1
$\begingroup$

Rapids and several other "white water" systems could be typical examples of turbulent motion. To describe such flows the Navier-Stokes equations must be extended by terms describing the surface tension and other features of a multiphase flow. I have no doubt that this can get really ugly at "still simple" approximation.

My question is: does the surface tension attenuate the turbulent behavior or amplify it? I assume that there isn't simple yes/no answer and that it would require a discussion of energies and scales.

$\endgroup$
  • $\begingroup$ It's clear how surface tension might attenuate turbulence, but I can't think of a mechanism by which it might amplify it. Did you have something in mind? $\endgroup$ – lemon Jul 11 '16 at 17:09
3
$\begingroup$

Turbulence is well defined in the bulk of the flow, hence in 3D in the examples you quote, while surface tension acts, well, at the surface! I believe a good reformulation of your question would be "will a flow be less chaotic with larger surface tension?"

Surface tension is highly dependent on geometry. It stabilizes some flows (e.g. a not-too-shallow layer on an incline) and destabilizes others (e.g. jets). Since you refer to rapids, then you are rather in the first situation when the free surface is a small perturbation of a planar surface. Surface tension will contribute to stabilize it (unless there is some splashing that will feature the jet case). But then will it decrease turbulence in the bulk below? That would need to be investigated with a precise geometry and numbers: indeed, by reducing the "freedom" of the free surface, it can confine the flow above an obstacle, which can have the effect of triggering turbulent transition.

$\endgroup$
  • $\begingroup$ Probably Pira is thinking of two-phase flows, say air bubbles entrained in turbulent flow. $\endgroup$ – Deep Jul 12 '16 at 12:03
  • $\begingroup$ @Zero: the question is vast, and the OP should narrow it down if there is some phenomenon of special interest to them. $\endgroup$ – Joce Jul 12 '16 at 13:32

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.