# RNG k-epsilon model

(Difference between revisions)
 Revision as of 13:13, 14 September 2005 (view source)Zxaar (Talk | contribs)← Older edit Revision as of 07:46, 4 October 2006 (view source)Newer edit → Line 55: Line 55: C_{1\epsilon} = 1.42, \; \; C_{2\epsilon} = 1.68 C_{1\epsilon} = 1.42, \; \; C_{2\epsilon} = 1.68 [/itex] [/itex] + + [[Category:Turbulence models]]

## RNG k-epsilon model

Transport equations for k and $\epsilon$ are

$\frac{\partial}{\partial t} (\rho k) + \frac{\partial}{\partial x_i} (\rho k u_i) = \frac{\partial}{\partial x_j} \left(\alpha_k \mu_{\rm eff} \frac{\partial k}{\partial x_j}\right) + P_k + P_b - \rho \epsilon$
$\frac{\partial}{\partial t} (\rho \epsilon) + \frac{\partial}{\partial x_i} (\rho \epsilon u_i) = \frac{\partial}{\partial x_j} \left(\alpha_{\epsilon} \mu_{\rm eff} \frac{\partial \epsilon}{\partial x_j}\right) + C_{1 \epsilon}\frac{\epsilon}{k} \left( G_k + C_{3 \epsilon} G_b \right) - C_{2\epsilon} \rho \frac{\epsilon^2}{k} - R_{\epsilon}$

## Turbulent viscosity is modelled as

$d \left(\frac{\rho^2 k}{\sqrt{\epsilon \mu}} \right) = 1.72 \frac{\hat{\nu}}{\sqrt{{\hat{\nu}}^3-1+C_\nu}} d{\hat{\nu}}$

$\hat{\nu} = \mu_{\rm eff}/\mu$

and

$C_\nu \approx 100$

## Term $R_{\epsilon}$

The term $R_{\epsilon}$ is modelled as

$R_{\epsilon} = \frac{C_\mu \rho \eta^3 (1-\eta/\eta_0)}{1+\beta\eta^3} \frac{\epsilon^2}{k}$
$\eta \equiv Sk/\epsilon$
$\eta_0 = 4.38$
$\beta = 0.012$

The transport equation for $\epsilon$ can be re-written as:

$\frac{\partial}{\partial t} (\rho \epsilon) + \frac{\partial}{\partial x_i} (\rho \epsilon u_i) = \frac{\partial}{\partial x_j} \left(\alpha_{\epsilon} \mu_{\rm eff} \frac{\partial \epsilon}{\partial x_j}\right) + C_{1 \epsilon}\frac{\epsilon}{k} \left( G_k + C_{3 \epsilon} G_b \right) - C_{2\epsilon}^* \rho \frac{\epsilon^2}{k}$
$C_{2\epsilon}^* \equiv C_{2\epsilon} + {C_\mu \eta^3 (1-\eta/\eta_0)\over 1+\beta\eta^3}$

$C_{1\epsilon} = 1.42, \; \; C_{2\epsilon} = 1.68$