0
$\begingroup$

For an irrotational, incompressible fluid we can solve Laplace's equation for the velocity potential in a fluid in order to obtain the velocity field. This can be done for flow around a cylinder to obtain $$u_r=U(1-\frac{a^2}{r^2})\cos\theta$$ $$u_{\theta}=-U(1+\frac{a^2}{r^2})\sin\theta$$ as the radial and tangential velocity components, with the far field velocity of magnitude U in the x direction. This seems to be a laminar flow solution. However what is there in the method of solution that causes this to be the case? My notes say that this method holds best in the inviscid limit (high Reynolds number) because in that case an irrotational fluid remains irrotational (Kelvin's circulation theorem holds), however high Reynolds number is actually the turbulent flow regime so this contradicts the above.

Can anybody offer any insight here? Thanks :)

$\endgroup$
3
$\begingroup$

There's an important distinction to be made here -- inviscid (which is required for irrotational flow) is at infinite Reynolds number. As in the highest possible number; so high, in fact, it cannot be mathematically defined.

Turbulence, on the other hand, requires viscosity and therefore is rotational in nature. This merely requires high enough Reynolds number. As in big, but finite. And as the Reynolds number increases towards infinity, a turbulent flow will "freeze" -- there is no dissipation at the smallest scales and so the energy cascade breaks down and it just doesn't work anymore.

So, the difference here is subtle but important -- high Reynolds number (even really, crazy, ridiculously high) is not the same as infinite Reynolds number. Neither physically, nor mathematically. In the former case, you still have a second-order PDE. In the latter case, it becomes a first-order PDE and the whole problem is different.

What is actually most interesting is that solutions to the potential equation often look almost identical to solutions of the creeping flow conditions (very very small Reynolds numbers). So the solution with no viscosity will look like the solution where viscosity is the most important force in the problem.

$\endgroup$
  • $\begingroup$ 'Turbulence, on the other hand, requires viscosity and therefore is rotational in nature' - how do these things follow from one another? $\endgroup$ – Watw Dec 8 '15 at 22:21
  • $\begingroup$ @Watw That's actually a very interesting question in it's own right -- if a velocity gradient exists and there is viscosity, the shear stress on one side of the fluid element is larger than the shear stress on the opposite side of it. This alone begins rotation. $\endgroup$ – tpg2114 Dec 8 '15 at 22:40
  • $\begingroup$ That's why potential flow must be irrotational -- it has to be inviscid, and thus there is no vorticity. $\endgroup$ – tpg2114 Dec 8 '15 at 22:41
  • $\begingroup$ See for example this link for pictures of the fluid element. $\endgroup$ – tpg2114 Dec 8 '15 at 22:45
  • $\begingroup$ Is this a fair summary? The fluid may or may not be inviscid (although it helps if it is in terms of maintaining an irrotational fluid), however the incompressibility and irrotationality mean the viscous term is always zero (vector identity to get laplacian of velocity in terms of its divergence and curl which are both zero) and the system behaves as if it were inviscid. So Re is in effect infinity which happens to give a laminar solution (despite being at high Re) as the lack of viscosity and the irrotationality does not allow turbulence. $\endgroup$ – Watw Dec 8 '15 at 23:34

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.