0
$\begingroup$

In Taylor-Couette flow, the interior fluid becomes fully turbulent if the relative angular velocity of the cylinders is high enough. The turbulent fluid has a vorticity distribution, and each of the (time-averaged) vorticity vectors must point along the rotation axis. If one increases the relative velocity even more, the strength of the turbulence (and its vorticity) also increases, as does the total angular momentum of the fluid. What is the quantitative relation between the two?

$\endgroup$
0
0
$\begingroup$

Around each vortex of vorticity $n$ the rotor of velocity is: $$ \oint v \ dl = 2 \pi \kappa $$ Where $\kappa$ is some constant ( in case of superfluids $\frac{n \hbar}{m}, n \in \mathbb{Z}$). At the same time, using $v = \Omega \times r $, $\Omega$ - angular velocity, and $\text{rot} v = 2 \Omega$. Let us consider some circle of unit area, on one hand taking the integral around this circle: $$ \oint v \ dl = 2 \pi \kappa \nu $$ Where we $\nu$ denotes the average vorticity of the fluid. And at the same time $$ \oint v \ dl = 2 \Omega $$ Where we have applied the Stokes formula. So finally one gets: $$ \Omega = \pi \kappa \nu $$

$\endgroup$
1
  • $\begingroup$ I understand there is a relation between vorticity and the local angular speed. I am asking about something else. Given the vorticity distribution in a 2D fluid, what is the total angular momentum content of that fluid? $\endgroup$ May 3 '20 at 22:35

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.