= Solution
\b[Meaning and microscopic origin.] For a <Newtonian fluid>, the <dynamic viscosity> $\eta$ measures tangential momentum transfer between fluid layers moving at different speeds. For the Newtonian shear considered here, $\tau_{xz}=\eta\,dv_z/dx$, with units $\mathrm{Pa\,s}$ for $\eta$. Molecular motion and collisions transport momentum across layers in a gas; intermolecular interactions contribute strongly in liquids. The resulting viscous forces oppose relative shear motion and convert organized kinetic energy to heat. The constitutive law is Newtonian and isotropic; incompressibility alone does not establish it.
\b[Vertical force balance.] Choose $z$ downward, as in the original diagram, and $x$ outward normal to the plate. In a uniform steady film, $\mathbf v=v_z(x)\widehat{\mathbf z}$ has zero advective acceleration and no along-plate <pressure gradient>. On a slice of thickness $dx$, the net shear force per unit face area is $\tau(x+dx)-\tau(x)$, while its weight per unit area is $\rho g\,dx$. Therefore
$$
\boxed{d\tau=-\rho g\,dx.}
$$
Here $\tau=\eta\,dv_z/dx$ denotes the downward traction on a face with positive $x$ normal. The opposite face has the opposite traction orientation, accounting for the sign in the force difference.
\b[Boundary conditions.] At the fixed solid wall, the <no-slip boundary condition> is $v_z(0)=0$, with no normal penetration. At the free surface $x=d$, neglect air shear, so $\tau(d)=0$ and hence $v_z'(d)=0$. The flat <free surface> has normal pressure $p_{\rm atm}$, neglecting surface-curvature effects. These are a velocity condition at the wall and a stress condition at the free surface, not zero velocity at both boundaries.
\b[The parabolic speed profile.] Integrating $\tau'=-\rho g$ from the shear-free surface gives $\tau(x)=\rho g(d-x)$. Thus
$$
\eta\frac{dv_z}{dx}=\rho g(d-x),
$$
and the wall condition yields
$$
\boxed{v_z(x)=\frac{\rho g}{\eta}\left(dx-\frac{x^2}{2}\right)
=ax+bx^2,\qquad
a=\frac{\rho gd}{\eta},\quad b=-\frac{\rho g}{2\eta}.}
$$
It is zero at the wall and largest at the free surface, where its gradient vanishes. The wall shear $\rho gd$ supports the downslope weight of the film per unit plate area.
\b[Volume flux per unit width.] Integrate through the depth:
$$
q=\int_0^d v_z(x)\,dx
=\frac{\rho g}{\eta}\left(\frac{d^3}{2}-\frac{d^3}{6}\right),
\qquad
\boxed{q=\frac{\rho gd^3}{3\eta}.}
$$
This is volume flow per unit length in $y$, with units $\mathrm{m^2\,s^{-1}}$.
\b[Inclining the plate.] Keep $x$ normal outward from the plate and use $s$ downslope. The gravitational components are $g\sin\theta$ downslope and $-g\cos\theta$ normal to the plate. For a uniform film of normal thickness $d$, the steady slice equations are
$$
0=-\frac{\partial p}{\partial x}-\rho g\cos\theta,\qquad
0=-\frac{\partial p}{\partial s}+\rho g\sin\theta
+\eta\frac{d^2v}{dx^2}.
$$
The free-surface pressure condition gives
$$
p(x)=p_{\rm atm}+\rho g\cos\theta(d-x).
$$
Uniform $d$ makes $\partial_sp=0$. The same no-slip and zero-surface-shear conditions then give the <uniform gravity-driven Newtonian film>,
$$
\boxed{v(x)=\frac{\rho g\sin\theta}{\eta}
\left(dx-\frac{x^2}{2}\right).}
$$
The component normal to the plate changes the hydrostatic pressure, while the tangential component drives the flow. The thickness is measured normal to the tilted plate.
\b[The inclined-film volume flux.] Integrating the new profile gives
$$
\boxed{q(\theta)=\frac{\rho gd^3}{3\eta}\sin\theta.}
$$
It recovers the vertical result at $\theta=\pi/2$ and vanishes on a horizontal plate at fixed uniform depth. A varying depth, imposed pressure gradient or significant air shear would require additional terms; the solution here uses the uniform film specified by the simple draining-plate model.
Back to article page
