Reynolds-Averaged Navier–Stokes (RANS) modelling makes turbulent flow tractable by ensemble-averaging the governing equations. Averaging introduces an unknown Reynolds stress, which the Boussinesq hypothesis collapses into a single turbulent viscosity — evaluated by a turbulence transport model.
1. Incompressible Navier–Stokes Equations
$$ \frac{\partial u_i}{\partial t} + u_j \frac{\partial u_i}{\partial x_j} = \frac{\partial \sigma_{ij}}{\partial x_j} \\[1.0em] \sigma_{ij} = -p\delta_{ij} + 2\nu S_{ij} \\[1.0em] S_{ij} = \frac{1}{2}\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) $$
2. Reynolds Decomposition
$$ \phi = \bar{\phi} + \phi', \qquad \phi \in \{u, v, w, p\} $$
3. Reynolds-Averaged Navier–Stokes Equations
$$ \frac{\partial \bar{u}i}{\partial t} + \bar{u}j \frac{\partial \bar{u}i}{\partial x_j} = \frac{\partial}{\partial x_j}\left(\bar{\sigma}{ij} - \overline{u_i' u_j'}\right) \\[1.0em] \bar{\sigma}{ij} = -\bar{p}\delta{ij} + 2\nu \bar{S}{ij} \\[1.0em] \bar{S}{ij} = \frac{1}{2}\left(\frac{\partial \bar{u}_i}{\partial x_j} + \frac{\partial \bar{u}_j}{\partial x_i}\right) $$
Boussinesq Hypothesis
6 unknown stress components $\overline{u_i' u_j'}$ collapse to 1 unknown turbulent viscosity $\nu_t$:
$$ -\overline{u_i' u_j'} = \underbrace{-\tfrac{2}{3} k \delta_{ij}}{\text{absorbed into } \bar{p}} + \underbrace{2\nu_t \bar{S}{ij}}_{\text{new unknown } \nu_t} $$
4. Turbulence Transport Equations
Transport a turbulence quantity $\psi$ to evaluate the turbulent viscosity $\nu_t$:
$$ \underbrace{\frac{\partial \psi}{\partial t}}{\text{temporal}} + \underbrace{\bar{u}j \frac{\partial \psi}{\partial x_j}}{\text{convection}} = \underbrace{\frac{\partial}{\partial x_j}\left[\Gamma\psi \frac{\partial \psi}{\partial x_j}\right]}{\text{diffusion}} + \underbrace{P\psi}{\text{production}} - \underbrace{\varepsilon\psi}_{\text{destruction}} $$
| Model | $\psi$ | $\nu_t$ |
|---|---|---|
| Mixing Length | – | $\nu_t = f(\lvert S \rvert)$ |
| Spalart–Allmaras | $\tilde{\nu}$ | $\nu_t = f(\tilde{\nu})$ |
| $k\text{-}\varepsilon$ | $k, \varepsilon$ | $\nu_t = f(k, \varepsilon)$ |
| $k\text{-}\omega$ | $k, \omega$ | $\nu_t = f(k, \omega)$ |