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.

enter image description here

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?

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

enter image description here

Using the above schematic we have the continuity equation


$$\frac{v_1^2}{2g}+y_1=\frac{v_2^2}{2g}+y_2+\Delta h,$$

and we get


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:

enter image description here

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.