Timeline for Navier-Stokes - Complete set under turbulent eddy viscosity hypothesis
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 9, 2013 at 18:58 | comment | added | Bernhard | I don't know it this far by hard, but check which terms vanish by taking the divergence. | |
| Jun 9, 2013 at 18:17 | comment | added | l3win | is it not:$-\rho \overline{u'_i u'_j} = \mu_t \, \left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i} - \frac{2}{3} \frac{\partial U_k}{\partial x_k} \delta_{ij} \right) - \frac{2}{3} \rho k \delta_{ij}$ which simplifies to $- \overline{u'_i u'_j} = \nu_t \, \left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i} \right) - \frac{2}{3} k \delta_{ij} = \nu_t \, \left( \frac{\partial U_i}{\partial x_j} + \frac{\partial U_j}{\partial x_i} \right) -\frac{1}{3} \left( \overline{u'^2} + \overline{v'^2} \right)\delta_{ij} $ | |
| Jun 9, 2013 at 18:13 | comment | added | l3win | only the term $\frac{\partial U_k}{\partial x_k}$ vanishes due to incompressibility? | |
| Jun 9, 2013 at 18:09 | comment | added | Bernhard | Your equations are incompressible, which simplifies it a bit. | |
| Jun 9, 2013 at 17:58 | comment | added | Bernhard | The Boussinesq hypothesis is no more than substituting $\nu_\tau\frac{\partial \overline{u_i}}{\partial x_j}$ for $\overline{u'_i u'_j}$ | |
| Jun 9, 2013 at 17:32 | comment | added | l3win | I understand how he gets to the conclusion of the equations on the middle of the second page. What I don't understand is how he is able to simplify them using the Boussinesq or eddy viscosity hypothesis to the last set of equations on page 2. I have updated my original post. | |
| Jun 9, 2013 at 11:41 | history | answered | Bernhard | CC BY-SA 3.0 |