 Solution

The hydrostatic approximation gives $p=\rho g(h-z)$ above the horizontal floor, up to atmospheric pressure. The horizontal Stokes flow equation is $\mu {\mathbf{u}}_{zz}=\rho g\nabla h$, with the no-slip boundary condition at $z=0$ and zero tangential stress on the free surface. Integrating twice gives $\mathbf{u}=(\rho g/\mu )\nabla h(z^2/2-hz)$ and the lubrication gravity-current fluxFor the squeegee lubrication model, use $h_0$ vertically, $\ell =\rho g{h}_0^3/(\mu U)$ horizontally, and $\ell /U$ in time. If $X$ is the laboratory coordinate in the direction of travel, set $x=(X-Ut)/\ell$, $y=Y/\ell$, $t=U{t}_{\mathrm{lab}}/\ell$, and $h=h_{\mathrm{lab}}/h_0$. The blade half-length is $\alpha =L/\ell$. The moving-frame flux is $\mathbf{q}=-h{\mathbf{e}}_x-h^3\nabla h/3$. Conservation of volume gives the dimensionless lubrication equationUsing these scales as representative of the actual flow, lubrication theory requires a small aspect ratio and negligible inertia relative to vertical viscous resistance:The second is the Reynolds number based on $h_0$, multiplied by $h_0/\ell$. Surface tension is neglected as assumed. Large piles, rounded gaps, and narrow end regions need their own local slope and inertia checks if these representative scales cease to describe them.