A "little" non-isotropic turbulence generation

lumley_triangle_production.png

As we discussed past week, at the moment the method is able to reproduce all the Reynolds tensor where $ \lambda_i - 2 \max{\lambda_i} > 0 $.This conditions is spotted in the picture though a Lumley triangle. The dots are the result of a DNS channel flow (Kim et alt. 1999). We can suddenly see that me majority part of a real turbulence cannot be captured by the DF-SEM. Is this a limit? Is this not a limit?

Bousinnesq tensor

   \begin{equation*}  \tau_{ij}=-\overline{\rho}\overline{u_{i}'u_{j}'}=\mu_{t}\left(\cfrac{\partial\overline{u}_{i}}{\partial x_{j}}+\frac{\partial\overline{u}_{j}}{\partial x_{i}}\right)-\cfrac{2}{3}\rho k\delta_{ij}  \end{equation*}  (1)

The Bousinneq tensor, used in all 2 equations models, is highly isotropic (because of the second term in equation (1). Their results are usually contained in the part of the Lumley triangle which we are able to reproduce then. Considering that, in a general simulation, there will be available only the RANS results, at the moment the method seems quite appropiate.


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Topic revision: r2 - 2010-07-12 - 10:54:30 - RuggeroPoletto
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25 Sep 2018

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