Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If the significance of the Reynolds number is:

Total Momentum Transfer / Molecular Momentum Transfer

Then what is the effect of angular momentum on the transition from laminar to turbulent as in a convective vortex? Waterspouts, in particular, seem to be obvious cases where a very large Rayleigh Number would be assigned, yet the systems are admitted by observational experts to be dominated by laminar flow!

share|cite|improve this question
What do You mean with that waterspout? The water streaming in the nozzle or the water jet in air? – Georg Nov 28 '11 at 20:56
Waterspouts as in tornadoes over water. Lots of videos of them on youtube and its hard to imagine turbulent flows dominate them. – James Bowery Nov 28 '11 at 21:32
Ah, in a tornado hose! Why would someone assign high Reynolds numbers for those? – Georg Nov 28 '11 at 21:40
up vote 4 down vote accepted

This is a reasonable question. At the scale of a waterspout, the inertial forces of fast-moving air should be large compared to the viscous forces (i.e., very large Reynolds number). Yet the inflow along the surface of the water is laminar, where we would ordinarily expect boundary-layer vorticity (i.e., turbulence). A detailed description of the expected properties of suction vortexes (including the boundary-layer turbulence) can be found here:

Thermodynamic Tornadoes?

An hypothesis that directly addresses the anomalous behavior is here:

Tornadic Inflow

This work includes plenty of references in case you want to research it further. For example, you could read up on two-fluid simulations, which accurately describe the fast-moving, laminar flow along the surface (but beg the question of how two-fluid behaviors are instantiated in a well-mixed fluid such as the air).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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