# ? fluctuating equations for homogenous shear turb.

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 September 19, 2002, 19:24 ? fluctuating equations for homogenous shear turb. #1 ff_fan Guest   Posts: n/a This is a question for those familiar with plane homogeneous shear turbulence. I have a reference [Rogallo (1981) 'Experiments in homogeneous turbulence'] that gives the governing equations for just the fluctuating part of the flow. To get these, first Rogallo decomposes the velocity and pressure into mean and fluctuating components and substitutes this into the original governing equations. In the new equation, the differentiation of the mean components is with respect to an orthogonal coordinate system but the differentiation of the fluctuating components is with respect to a coordinate system that moves with the mean flow. [The point of this is to be able to apply periodic boundary conditions to the fluctuating components]. The next step he does is what I'm confused on. He states: "To separate the mean from the fluctuating parts we take the ensemble average of the Navier-Stokes equations, using the ergodic hypothesis to replace ensemble averages of spatially homogeneous terms by their volume averages" In his equations of the mean flow, there are no Reynolds stresses. Correct me if I'm wrong, but doesn't the ergodic hypothesis state that we should get the same result for ensemble / time / space averages? I just don't understand when trying to follow his work how he gets rid of the averaged '' terms his mean equations. I'm probably missing something obvious, but after working on it for hours I'm ready to see the 'light'. Feel free to e-mail me at ff_fanX@hotmail.com if you would wish to discuss this outside the forum. I'd be happy to explain my problem in more detail if it isn't clearly stated enough. Thanks.

 September 20, 2002, 07:39 Re: ? fluctuating equations for homogenous shear t #2 Tom Guest   Posts: n/a The volume averaged Reynolds stresses are not zero, but the gradients of the mean Reynolds stresses are zero because the flow is homogeneous. By the way, time and volume averaged Reynolds stresses are not the same in this case because the mean flow is a function of time. Tom

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