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How do we calculate the normal Reynolds stresses? (Linear eddy viscosity models)

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August 18, 2017, 12:19	How do we calculate the normal Reynolds stresses? (Linear eddy viscosity models)	#1
userid42 New Member Daniel Join Date: Aug 2017 Posts: 7 Rep Power: 8	I have read in multiple places that the normal Reynolds stresses are equal when the linear eddy viscosity model is used. If they are equal, per the definition of turbulent kinetic energy: $k = \frac{1}{2} (\overline{u'u'} + \overline{v'v'} + \overline{w'w'})$ $\overline{u'u'} = \frac{2}{3}k$ . From what I understand, we can calculate the Reynolds stresses from the Boussinesq approximation. In this approximation, the eddy viscosity is assumed to be isotropic (not turbulence). The Boussinesq approximation is: $-\rho\overline{u^{'}_{i}u^{'}_{j}}=\mu_t\left(\frac{\partial\overline{u_i}}{\partial{x_j}}+\frac{\partial\overline{u_j}}{\partial{x_i}}-\frac{2}{3}\frac{\partial\overline{u_k}}{\partial{x_k}}\delta_{ij}\right)-\frac{2}{3}\rho k\delta_{ij}$ . From this equation, the normal stresses (for incompressible flows) can be calculated by: $\overline{u^{'}u^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{u}}{\partial{x}}+\frac{\partial\overline{u}}{\partial{x}}\right),$ $\overline{v^{'}v^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{v}}{\partial{y}}+\frac{\partial\overline{v}}{\partial{y}}\right),$ and $\overline{w^{'}w^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{w}}{\partial{z}}+\frac{\partial\overline{w}}{\partial{z}}\right)$ . From the expressions above, I see that the normal stresses are not equal because the velocity gradients will be different. Also, this approach introduces an extra term " $-\frac{\mu_t}{\rho}(...)$ ". Which approach is correct? Am I missing something? Or is this a misconception of what is isotropic? Feel free to correct me if I stated something wrong. Thanks! Ashkan Kashani likes this. Last edited by userid42; August 18, 2017 at 21:07.

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