Velocity potential of flow under rigid disk

380 Views Asked by At

Determine velocity potential of the flow in this system: Rigid disk of radius R at a heigh h(t) above horizontal plane z=0 with incompressible, inviscid flow between them, and h<

The flow is axisymmetric and has horizontal component independent of z.

(Answer is $\phi=\frac{\frac{dh}{dt}}{4h}(2z^2-r^2)$)

Also, what is the pressure distribution under the disk and the force on the plane?

Working so far:

So we are solving the Laplace equation for $\phi$. In cylindrical co-ords, $\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial\phi}{\partial r})+\frac{\partial^2\phi}{\partial z^2} = 0$

Does "horizontal component independent of z" mean that $\frac{\partial\phi}{\partial r}$ is independent of z (since flow is axisymmetric?

In this case, $\phi=R(r)+Z(z)$.

Solving this, I think I get $\phi=\lambda (2z^2 - r^2) + Az + B$ but I'm not sure. Now I'm not sure what b.c. to apply.

Once I find the pressure distribution (do I use Bernoulli?), what surface am I integrating over?