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Old   December 18, 2000, 05:20
Default Re: Turbulent length-scale
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I believe that it could be a physical meaning for the Taylor microscale.

If you take a look to my previous mail in this thread, you will see that, in the case of decaying turbulence, what I call K_tra is on the same ratio with the Kolmogorov wave-length (K_eta) than the inverse of the Taylor Microscale. So the inverse of this wave-length and the Taylor microscale could be see as the same thing. This wave length corresponds to the point where the transfert term is equal to zero. Since the integral of the transfert term with respect to the wave length is equal to zero and due to the shape of this transfert term, this point has a physical meaning : the energy flux of the Kolmogorov cascade goes from the scales greater than the Taylor microscale to the scales smaller than the Taylor microscale.

Does anybody could confirm this feeling ?
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Old   December 18, 2000, 07:20
Default Re: Turbulent length-scale
Ghabi Chekib
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Old   December 19, 2000, 07:24
Default Re: Turbulent length-scale
B. Anders Pettersson Reif
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Some researchers think of the Taylor microscale as the scale at which maximum dissipation occurs. The rate of viscous dissipation increases with wavenumber through the inertial subrange before it eventually decreases as the wavenumber gets large enough. At very high Re, the Taylor microscale is approximately equal to the Kolmogorov scale.
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Old   January 6, 2011, 22:28
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Originally Posted by Ghanshyam Singh
...... I am sorry to disagree with Lars, who says that the "most of the dissipation occurs" at the Taylors micro scale.
Singh is King! You got that right!
I hope all you guys, who participated in this thread a decade back, are doing well...


Last edited by pban92; January 6, 2011 at 23:43.
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