Von-Karman vortex street - Reynolds number range
 

Dear Friends,

I am trying to model flow around a 2D cylinder in Fluent 6.3 and capture vortices shedding. I did the same project in Reynolds number ranging 100-4500 with reasonable results. But now I am doing a project with Air as fluid and cylinder diameter of 0.05 m. other specifications are:

V = 10 m/s
Solver = unsteady 2nd order
Viscous = k-e standars
solution=second order upwind
time step=0.005
time = up to 100 s (even more sometimes)

My reynolds number is around 10000. I am not sure if I should expect von karman vortex street in this reynolds number. my result is only a fully developed and symmetric flow behind the cylinder without any vortex.
attached picture is contour of vorticity magnitude

Help me if there is a reynold range for vortex shedding in cylinder case.

Thanks

Well, a couple of comments from the picture ... you are modelling the turbulence via (U)RANS approach but the solution you computed is statistically steady. That means you would accept that turbulence fluctuations at high Re number do not appear in the stationary averaged field. However, URANS allows potentially for some unsteady energy component to be described if the frequency is low.

Try to eliminate any turbulence modelling, using second order upwind you can perform an ILES simulation that should allow you to see some unsteady shedding.

FMDenaro,

I did it again with choosing Laminar model for viscous model. Vortices are generated. but is it reasonable to use laminar model for a flow with RE=10000. Is that your mean to use laminar model instead of URANS?

if the grid is sufficiently fine (cell Reynolds O(1)) you can consider that the DNS apporoach that does not require any turbulence model.

Have a look at this link -

http://www.thermopedia.com/content/1...id=104&sn=1410

This should give you some information about what you should expect to see at different Reynolds numbers.

But for that one should to do 3d simulations instead 2d. Do you think it is realistic to run on PC at Re=10^4 ?

Of course, DNS is always 3D apart some specific geophysical 2D DNS simulations.

As the feasibility of a DNS on a cylinder at Re=10^4 is concerned, I based my assumption on the fact that DNS of plane channel flow is now realized at Re_tau more than 10^3, that means the Re based on the channel height is about 20 times greater. But that can be performed on specific HPC platform.

the link does not work...

Have you managed to access the link yet? I had trouble with it last night but it seems to be fine this morning.

Yes it works now. thank you