Von Karman 2D Stable, no shedding
 

I am trying to simulate the shedding of Von Karman Vortices from a 2d cylinder. However I am only seeing steady no shedding results. I have read the posting on this site, which appear to indicate that a disturbance is required before the unsteady result will be achieved. If this is the case could someone give me an Idea of what disturbance is required and how I could go about it.

Cheers!! in advance!!

The problem is set up as follows Re = 70

V = 0.001

D = 1

Fluid = Air

Domain is ten cords with Circular shaped

Velocity inlet, and velocity outlet (No external walls)

Model = Laminar

Time step = 0.00001

Have tried altering the relaxation factors

Cheers!! in advance!!

Re: Von Karman 2D Stable, no shedding
 

Stephen

If you have a look at the validations that are posted up on the Fluent website for Fluent 6.0, there is a validation that uses the vortex shedding around a cylinder. They give details about the disturbance required to initiate the vortex street.

Hope this helps

Paul

Re: Von Karman 2D Stable, no shedding
 

Hi Stephen

The use of a disturbance may not be necessary. At least I runned this flow test some time ago for Re=100, and obtained the transient flow with good accuracy.

If your mesh is coarse or you are using first order upwind for the discretization of convective terms you may not obtain the shedding as expected.

Try using QUICK in the discretization of convection. I think it will be enough (Solve->Controls->Solution->Momentum->QUICK).

Of course you should use the unsteady solver, preferably with second order accuracy. Your time step should also be well chosen.

For Re=70 -> Strouhal = 0.15 (approx) so:

T (period) =1/f=D/(U*0.15)=1/(0.001*0.15)=6670 s

So using dt=1E-=5 you will need 667000000 time steps per shedding cicle !!!

You should use a time step of order of T/100, i.e. around 5 sec.

Hope this help, Cheers Manuel

Re: Von Karman 2D Stable, no shedding
 

Thanks! Paul and Manual,

The help was much appreciated

Cheer!!!!!!