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May 22, 2006, 09:34 
length scales in turbulence

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May 22, 2006, 11:01 
Re: length scales in turbulence

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Read the introduction of tuebulence.


May 22, 2006, 11:24 
Re: length scales in turbulence

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What *causes* turbulence? What *sustains* turbulence? What *aggrivates* turbulence?
What *is* turbulence? Answer those questions & you may well have answered the original lengthscale question... diaw... (Des Aubery) 

May 22, 2006, 11:47 
Re: length scales in turbulence

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thanks friend's but can u suggest where to read introduction by the way my question is what is the advantage of using length scales cant we proceed or describe turbulence phenomenon without mentioning scales .


May 22, 2006, 13:36 
Re: length scales in turbulence

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May 23, 2006, 01:24 
Re: length scales in turbulence

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Turbulence is characterised by "eddies" of various sizes, ranging from large, determined by the physical boundaries of the flow, to very small, determined by the viscocity of the fluid. The kinetic energy of the turbulence travels from the large eddies pretty much conserved until it reaches the smallest eddies where it is dissipated by viscous effects.
Under equilibrium, the turbulent kinetic energy produced is equal to that dissipated. So, if we can estimate the scale of te large energycontainig eddies, we can use that scale (under equilibrium) to estimate how much is dissipated. I think one reason why one gets an overestimated dissipation (length scale) is that there is really a lag between energy showing up at the larger scales and then dissapearing (to heat) at the small scales. So, if you use the "lengthscale" (big) to estimate the dissipation (small) you get an overestimate. I hope my rambling is not confusing. 

May 23, 2006, 05:25 
Re: length scales in turbulence

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What *causes* the eddies to form in the first place?


May 23, 2006, 07:08 
Re: length scales in turbulence

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The exact production term in k equation is uu.dU/dX. It's show that turbulence grows under the effect of shear stress.
If the question is : why uu is not always equal to zero  which is a solution of Reynolds Average Navier Stokes Equation  I am afraid that no one can answer. ZubenUbi 

May 23, 2006, 07:44 
Re: length scales in turbulence

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To complet Pennysworth's message :
kEps, as any other 1 point model, is build over the hypothesis of turbulence under equilibrium. In fact, at the beginning, turbulence energy is produce at large scale by the velocity gradients. Then this energy is transfer to small scall, throught what is called the Energy cascade. After that, the energy is dissipated by small scales. 1 point models used to consider that the amount of transfered energy by time unit and the amount dissipated energy by time unit are equals  which happen at equilibrium. This is obvious if we consider the production term in the Eps equation : it is proportional to the k production term. From another point of vue, that's mean that the dissipation in 1 point model is not the amount of energy dissipated by time unit, but is the amount of energy transfered by time unit. In shear region, the amount of energy transfered by time unit overestimates the dissipation, in shearless region, it underestimates the dissipation. 3 equations "k_Eps_Phi" models have been developped to correct this, where Phi could be a time scale build over k/Eps and dUdX. Multiscales models also correct this. As Pennysworth, hope this is not to confusing, ZubenUbi 

May 24, 2006, 02:19 
Re: length scales in turbulence

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Thanks Pennsworth ,ZubenUbi , Jonas , diaw and Q thanks for u r participation in making this clear to me ...regards bajjal


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