Codex @codex  0

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.

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.

Let $Y(t)$ be the porous thermal front and $X(t)$ the polymer front. With the literal stated convention for Darcy velocity,Therefore $Y=\beta X$ with $\beta =\Gamma \varphi$. This is thermal-front retardation in Darcy flow. The region $0<x<Y$ is cold injected fluid of dynamic viscosity $\mu$; $Y<x<X$ is heated injected fluid of dynamic viscosity $b\mu$; $X<x<L$ is oil of dynamic viscosity ${\mu }_h$. The pressure drop is consequentlywhere ${\mu }_{\mathrm{eff}}=\mu [\beta +b(1-\beta )]$. Apply the preceding fixed-pressure planar Darcy displacement calculation with $\mu$ replaced by this effective dynamic viscosity. The polymer-front speed isIts position solves ${\mu }_{h}LX+({\mu }_{\mathrm{eff}}-{\mu }_h)X^2/2=k\Delta Pt/\varphi$, and $Y=\Gamma \varphi X$. These expressions hold while both interfaces remain in the field, the porous thermal front is sharp and stable, and $0\le \Gamma \varphi \le 1$. For $b=1$ the original constant-viscosity answer is recovered.

Sometimes a thermal retardation factor is defined relative to the pore velocity instead. Under that alternative definition one substitutes $\beta =\Gamma$ throughout. The factor of $\varphi$ in the boxed answer follows specifically from the PDF's phrase “fraction of the Darcy speed.”

Let $X(t)$ denote the polymer–oil material interface, measured from the injection well, and let $U(t)$ be the Darcy velocity. Incompressibility and uniform cross-sectional area make $U$ independent of $x$ in this planar model. Integrating Darcy's law through the injected layer and the oil layer givesThe material interface moves at the pore velocity, so $X^{\prime }=U/\varphi$. Put $a=\mu -{\mu }_h$ and $A={\mu }_{h}L$. Then $(A+aX)X^{\prime }=k\Delta P/\varphi$, which integrates from $X(0)=0$ toThus the fixed-pressure planar Darcy displacement hasFor $a\ne 0$, $X=[\sqrt{A^2+2ak\Delta Pt/\varphi }-A]/a$; its continuous $a=0$ limit is $X=k\Delta Pt/(\varphi {\mu }_{h}L)$. Positive layer dynamic viscosities ensure the resistance stays positive until breakthrough, which occurs at $t_b=\varphi L^2(\mu +{\mu }_h)/(2k\Delta P)$. The formulas stop there, since the two-layer geometry then changes.

Take constant Darcy velocity $U$ and work locally near a planar porous thermal front moving at $V_T=\Gamma U$. Let $\xi =x-V_{T}t$ and perturb its position by $\eta ={\eta }_0{e}^{\sigma t+iay}$, with wavelength $\lambda =2\pi /|a|$. Neglect thermal conduction and regard the adjoining layers as semi-infinite on this disturbance scale. In each region, Darcy's law and incompressibility imply that the pressure perturbation is harmonic, so the bounded normal modes areHere the minus side is cold with dynamic viscosity $\mu$, and the plus side is hot with dynamic viscosity $b\mu$. The base pressure gradients are $p_{-,x}^0=-\mu U/k$ and $p_{+,x}^0=-b\mu U/k$. Linearizing pressure continuity at the displaced front yieldsContinuity of the perturbed normal Darcy velocity givesso $A_{+}=-b{A}_{-}$. Therefore $\delta U=U|a|(b-1){\eta }_0/(b+1)$. The thermal-front kinematic law is $\sigma {\eta }_0=\Gamma \delta U$, proving the thermal-front viscous-fingering dispersion relationIt is positive for $b>1$, zero for $b=1$, and negative for $b<1$. Without thermal conduction the model has no finite fastest wavelength: the positive growth rate increases without bound as $\lambda \to 0$. A resolved thermal transition and diffusion regularize that short-wave limit.

For a material interface the kinematic coefficient would be $1/\varphi$ rather than $\Gamma$. Thus, if the interface of interest were instead the hot polymer–oil interface, the same calculation would give $\sigma =2\pi (U/\varphi )({\mu }_h-b\mu )/[\lambda ({\mu }_h+b\mu )]$, the usual planar viscous-fingering dispersion relation. This distinguishes the two interfaces and their speeds. The boxed thermal result uses the local planar approximation; distant wells or the other front modify the pressure eigenfunctions for wavelengths comparable to those separations.

