# timestep for time dependent RANS

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 April 28, 2005, 04:46 timestep for time dependent RANS #1 fab Guest   Posts: n/a 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

 April 28, 2005, 07:22 Re: timestep for time dependent RANS #2 Tom Guest   Posts: n/a 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.

 April 29, 2005, 01:41 Re: timestep for time dependent RANS #3 fab Guest   Posts: n/a 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

 April 29, 2005, 03:18 Re: timestep for time dependent RANS #4 Halim Choi Guest   Posts: n/a 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.

 May 2, 2005, 01:29 Re: timestep for time dependent RANS #5 fab Guest   Posts: n/a Hi, thanks! I will try to get a copy. Greetings! fab

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