# URANS: questions

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 April 9, 2005, 15:11 URANS: questions #1 FELIPE Guest   Posts: n/a I've use k-e model for calculation of flow in a 2 - D rectangular cavity with differentially heated walls and it gives very good results comparing to the reference bibliography solutions. I have the questions: I'll obtain unsteady solutions for the problems of rectangular cavity with boundary conditions dependents of time, if I use the URANS model?

 April 10, 2005, 22:09 Re: URANS: questions #2 Mani Guest   Posts: n/a yes.

 April 11, 2005, 11:15 Re: URANS: questions #3 FELIPE Guest   Posts: n/a Hi Mani Tanks a lot. What is the difference in RANS and URANS methods?

 April 11, 2005, 11:42 Re: URANS: questions #4 Harry Fulmer Guest   Posts: n/a URANS should resolve periodic flow features, RANS will assume they are time averaged out with all the other aperiodic turbulent time varying fluctuations. That's my take on it anyway.

 April 11, 2005, 12:41 Re: URANS: questions #5 Mani Guest   Posts: n/a That's correct. With the URANS equations you should be able to resolve unsteadiness of certain time scales. You will, in general, not use a resolution fine enough to predict turbulent length- and time-scales (that's why Reynolds-Averaged). However, you'll be able to resolve larger scales, depending on your time-step and grid resolution. You should certainly be able to see unsteady behavior in case of your unsteady boundary conditions. The unsteadiness does not have to be periodic, if you are using a time-marching method. Mathematically, the URANS equations include the time derivatives of the Reynolds-averaged flow variables. These terms are not included in the steady RANS equations. Maybe somebody can comment on the use of the k-eps turbulence model in URANS. Is it fully unsteady or quasi-steady?

 April 11, 2005, 14:47 Re: URANS: questions #6 jasond Guest   Posts: n/a When talking about URANS, it is perhaps better to think of the equations as being ensemble-averaged rather than time-averaged. Then a time-dependent bc is not inconsistent in any way. Any unsteadiness must necessarily be interpreted as large-scale unsteadiness. Generally speaking, most of the 1-eqn and 2-eqn models include a time derivative term, so initial conditions are required just as for the governing equations.

 April 11, 2005, 14:59 Re: URANS: questions #7 FELIPE Guest   Posts: n/a Tanks you I READ THE BOOK OF TUCKER 2001 (COMPUTATION OF UNSTEADY INTERNAL FLOWS) THE ENSAMBRE AVERGE FOR THE NAVIER STOKES EQUATIONS, THE AUTOR PRESENT THE EQUATIONS. I THINK THE EQUATIONS FOR THE TIME AVERAGE ARE EQUAL FOR EMSAMBLE AVERAGE, OR NO???

 April 11, 2005, 16:01 Re: URANS: questions #8 Guillaume Guest   Posts: n/a No. (unless ergodicity ~ steadyness)

 April 11, 2005, 17:13 Re: URANS: questions #9 jasond Guest   Posts: n/a I would add that while the equations look the same, they are not unless you involve ergodicity. They are not the same because the interpretation of the various varables are very different. I suppose it may look like semantics, but it isn't.

 April 29, 2005, 13:08 earth #10 stephanie Guest   Posts: n/a What is the distance from the sun

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