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Old   July 28, 2008, 17:06
Default how can I determine the vortex shedding time step
Zhe Liu
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I am using the CFX to simulate the vortex shedding based on the unsteady solver, but I meet some problems,

1 you know St=fB/U, f is the vortex shedding frequency, but in unsteady simulation, we should know the time step, how can I determine it, I make sure it based on the min grid, for example, if the min grid is known, the time step= min grid/velocity, is that correct, but I read some document they did not consider that, so is there any guy to give me some suggestions about the vortex shedding time step, thanks
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Old   July 28, 2008, 18:11
Default Re: how can I determine the vortex shedding time s
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You need to show your results depend on the physics only, and do not depend on the computational timestep you use.

So, choose a sensible timestep, then run the simulation. Then divide your timestep by two, and re-run the simulation using this smaller timestep. Compare the results. If the results agree, then the timestep is adequate. If the results do not agree, then you need to investigate why, and explore whether an even smaller timestep is neccessary....

I have not done much work on this myself, but colleagues have told me they successfully modelled this using 50 timesteps to cover one complete oscillation. Ie if T = 1/Fb = St/U, then they used dt = T/50. That's only a rough suggestion - you should experiment yourself!

I hope that gives you some ideas. But the main idea is you must perform some experiments until you are convinced you have enough resolution to make your results independent of the timestep.

Good luck. andy
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Old   July 29, 2008, 21:25
Default Re: how can I determine the vortex shedding time s
Zhe Liu
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hi andy

firstly thanks your help, but i think there is something wrong, you said, T=1/Fb=St/U, but the strouhal number St=F*B/U, so i think the equation is T=1/F=1/(St*U/B), is that right,thanks
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Old   July 30, 2008, 17:16
Default Re: how can I determine the vortex shedding time s
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you're right! sorry - i typed it all very fast and made a silly error... i think the idea i described is still generally sensible though.

good luck. andy
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