= Solution
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> $C_f=\rho_0c_f$. Define $B=\rho_0g\alpha K/\mu$ 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 $F\,dz$ 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>:
$$
w=B\theta,\qquad v_y+w_z=0,\qquad
v\theta_y+w\theta_z=\kappa\theta_{yy}.
$$
At the plate, $v=0$ and $-k_e\theta_y=F$; in the far field $w,\theta\to0$. <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 obtain
$$
\boxed{\delta(z)=\left(\frac{\kappa k_e z}{BF}\right)^{1/3},\quad
\theta\sim\frac{F\delta}{k_e}\propto z^{1/3},\quad w\propto z^{1/3}.}
$$
Thus the warmed layer thickens and accelerates upward. Define a <streamfunction> by $w=\psi_y$, $v=-\psi_z$ and set
$$
\eta=\frac y\delta,\qquad
\psi=\frac{\kappa z}{\delta}f(\eta),\qquad
w=\frac{\kappa z}{\delta^2}f'(\eta),\qquad
\theta=\frac{F\delta}{k_e}f'(\eta).
$$
The last two expressions agree because $\delta^3=\kappa k_e z/(BF)$. Differentiation gives $v=-(\kappa/3\delta)(2f-\eta f')$. Substituting in the thermal equation leaves the single nonlinear equation for the <uniformly heated vertical plate in a porous medium>:
$$
\boxed{f'''+\frac23ff''-\frac13(f')^2=0,\qquad
f(0)=0,\quad f''(0)=-1,\quad f'(\infty)=0.}
$$
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 $f'\geq0$ and finite $f(\infty)$. Decay also gives $f''(\infty)=0$. 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> give
$$
0=[f'']_0^\infty+\frac23[ff']_0^\infty-\int_0^\infty(f')^2d\eta
=1-\int_0^\infty(f')^2d\eta.
$$
Thus the exact thermal budget is $\int_0^\infty(f')^2d\eta=1$. 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> is
$$
\boxed{f(\eta)\simeq2^{2/3}(1-e^{-\eta/2^{1/3}}),\qquad
\theta(y,z)\simeq\frac{2^{1/3}F\delta}{k_e}e^{-y/(2^{1/3}\delta)}.}
$$
It 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 by
$$
Q(z)=\int_0^\infty w\,dy,\qquad
\mathcal M(z)=\int_0^\infty w^2\,dy,\qquad
\mathcal B(z)=g\alpha\int_0^\infty w\theta\,dy.
$$
With these standard discharge conventions the <mass flux> is $\dot m=\rho_0Q$. Their similarity expressions are
$$
Q=\frac{\kappa z}{\delta}f(\infty),\qquad
\mathcal M=\frac{\kappa^2z^2}{\delta^3}\int_0^\infty(f')^2d\eta,\qquad
\mathcal B=\frac{g\alpha}{B}\mathcal M.
$$
The <integral> identity evaluates two of them exactly:
$$
\boxed{\mathcal M(z)=\frac{B\kappa Fz}{k_e}=\frac{BFz}{C_f},\qquad
\mathcal B(z)=\frac{g\alpha\kappa Fz}{k_e}=\frac{g\alpha Fz}{C_f}.}
$$
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> is
$$
\boxed{\dot m(z)\simeq2^{2/3}\rho_0\frac{\kappa z}{\delta(z)}
=2^{2/3}\rho_0\kappa^{2/3}\left(\frac{BF}{k_e}\right)^{1/3}z^{2/3}.}
$$
For clarity, if “<momentum flux>” means actual pore-fluid momentum rather than the commonly quoted kinematic Darcy diagnostic, constant <porosity> $\phi$ and pore speed $w/\phi$ give $\dot P=(\rho_0/\phi)\mathcal M$. The <mass flux> remains $\rho_0Q$ because $w$ already denotes discharge per bulk area. Stating both prevents an implicit <porosity> error. Validity also requires $\alpha\theta\ll1$ and a sufficiently slender, low-inertia porous <boundary layer>; the approximation is not the immediate leading-edge conduction field.
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