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 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.

For Taylor dispersion in a linear porous-layer velocity profile, take isotropic pore-scale mass diffusivity $D_p>0$ and reflecting boundaries at $y=0,H$. Write $v-U=\gamma (y-H/2)$, where $\gamma =\Delta P{k}_1/(\mu \varphi LH)$. Let $C(x,t)$ be cross-sectionally averaged concentration and write the leading transverse correction as $c=C-\chi (y)C_x$. Substitution into scalar transport $c_t+v{c}_x=D_p(c_{xx}+c_{yy})$ yields the cell problemThus ${\chi }^{\prime }=\gamma y(H-y)/(2D_p)$, and averaging the axial solute flux gives $UC-(D_p+D_m)C_x$, withThe integration by parts has no boundary term. It also proves nonnegativity, even when $k_1<0$. The total effective mass diffusivity is $\mathcal{D}=D_p+D_m$. For anisotropic pore dispersion, the denominator uses the transverse coefficient and the added molecular term uses the axial coefficient.

There are two separate tests for significance. The shear Péclet number ${\mathrm{Pe}}_s=|\gamma |H^2/D_p$ gives $D_m/D_p={\mathrm{Pe}}_s^2/120$. A large value implies substantial enhancement of axial spreading. But the Taylor dispersion limit additionally needsOtherwise particles can cross the rock before transverse mixing occurs, and a constant long-time $D_m$ is not an adequate breakthrough model. The relative front width is of order $\sqrt{\mathcal{D}/(UL)}$ once the long-time model applies.

For a constant tracer input flux, normalize $j_0$ as solute flux per pore cross-sectional area, and use the initially tracer-free half-line modelThis is a constant-flux tracer inlet solution; prescribing a flux is distinct from prescribing the inlet concentration. A Laplace transform in time has decaying spatial root $r=(U-\sqrt{U^2+4\mathcal{D}p})/(2\mathcal{D})$ and givesDefine $\xi =UL/\mathcal{D}$, $\vartheta =U^{2}t/\mathcal{D}$, and $z_{\pm }=(\xi \pm \vartheta )/(2\sqrt{\vartheta })$. Inverting, or differentiating the following expression to check the equation and flux boundary condition, gives the requested arrival history:It tends from zero to $j_0/U$. For $UL/\mathcal{D}\gg 1$ near breakthrough, its leading front isIf a well-stirred inlet instead holds $C(0,t)=C_0=j_0/U$, the exact constant-concentration inlet solution is $C(L,t)=\frac{C_0}{2}\{\mathrm{erfc}(z_{-})+e^{\xi }\mathrm{erfc}(z_{+})\}$. The two inlet models have the same leading advective front, but are not identical at finite axial Péclet number.

Take $x$ increasing upslope and let $u$ be the superficial Darcy velocity, so the mobile discharge is $uh$. Its pore transport speed is $u/\varphi$. This interpretation is required by the factors of porosity in the printed equations; a literal pore-speed convention would instead replace $u$ there by $\varphi u$.

For capillary residual trapping, let $M(x,t)={\max }_{0\le \tau \le t}h(x,\tau )$ be the maximum invaded thickness, including the initial state. The stored carbon-dioxide volume per plan area isDuring advance, $M=h$ and new pores fill with mobile carbon dioxide; during recession, $M$ is fixed and the newly vacated pores retain saturation $s$. Conservation of mobile plus trapped fluid isConsequentlyAssume $a,L,u,\varphi >0$ and $0<s<1$. The two speeds differ because the advancing front fills a whole pore volume while the receding tail removes only its mobile fraction.

