Flow over an indentation I have a question regarding classical fluid dynamics:
I want to describe the shape of fluid flowing over an indentation as shown in grey the figure below.

I would like to understand which result is observed at the surface of the fluid. Is it going:


*

*up

*down

*stays the same


(As shown by the three lines in the figure.) And by how much!? 
Assuming an incompressible, non viscid, stationary, slow flow over a small indentation. 
My approach was first using the bernoulli equation:
\begin{equation}
\frac{v^2}{2}+{p \over \rho} +\phi=const
\end{equation}
As well as the continuity equation:
\begin{equation}
\rho v A =const
\end{equation}
This ended up solving:
\begin{equation}
\frac{v_0^2}{2}+g z_0=\frac{v_1^2}{2}+g z_1\\
v_0 z_0=v_1(z_1-h)
\end{equation}
for the height $z_1$ of the indentation of the surface. $h$ is the height of the grey area.
I am confused since for some example values, as $v_0 = 1$, $z_0 = 10$, $g = 10$, $h = 0.1$ this gave me $z_1=9.99$, $z_1=0.83$, $z_1=-0.58$.
Surely the last solution is rubbish but which one of the other ones is true? Probably the first one but why not the second?
So my clear questions are:


*

*Are Bernoulli and the continuity equation applicable here?

*If yes which of the solutions is correct and why?

*The solutions behaves very weird at greater $v_0$ values ($v_0$>8 for the example values above), why is this approach not valid for higher velocities?

 A: Your analysis is fine, and the different solutions you found correspond to physically observable cases. 

Using the above schematic we have the continuity equation
$$v_1\,y_1=v_2\,y_2,$$
$$\frac{v_1^2}{2g}+y_1=\frac{v_2^2}{2g}+y_2+\Delta h,$$
and we get
$$y_2^3-E_2\,y_2^2+\frac{v_1^2\,y_1^2}{2g}=0,$$
where $E_2$ is the specific flow energy at the bump,
$$E_2=\frac{v_1^2}{2g}+y_1-\Delta h.$$
The equation for $y_2$ has two positive and one negative solution if $\Delta h$ is not too large. Now the behavior of the free surface depends on whether the approaching flow (index $1$) corresponds to subcritical or supercritical flow:

The behavior depends on the Froude number $$Fr=\frac{v}{\sqrt{g\,y}}$$.


*

*If $Fr_1 < 1$ (subcritical approach) the water level will decrease
at the bump

*If $Fr_1 > 1$ (supercritical approach) the water level will decrease
at the bump

*If the bump height reaches $\Delta h_{max}=E_1 – E_c$, the flow at the crest will be exactly critical, $Fr = 1$.

*If the bump $\Delta h>h_{max}$, there exist no solutions satisfying our assumptions. Such a large bump will "choke" the channel and cause frictional effects, typically a hydraulic jump.


The material in the Wikipedia article on hydraulic jumps applies to this situation, mutatis mutandis, and has some interesting additional material.
