- #1
Joshua Pham
- 14
- 0
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?
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?