# Favre averaged Navier-Stokes equations

(Difference between revisions)
 Revision as of 08:09, 5 September 2005 (view source)Jola (Talk | contribs)← Older edit Revision as of 08:09, 5 September 2005 (view source)Jola (Talk | contribs) Newer edit → Line 6: Line 6: \frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x_j}\left[ \rho u_j \right] = 0 \frac{\partial}{\partial x_j}\left[ \rho u_j \right] = 0 - [/itex] + [/itex] (1) - + - $+ :[itex] \frac{\partial}{\partial t}\left( \rho u_i \right) + \frac{\partial}{\partial t}\left( \rho u_i \right) + \frac{\partial}{\partial x_j} \frac{\partial}{\partial x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right] = 0 \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right] = 0 -$ + [/itex] (2) - $+ :[itex] \frac{\partial}{\partial t}\left( \rho e_0 \right) + \frac{\partial}{\partial t}\left( \rho e_0 \right) + \frac{\partial}{\partial x_j} \frac{\partial}{\partial x_j} \left[ \rho u_j e_0 + u_j p + q_j - u_i \tau_{ij} \right] = 0 \left[ \rho u_j e_0 + u_j p + q_j - u_i \tau_{ij} \right] = 0 -$ + [/itex] (3) For a Newtonian fluid, assuming Stokes Law for mono-atomic gases, the viscous For a Newtonian fluid, assuming Stokes Law for mono-atomic gases, the viscous

## Revision as of 08:09, 5 September 2005

The instantaneous continuity equation, momentum equation and energy equation for a compressible fluid can be written as:

$\frac{\partial \rho}{\partial t} + \frac{\partial}{\partial x_j}\left[ \rho u_j \right] = 0$ (1)
$\frac{\partial}{\partial t}\left( \rho u_i \right) + \frac{\partial}{\partial x_j} \left[ \rho u_i u_j + p \delta_{ij} - \tau_{ji} \right] = 0$ (2)
$\frac{\partial}{\partial t}\left( \rho e_0 \right) + \frac{\partial}{\partial x_j} \left[ \rho u_j e_0 + u_j p + q_j - u_i \tau_{ij} \right] = 0$ (3)

For a Newtonian fluid, assuming Stokes Law for mono-atomic gases, the viscous stress is given by:

$\tau_{ij} = 2 \mu S_{ij}^*$

Where the trace-less viscous strain-rate is defined by:

$S_{ij}^* \equiv \frac{1}{2} \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right) - \frac{1}{3} \frac{\partial u_k}{\partial x_k} \delta_{ij}$

The heat-flux, $q_j$, is given by Fourier's law:

$q_j = -\lambda \frac{\partial T}{\partial x_j} \equiv -C_p \frac{\mu}{Pr} \frac{\partial T}{\partial x_j}$

Where the laminar Prandtl number $Pr$ is defined by:

$Pr \equiv \frac{C_p \mu}{\lambda}$

To close these equations it is also necessary to specify an equation of state. Assuming a calorically perfect gas the following relations are valid:

$\gamma \equiv \frac{C_p}{C_v} ~~,~~ p = \rho R T ~~,~~ e = C_v T ~~,~~ C_p - C_v = R$

Where $\gamma, C_p, C_v$ and $R$ are constant.

The total energy $e_0$ is defined by:

$e_0 \equiv e + \frac{u_k u_k}{2}$

Note that the corresponding expression~\ref{eq:fav_total_energy} for Favre averaged turbulent flows contains an extra term related to the turbulent energy.

$\frac{\partial \overline{\rho}}{\partial t} + \frac{\partial}{\partial x_i}\left[ \overline{\rho} \widetilde{u_i} \right] = 0$

$\frac{\partial}{\partial t}\left( \overline{\rho} \widetilde{u_i} \right) + \frac{\partial}{\partial x_j} \left[ \overline{\rho} \widetilde{u_j} \widetilde{u_i} + \overline{p} \delta_{ij} - \widetilde{\tau_{ji}^{tot}} \right] = 0$