is upslope advection driven by the mass density difference; the second is nonlinear hydrostatic spreading. The Darcy velocity $u$ and the corresponding pore velocity $u/\varphi$ must be distinguished.

The mobile fluid volume per unit well length includes the porosity:Integrating the residual-trapping attenuation of a porous current gives, for $0\le s<1$,This is mobile volume in the current; retained fluid outside its mobile thickness is not counted. Since $h_t=-h/\tau$ in the thinning region, the rate of accumulation of immobile retained volume is ${\dot{V}}_r=sV/\tau$. Direct differentiation verifies $\dot{V}+{\dot{V}}_r=Q{e}^{-t/\tau }$, so the mobile-volume expression obeys the overall injection balance.

Without capillary residual trapping, $s=0$ gives $V=Q\tau (1-e^{-t/\tau })$. The limiting expression as $s\to 1$ is $V=Qt{e}^{-t/\tau }$, though the local thinning equation degenerates at exactly $s=1$. Both limits provide useful consistency checks.

Take the boundary to be horizontal, put $z=0$ on it, and let $z=h(x,t)$ be the interface. Write $\Delta \rho =\rho -{\rho }_2>0$. Under the long-thin hydrostatic approximation, the deep ambient fluid has pressure $p_2=p_{\ast }-{\rho }_{2}gz$ to leading order, with negligible horizontal pressure gradient. Matching pressure at the interface gives, inside the dense porous gravity current,The Darcy velocity is superficial volume flux per total area, not pore speed. The horizontal Darcy law and depth-integrated flux therefore giveThe porosity $\varphi$ multiplies storage, so mass conservation is $\varphi h_t+q_x=0$. ConsequentlyThe deep-ambient and long-thin assumptions are what permit neglecting the ambient return-flow pressure variation, vertical inertia, and nonhydrostatic pressure corrections.

For a localized finite release, first use the reflecting half-line $x\ge 0$, so $h_x(0,t)=0$. Let $S=V/\varphi$ be the conserved cross-sectional area of the current; $V$ here is actual pore-fluid volume per unit cross-flow distance. A similarity solution of the porous medium equation has thickness proportional to $t^{-\alpha }$ and extent proportional to $t^{\beta }$. Conservation gives $\alpha =\beta$, and balancing $h_t$ against $(h{h}_x{)}_x$ gives $\alpha +2\beta =1$. Thus $\alpha =\beta =1/3$. PutSubstitution gives $-(F+\xi F^{\prime })/3=(F{F}^{\prime }{)}^{\prime }$. Zero flux at $\xi =0$ allows one integration: $F{F}^{\prime }=-\xi F/3$. In the positive-current region, $F^{\prime }=-\xi /3$, so $F=({\xi }_f^2-{\xi }^2)/6$. The normalization ${\int }_0^{{\xi }_f}Fd\xi =1$ gives ${\xi }_f^3=9$. The one-sided constant-volume porous gravity current is thereforeHere $[a{]}_{+}=\max (a,0)$. Its Darcy flux vanishes at the moving front because $h=0$ there, and its physical volume is $\varphi {\int }_0^{L}hdx=V$. This is the exact concentrated-release Barenblatt solution of the depth equation, physically used after the short initial adjustment to a long-thin current; an arbitrary finite initial patch requires its initial profile as well as its volume, although the same profile describes the usual late spreading regime. If a line release spreads symmetrically in both directions and $V$ denotes the total volume, use the same parabola on $-L<x<L$ with $L^3=9\lambda Vt/(2\varphi )$. If instead volume has already been divided by porosity, replace $V/\varphi$ by that geometric volume.

Drainage at superficial rate $\varphi \Gamma h$ changes mass conservation to $\varphi h_t+q_x=-\varphi \Gamma h$, soFor constant $\Gamma >0$, use the exponential drainage transform for porous-medium diffusionIndeed, $h_t+\Gamma h=e^{-\Gamma t}{\tau }^{\prime }{H}_{\tau }$ and $(h{h}_x{)}_x=e^{-2\Gamma t}(H{H}_x{)}_x$. Choosing ${\tau }^{\prime }=e^{-\Gamma t}$ leaves $H_{\tau }=\lambda (H{H}_x{)}_x$. Substituting the undrained Barenblatt solution givesUse the factor $1/2$ in $L_d^3$ for a total-volume symmetric release, as above. For general initial data the same transform is exact, but $H$ is the corresponding undrained solution rather than automatically a source parabola.

