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Simple examplew of viscous flow

Channel flow

The book goes through many examples: let me discuss a particularly simple one, the steady flow of viscous fluid in two dimensions through a straight channel. Let the walls of the channel be at $y=0,L$, extending in the $x$ direction, while the fluid is flowing entirely along the $x$ direction, and is $t$-independent. Incompressibility sets ${\partial }_{x}u=0$, so that $u=u(y)$. We will look for a solution fo which $p=p(x)$, so that the fluid is being pushed down the channel. Plugging these solutions into he Navier-Stokes equation, we find

$$\frac{1}{\rho }{\partial }_{x}p(x)=\nu {\partial }_y^{2}u(y)$$

Each side depends on an independent variable, so they must be equal to a constant $\nu c$. We can write $u(y)=c{y}^2+dy+e$. We set $u=0$ at $y=0$, so that $e=0$, and also at $y=L$, so that $c{L}^2+dL=0$, or $d=-cL$, so that $u(y)=c(y^2-Ly)$. $c$ is thus set by the pressure gradient. Since ${\partial }_{x}p=\nu \rho c=\mu c$, $p=\mu cx+p_0$.

Flow down a plane.

Anther example, discussed in the book, is one of a layer of water with height $h$ undergoing steady flow down an incline at angle $\alpha$ from the ground, with atmospheric pressure $p_0$ at the top of the fluid. Take the problem to be two-dimensional with $x,z$ the directions parallel and perpendicular to the incline, and $z=0$ the incline and $z=h$ the upper boundary. Let $p=p(z)$, and $\overrightarrow{u}=u(z)\hat{x}$, where incompressibility kills the $x$ dependence.

The Navier-Stokes equations in components become:

(22)$$\begin{array}{rlr}& & \frac{1}{\rho }{d}_{z}p+g\cos \alpha =0\\ & & \nu d_z^{2}u+g\sin \alpha =0\end{array}$$

NOte that $\overrightarrow{u}\cdot \overrightarrow{\mathrm{\nabla }}\overrightarrow{u}=0$ automatically. At $z=0$ the boundary conditions are $u(0)=0$, At $z=h$, the no-slip condition is clearly false. At such a fluid interface, we instead impose the “no stress” condition on the tangential stress The tangential stress is taken to be proportional to ${\partial }_{z}v$, so this must vanish. The pressure, as stted above, is $p(h)=p_0$. The resoluting solution is

(23)$$\begin{array}{rlr}p(z)& =& p_0+\rho g(h-z)\\ u(z)& =& \frac{g}{2\nu }\sin \alpha (2hz-z^2)\end{array}$$

This is an idealized situation with a flat bottom. In practice, deviations fom flatness will start to lead to nonvanishing intertial terms, and turbulent flow is possible. If we assume that the hill has a rise angle $\beta$, and a length scale $L\sim h\beta$, the natural Reynolds number is

$$Re=\frac{u\beta h}{\nu }=\frac{g\alpha \beta H^3}{{\nu }^2}$$

For water $\nu \sim {10}^{-4}{m}^2/s$. If we cnsider a puddle with $h\sim 1mm$, and $\alpha \sim .01$, we find for laminar flow $v\sim 5cm/sec$. Add an imperfection and then $Re\sim 50\beta$.

For a large slow river with $h\sim 10m$, $\alpha \sim {10}^{-4}$, we would find $v\sim 100km/sec$ which is crazy. Here however $Re\sim {10}^{12}\beta$. Thus, the slightest imperfection leads to turbulence, and our solution above is invalid.

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Introduction to diffusion