Inviscid flows and the turbulent (eddy) viscosity

[Joshua Pham]

Hello,
After Favre averaging the momentum equation for an inviscid flow, the following can be obtained:

$$\frac{\partial}{\partial t} \left(\overline{\rho}\tilde{u}_i \right) + \frac{\partial}{\partial x_j}\left( \overline{\rho}\tilde{u}_i \tilde{u}_j \right) + \frac{\partial \overline{p}}{\partial x_i} = \frac{\partial}{\partial x_j} \left(-\overline{\rho u_i'' u_j''}\right)$$

The term on the right hand side is the partial derivative of the Reynolds stress tensor with respect to $x$.

By the Eddy viscosity concept,

$$\overline{\rho} \overline{u_i'' u_j''} = \mu_T \left( \frac{\partial \tilde{u}_i}{\partial x_j} + \frac{\partial \tilde{u}_j}{\partial x_i} \right) - \frac{2}{3} \left( \mu_T \frac{\partial \tilde{u}_i}{\partial x_j} + \overline{\rho} k \right) \delta_{ij}$$

For an inviscid flow do the $\mu_T$ and $k$'s go to 0?

 

[olivermsun]

Three relevant questions:
1) Where does the Reynolds stress come from in the equations of motion?
2) Which term in the Navier-Stokes equation is zero in the inviscid case?
3) What are $\mu_T$ and $k$, and do they go to zero in case 2)?