Realisable k-epsilon model

Transport Equations

$\frac{\partial}{\partial t} (\rho k) + \frac{\partial}{\partial x_j} (\rho k u_j) = \frac{\partial}{\partial x_j} \left [ \left(\mu + \frac{\mu_t}{\sigma_k}\right) \frac{\partial k} {\partial x_j} \right ] + P_k + P_b - \rho \epsilon - Y_M + S_k$

$\frac{\partial}{\partial t} (\rho \epsilon) + \frac{\partial}{\partial x_j} (\rho \epsilon u_j) = \frac{\partial}{\partial x_j} \left[ \left(\mu + \frac{\mu_t}{\sigma_{\epsilon}}\right) \frac{\partial \epsilon}{\partial x_j} \right ] + \rho \, C_1 S \epsilon - \rho \, C_2 \frac{{\epsilon}^2} {k + \sqrt{\nu \epsilon}} + C_{1 \epsilon}\frac{\epsilon}{k} C_{3 \epsilon} P_b + S_{\epsilon}$

Where

$C_1 = \max\left[0.43, \frac{\eta}{\eta + 5}\right] , \;\;\;\;\; \eta = S \frac{k}{\epsilon}, \;\;\;\;\; S =\sqrt{2 S_{ij} S_{ij}}$

In these equations, $P_k$ represents the generation of turbulence kinetic energy due to the mean velocity gradients, calculated in same manner as standard k-epsilon model. $P_b$ is the generation of turbulence kinetic energy due to buoyancy, calculated in same way as standard k-epsilon model.

Modelling Turbulent Viscosity

$\mu_t = \rho C_{\mu} \frac{k^2}{\epsilon}$

where
$C_{\mu} = \frac{1}{A_0 + A_s \frac{k U^*}{\epsilon}}$
$U^* \equiv \sqrt{S_{ij} S_{ij} + \tilde{\Omega}_{ij} \tilde{\Omega}_{ij}}$  ;
$\tilde{\Omega}_{ij} = \Omega_{ij} - 2 \epsilon_{ijk} \omega_k$ ;
$\Omega_{ij} = \overline{\Omega_{ij}} - \epsilon_{ijk} \omega_k$

where $\overline{\Omega_{ij}}$ is the mean rate-of-rotation tensor viewed in a rotating reference frame with the angular velocity $\omega_k$. The model constants $A_0$ and $A_s$ are given by:
$A_0 = 4.04, \; \; A_s = \sqrt{6} \cos \phi$

$\phi = \frac{1}{3} \cos^{-1} (\sqrt{6} W), \; \; W = \frac{S_{ij} S_{jk} S_{ki}}{{\tilde{S}} ^3}, \; \; \tilde{S} = \sqrt{S_{ij} S_{ij}}, \; \; S_{ij} = \frac{1}{2}\left(\frac{\partial u_j}{\partial x_i} + \frac{\partial u_i}{\partial x_j} \right)$

Model Constants

$C_{1 \epsilon} = 1.44, \;\; C_2 = 1.9, \;\; \sigma_k = 1.0, \;\; \sigma_{\epsilon} = 1.2$