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timestep for time dependent RANS

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Old   April 28, 2005, 04:46
Default timestep for time dependent RANS
  #1
fab
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Hi,

I wonder if there is a size limit of the smallest time step for unsteady RANS models. I read that the averaging time T should be T>> Tf , where the turbulent fluctuations have the time scale Tf.

Where is the problem setting T=Tf?

Maybe, someone has an explanation.

Regards! fab
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Old   April 28, 2005, 07:22
Default Re: timestep for time dependent RANS
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Tom
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The problem with T=O(Tf) is that in performing the Reynolds averaging you have "averaged out" these time scales. Furthermore in oder to justify retaining the d/dt term in the averaged equations there must be scale separation (i.e. T >> Tf ). In unsteady RANS the unsteadiness should be viewed as a slow modulation of the more rapid turbulent fluctuations.
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Old   April 29, 2005, 01:41
Default Re: timestep for time dependent RANS
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fab
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Hi,

thanks! I found for Tf: Tf= max (k/epsilon, alpha sqrt(nu/epsilon)) with alpha=6.

So should the limit for T be: T > 10*Tf ?

Regards! fab
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Old   April 29, 2005, 03:18
Default Re: timestep for time dependent RANS
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Halim Choi
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Read the following paper, G. Iaccarino,A.Ooi, P. A. Durbin, M. Behnia, "Reynolds averaged simulation of unsteady separated flow", International Journal of Heat and Fluid Flow,Volume 24, 2003, pp.147-156. Then you can get answer for your question.
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Old   May 2, 2005, 01:29
Default Re: timestep for time dependent RANS
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fab
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Hi,

thanks! I will try to get a copy.

Greetings! fab
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