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Why only the anisotropic part of the Reynolds-stress tensor affects production. |
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#1 |
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Js
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Why only the anisotropic part of the Reynolds-stress tensor affects production?
PS: I am learning turbulence. I am confused why only the anisotropic part of Reynolds stress tenosr will affect production of TKE (for incompressible flow) Last edited by wertyfg; October 3, 2016 at 15:20. |
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#2 |
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Filippo Maria Denaro
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#3 |
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Js
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Yes I am talking bout production of TKE in incompressible flow
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#4 |
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Filippo Maria Denaro
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I give you a more general answer that you can find from the original NS equations :
A stress tensor is decomposed in isotropic and deviatoric parts. The isotropic part acts by means of the action of only normal component (only the entries in the main diagonal are not zero). That resembles the decomposition used in the momentum equation: T = -pI + Td Thus, when you consider the averaging of the NS, after the decompostion of the Reynolds stress the isotropic part can be recasted like a modified pressure term p'. Now, considering the kinetic energy equation, the work of the normal stress p' can be effective if you have a volume variation, measured by div v, which is zero for incompressible flow. You can organize the energy equation in a way to highlight production, dissipation and time variation as illustrated in many fundamental textbooks (Pope, Wilcox, Sagaut). |
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#5 | |
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Js
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