The method of characteristics keeps height constant on $x=x_0+v_{A}t$ on the leading face and $x=x_0+v_{R}t$ on the trailing face. Matching the two linear profiles gives the triangular current with capillary retention:Before extinction, the rear, front, crest position and crest height areEvery positive height has one trailing and one leading characteristic; their intersections trace the crest. The mobile current disappears at

The final trapped region is the maximum-invasion envelope of a retained current. For $0<x<L$, the initial height is already the maximum, $M_{\infty }=ax$. For $L<x<x_{\ast }$, the maximum occurs when the crest passes $x$, at $t_m=2(x-L)/(v_R+v_A)$. Substituting into the trailing profile givesIf $z$ is distance into the aquifer measured normally from its upper boundary, trapped carbon dioxide occupies $0<z<M_{\infty }(x)$ at residual pore saturation $s$. The occupied pore volume per unit transverse width isexactly the initial mobile volume. This is an independent mass-conservation check of both the envelope and extinction distance. In the limit $s=0$, the current translates without shrinking and nothing is trapped; the finite-extinction formula is not used in that limit.

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 $H/L$. DefineDarcy's law and the fixed two-dimensional volume flux giveHere $q_i$ is Darcy velocity; the parcel speed is $q_i/\varphi$. Integrating the pressure gradient over the length and defining $Q=H{k}_{\mathrm{eff}}\Delta p/(\mu L)$ yieldsThe equal-porous permeability limit is $k_{\mathrm{eff}}=k_1=k_2$. 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.

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 isLet $r=y_0/H$ 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 atBefore 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 ${\tau }_V=\varphi HL/Q$, the pore-volume throughput time. Integrating $dt=\varphi dx/q_i$ on the two portions of the path givesIts derivative with respect to ${\xi }_c$ is ${\tau }_{V}dK/(k_1{k}_2)$, 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. HenceThe printed expression assumes $k_1\ge k_2$. The absolute value is needed for a nonnegative maximum difference without that ordering. For equal porous permeabilities every parcel has time ${\tau }_V$.

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.

Choose $U$ and $Q$ as the speed and source strength in the same kinematic velocity convention. With $r=(x^2+y^2{)}^{1/2}$ and $\vartheta =\arg (x+iy)$, superposition of uniform potential flow and a two-dimensional point source giveswhere $(u,v)=\nabla \Phi =({\psi }_y,-{\psi }_x)$. The angular branch cut records the volume flux from the source; the logarithm's arbitrary reference length only changes an additive constant in the velocity potential.

The stagnation point lies on the negative axis, where $u=U+Q/(2\pi x)=0$. Thus the limiting upstream reach of source fluid isThe stagnation point is approached asymptotically, rather than reached in finite time by a parcel on the upstream axis. The dividing streamlines through it have $\psi =\pm Q/2$ when the angle is taken in $(-\pi ,\pi )$. Far downstream $\vartheta \to 0$, so these streamlines approach $y=\pm Q/(2U)$. Consequently the source fluid occupies the Rankine half-body between them, with downstream width $2b=Q/U$. In the upper half-plane its boundary can also be parametrized by $y=b(1-\vartheta /\pi )$, $x=y\cot \vartheta$, $0<\vartheta <\pi$.

For arrival at the downstream axis, the direct streamline $y=0$, $x>0$ has speed $U+Q/(2\pi x)$. Its transit time from the ideal point source isThe first source fluid at a downstream cross-section lies on this symmetry axis: for positive $x$, the horizontal source contribution $Qx/[2\pi (x^2+y^2)]$ is maximal at $y=0$, and off-axis paths cannot advance through positive $x$ more quickly. A far-away parcel passing the inlet line at the same release time has transit time $x/U$. The time advance is thereforeThis is an advance, not the source parcel's absolute transit time. If a source parcel is released at time $t_0$ while the distant comparison parcel crosses $x=0$ at time zero, the advance relative to that parcel is $\tau -t_0$; the printed formula uses simultaneous releases. If $U,Q$ instead denote Darcy velocity and bulk-area source flux in a medium of uniform porosity $\varphi$, their ratio still gives the same $b$, but both parcel times acquire a factor $\varphi$. No porosity is specified here, so the displayed time formula fixes the kinematic convention.

The imposed pressure gradient and Darcy's law give a linear velocity profile. Use $U$ as the mean pore velocity in the contaminant advection-diffusion equation; thenThe last conditions express zero transverse solute flux under the standard closed-layer interpretation. Transverse exchange through the layer boundaries would change this cell problem and requires additional boundary data. If the prescribed mean speed is a Darcy velocity, its corresponding pore velocity is that speed divided by porosity.