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lizzy0zhou February 9, 2011 14:55

How to calculation TKE in LES
Hi Dear,

I'm a real newbie in CFDs. My questions might be very stupid::confused:

1. How to calculate the velocity fluctuations (u', v', and w') based on velocities (u, v, and w)? I know u'=u-u_bar, but what does the u_bar represent? Is that time averaged? How to calculate the u_bar?

2. What should I do if I wanna compute the tke from a LES simulation results conducted by Flow3D? Find the velocity fluctuations and then calculate the tke based on tke=0.5(u'^2+v'^2+w'^2)?


julien.decharentenay February 9, 2011 19:45

Indeed u_bar is the time-average of u at the point of interest. A couple of additional comments:

1) You need to start the time averaging after the establishment of the flow patterns;
2) You need to average for a time long enough to capture the mean flow velocity appropriately;
3) I would strongly recommend that you check the time independence of u_bar - either by selecting a different time period for the averaging (best) or doubling the length of the averaging period.

Best of luck, Julien

lizzy0zhou February 10, 2011 00:28

Thank you so much man. And I have an additional concern about the tke. Seems like the velocity fluctuation (u') would be very small (and not stable) compare to the average velocity (u_bar) when time goes to infinity. However, the tke value given by Flow3D looks big (about 7-10% of average velocity) and stable. Can you help me to understand this?

julien.decharentenay February 10, 2011 19:09

Hi Tian,

I am not sure what you are referring to as "not stable".

The tke values given by Flow3D are likely to be time average: tke = 1/2*sum_N(u'^2 + v'^2 + w^2), where N is the number of sample (time-step). There is a formulation to calculate u_bar and tke on the fly (or at least an approximation of it).

Comparing the ratio u'/u_bar or sqrt(tke)/u_bar is equivalent to comparing a unsteady and a time average quantities. The ratio values will depend on the particular case under investigation, how the time averaging was done (particularly if the time averaging was started before the establishment of the unsteady solution).

lizzy0zhou February 11, 2011 14:05

Thank you Julien!I think I have better sense about this now.:)

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