A steady, fully developed liquid film of constant thickness h drains down a wide plane…

A steady, fully developed liquid film of constant thickness $h$ drains down a wide plane inclined at angle $\theta$ to the horizontal. The flow is unidirectional, $u=u(y)$ with $y$ measured normal to the wall from the wall toward the free surface. The liquid has constant $\rho$ and $\mu$. The free surface $y=h$ is stress-free and exposed to constant atmospheric pressure. There is no applied streamwise pressure gradient beyond hydrostatics; the driving body-force component is $\rho g\sin\theta$. 1. Reduce the streamwise momentum equation to $\mu u''+\rho g\sin\theta=0$. Integrate with $u(0)=0$ and $u'(h)=0$ to obtain \[ u(y)=\frac{\rho g\sin\theta}{\mu}\Bigl(hy-\frac{y^2}{2}\Bigr). \] 2. Compute the volume flow rate per unit width $q'=\int_0^h u\,\mathrm dy$, the bulk speed $\bar U=q'/h$, the surface speed $u_s=u(h)$, and the wall shear $\tau_w=\mu u'(0)$. Show \[ q'=\frac{\rho g\sin\theta\,h^3}{3\mu},\quad \bar U=\frac{\rho g\sin\theta\,h^2}{3\mu},\quad u_s=\frac32\bar U,\quad \tau_w=\rho g h\sin\theta. \] 3. Evaluate the four quantities for \[ \rho=1000\ \mathrm{kg/m^3},\quad \mu=0.1\ \mathrm{Pa\cdot s},\quad h=0.001\ \mathrm{m},\quad \theta=30^\circ,\quad g=9.81\ \mathrm{m/s^2}. \] 4. Confirm $u_s=(3/2)\bar U$ from the profile, not from a Couette guess. 5. Audit both sentences: (i) “the plane is steep enough that $\sin\theta$ may be replaced by $1$, doubling every numerical value”; (ii) “a free surface behaves like a moving lid, so $u_s=2\bar U$ as in plane Couette flow.” Do not replace the film by a pipe-loss problem, a Stokes sphere, or a boundary-layer integral.

$\rho g\sin\theta=1000\cdot 9.81\cdot(1/2)=4905\ \mathrm{N/m^3}$. Then $q'=4905\cdot 10^{-9}/0.3=1.635\times 10^{-5}\ \mathrm{m^2/s}$.