The remaining physical volume is $V(t)=V{e}^{-\Gamma t}$. Early times have $\tau \sim t$, recovering undrained spreading. Late times have $\tau \to 1/\Gamma$, so the front approaches a finite runout while the current drains exponentially:The front slows because drainage reduces the hydrostatic driving head. The finite transformed time also explains why drainage does not guarantee that an arbitrary initial profile reaches its undrained universal large-time shape.

Use the planar geometry indicated by the supplied $x,z$ equation and streamfunction; the source heat input is then measured per unit out-of-plane span. Let $\Theta =T-T_0$ and defineThis is the buoyancy part of Darcy law. Let $C_h$ denote the constant heat capacity per volume consistent with the given transport equation; for the fluid heat-flux convention $C_h={\rho }_0{c}_p$. Porosity or matrix heat-storage factors, if retained, must be used consistently in this coefficient and the effective transport parameters.

For plume width $b\ll z$, continuity gives $u\sim wb/z$. Both $u{\Theta }_x$ and $w{\Theta }_z$ are consequently of order $w\Theta /z$: lateral advection must not be discarded. Lateral thermal diffusion is of order $\kappa \Theta /b^2$, while vertical diffusion is smaller by $(b/z{)}^2$. The slender steady porous thermal plume model is thereforeWriting it in conservative form,and integrating across the plume, where $\Theta$ and lateral diffusive/advective heat transport vanish at large $|x|$, provesIn steady state with no lateral heat loss and negligible vertical conductive flux, this equals the heat supplied at the source. Define $Q=\beta F_h/C_h=\int w^{2}dx>0$.

For characteristic velocity $W$ and width $b$, flux conservation gives $W^{2}b\sim Q$ and the advection-diffusion balance gives $W{b}^2\sim \kappa z$. ConsequentlyThe dimensional scales are $b\sim ({\kappa }^2{z}^2/Q{)}^{1/3}$ and $W\sim (Q^2/(\kappa z){)}^{1/3}$. The plume becomes relatively more slender, $b/z\propto z^{-1/3}$, away from the source.

To find the profile, put $\eta =x/b(z)$ and $\psi =-W(z)b(z)f(\eta )$, so $w=W{f}^{\prime }$ andInserting into $u{w}_x+w{w}_z=\kappa w_{xx}$ givesIts coefficient is constant under the derived scaling. Choose the width normalization $W{b}^2=6\kappa z$. Integrating once with decaying velocity derivatives gives $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=1$ after normalizing the centreline velocity to $W$; symmetry has $f(0)=0$. The decaying, positive solution is $f=\tanh \eta$. Hence the hyperbolic-secant porous plume profile isSince ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, the precise amplitudes areThey satisfy both $W{b}^2=6\kappa z$ and $(4/3)W^{2}b=Q$. The derived $u$ supplies lateral entrainment; setting $u=0$ would not reproduce this profile or satisfy continuity.

A heat-weighted head speed of a porous plume estimate follows by filling the column below a head height $Z_h(t)$ with this steady profile. Its excess heat per unit height isConservation of the total injected heat gives $F_{h}t\simeq {\int }_0^{Z_h}\mathcal{H}(z)dz$. Differentiating,Thus a heat-weighted head advances at a speed of order the local centreline velocity and decelerates as $Z_h^{-1/3}$. With $K=(3Q^2/(32\kappa ){)}^{1/3}$,The factor $2/3$ is the energy-conserving estimate for a truncated steady column; the foremost centreline parcels have characteristic speed $W(Z_h)$ instead. The steady calculation does not resolve the transient nose or define an exact sharp temperature front. A truly axisymmetric point-source plume would require different geometry and cannot use this planar profile unchanged.

Use a constant porosity $\varphi$ and let $v$ denote pore velocity. Darcy law gives the Darcy velocity $q=\varphi v$. If the velocity convention has already absorbed porosity, set $\varphi =1$ in the following formulas. Assume positive permeability of a porous medium throughout the layer, so $k_0>0$ and $k_0+k_1>0$.Here $\mu$ is dynamic viscosity. The flux-weighted residence time in a porous layer is the pore volume divided by throughput. Equivalently, weighting each streamline's transit time $L/v(y)$ by its inlet flux $v(y)dy$ givesThis is also the advective travel scale after cross-layer mixing has homogenized the concentration. For resident-weighted transit time in a layered flow, sampling the initial fluid uniformly by resident volume, while suppressing cross-stream exchange, is a different experiment: its mean iswith the continuous $k_1=0$ limit $\mu \varphi L^2/(\Delta P{k}_0)$. Specifying the sampling convention avoids confusing the reciprocal of the mean speed with the mean reciprocal speed.