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Old   November 1, 2008, 20:18
Default CFD guy
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cfdguy
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I have solved the isotropic turbulence decay in 3D but the decay does not look like the one that I saw in 2D.

Can some one explain?

Thanks in advance
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Old   November 7, 2008, 08:54
Default Re: CFD guy
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Amod
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It may be because Turbulence is inherently a 3D phenomena. Hence 3D simulation would be more realistic as compared to 2D one.

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Old   November 7, 2008, 10:44
Default Re: CFD guy
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Tom
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You need to read up on this - try, for example,

"The Theory of Homogeneous Turbulence" by G.K. Batchelor

or

"Turbulence: An Introduction for Scientists and Engineers" by P.A. Davidson.

Basically the energy and enstrophy cascades are different in 2D and 3D.

The main reason why 2D and 3D flows are different is, and is actually why it's fair to say that there is no such thing as 2D turbulence in the strictest sense, the absence of the vortex stretching term in the 2D vorticity equation.
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