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\textbf{Question}

A viscous fluid of constant density $\rho$ and constant dynamic
viscosity $\mu$ flows steadily under gravity down a rigid
impermeable plane. The acceleration due to gravity is denoted by
$g$. The plane is inclined at an angle $\alpha$ to the horizontal.
The flow is two-dimensional and the coordinate origin is in the
plane. The $x$-axis is taken to be parallel to the plane (along
the line of greatest slope) and the $y$-axis is normal to the
plane. The fluid velocity is denoted by $\un{q}=(u,v)$. Show that,
at any point in the fluid, the shear stress $\tau$ (i.e. the
stress in the $x$-direction exerted on a plane $y=$constant) is
given by $$\tau = \mu (u_y + v_x)$$

On the top surface of the fluid the pressure is given by $p_a$ and
there is no shear stress. By assuming a flow velocity $\un{q}$ of
the form $$\un{q}=(u(y),0)^T,$$ determine a solution to the
Navier-Stokes equations that represents unidirectional flow down
the plane in a layer of constant thickness $h$.

Show that the momentum flux M of fluid flowing down the plane is
given by $$M=\frac{2h^5\rho^3g^2\sin^2\alpha}{12\mu^2}$$


\textbf{Answer}

$$\\$$
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The stress vector $\underline{t}$ is given by
$T\underline{\hat{n}}$.

Now $T=-p\delta_{ij}+2\mu d_{ij}$ i.e. $T=\left (
\begin{array}{cc} -p+2\mu u_x &amp; \mu(u_y+v_x)\\ \mu(u_y+v_x) &amp;
-p+2\mu v_y \end{array} \right )$

Now the normal to a plane $y=constant$ is $(0,1)$, so

$\underline{t}=\left ( \begin{array}{cc} -p+2\mu u_x &amp;
\mu(u_y+v_x)\\ \mu(u_y+v_x) &amp; -p+2\mu v_y \end{array} \right )
\left( \begin{array}{c} 0\\ 1 \end{array} \right ) =\left (
\begin{array}{c} \mu (u_y+v_x) \\ -p+2\mu v_y \end{array} \right )$

And thus the shear (i.e. in the x-direction) stress is
$\mu(u_y+v_x)$.

Now by Navier-Stokes \begin{eqnarray*} uu_x+vu_y &amp; = &amp; -p_x/\rho
+\nu(u_{xx}+u_{yy})+g\sin\alpha \\ uv_x+vv_y &amp; = &amp; -p_y/\rho
+\nu(v_{xx}+v_{yy}-g\cos\alpha \\ u_x+v_y &amp; = &amp; 0 \end{eqnarray*}

Now assume that $\underline{q}=(u(y),0)^T$. Then $u_x+v_y=0$ and
we get \begin{eqnarray*} 0 &amp; = &amp; -p_x/\rho
+\nu(u_{yy})+g\sin\alpha\\ 0 &amp; = &amp; -p_y/\rho+\nu(0)-g\cos\alpha
\end{eqnarray*}

so $p_y=-\rho g\cos\alpha$, $\Rightarrow p=-\rho gy\cos\alpha +K$.
Now on $y=h$ we have $p=p_a$ $\Rightarrow p=p_a+\rho
g(h-y)\cos\alpha$.

Thus $p_x=0$ and so

$u_{yy}=\displaystyle \frac{-g\sin\alpha}{\nu} \Rightarrow
u_y=\frac{-gy\sin\alpha}{\nu}+C$

Now since $\tau=\nu(u_y+v_x)$ and $v=0$, we have (since the shear
stress is 0 on $y=h$)

$u_y=0$ on $y=h$

$\Rightarrow \displaystyle u_y=\frac{g(h-y)\sin\alpha}{\nu}$

$\Rightarrow \displaystyle u=\frac{g(hy-y^2/2)\sin\alpha}{\nu}+C'$

But by no slip $u=0$ on $y=0$

$\displaystyle \Rightarrow u=\frac{g(hy-y^2/2)\sin\alpha}{\nu}$,
$p=p_a+\rho g(h-y)\cos\alpha$

Now momentum $=\rho u \ \ \Rightarrow$ momentum flux $=\rho u^2$

$\Rightarrow \displaystyle M=\int_0^h \rho u^2 \,dy = \int_0^h
\frac{\rho g^2 \sin^2\alpha}{\nu^2} \left (
h^2y^2-hy^4+\frac{y^4}{4} \right ) \,dy$

$\Rightarrow$ \begin{eqnarray*} \displaystyle M &amp; = &amp; \frac{\rho
g^2\sin^2\alpha}{\nu^2} \left [
\frac{h^2y^2}{3}-\frac{hy^4}{4}+\frac{y^5}{20} \right ]_0^h\\ &amp; =
&amp; \displaystyle \frac{\rho^3g^2\sin^2\alpha}{\mu^2}[h^5] (1/3
-1/4+1/20) \end{eqnarray*}

$\displaystyle \Rightarrow
M=\frac{2\rho^3g^2\sin^2\alpha}{15\mu^2}$


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