# Realizable k-ε Model

The standard k-ε turbulence model and RNG k-ε turbulence model do not satisfy mathematical constraints on the Reynolds stresses for the consistency with physics of turbulence.

These constraints include the positivity of normal stresses ( $\overline{{u}_{\alpha}^{\text{'}}{u}_{\alpha}^{\text{'}}}\ge 0$ ) and Schwarz's inequality ( $\overline{{u}_{\alpha}^{\text{'}}{u}_{\alpha}^{\text{'}}}\overline{{u}_{\beta}^{\text{'}}{u}_{\beta}^{\text{'}}}\ge \overline{{u}_{\alpha}^{\text{'}}{u}_{\beta}^{\text{'}}}\overline{{u}_{\alpha}^{\text{'}}{u}_{\beta}^{\text{'}}}$ $$ ). The realizable k-ε model proposed by Shih et al. (1995) employs a formulation of turbulent viscosity with a variation of ${C}_{\mu}$ to satisfy these constraints. This is why this model is referred to as realizable. The second improvement made with the realizable model is to have a source term in the dissipation rate transport equation, which is derived from the transport equation of the mean square vorticity fluctuation. Compared to the standard k-ε turbulence model, it performs better when simulating planar jet flows, round jet flows and flows under adverse pressure gradients. Although the realizable model has significant performance improvement over the standard model, it has a problem in predicting impinging flow, swirling flow and secondary flows in a square duct since this model is based on the Boussinesq assumption. In addition, this model needs additional functions to simulate the near wall effects.

## Transport Equations

## Production Modeling

## Dissipation Modeling

## Modeling of Turbulent Viscosity ${\mu}_{t}$

## Model Coefficients

${C}_{\epsilon 1}$ = 1.44, ${C}_{\epsilon 2}$ = 1.9, ${\sigma}_{k}$ = 1.0, ${\sigma}_{\epsilon}$ = 1.22.

where ${U}^{*}$ = $\sqrt{{S}_{ij}{S}_{ij}+\tilde{{\text{\Omega}}_{ij}}\tilde{{\text{\Omega}}_{ij}}}$ , $\tilde{{\text{\Omega}}_{ij}}={\text{\Omega}}_{ij}-2{\epsilon}_{ijk}{\omega}_{k}$ , ${\text{\Omega}}_{ij}=\overline{{\text{\Omega}}_{ij}}-2{\epsilon}_{ijk}{\omega}_{k}$ , ${A}_{0}=4.04$ , ${A}_{s}=\sqrt{6}cos\varphi $ , $\varphi =\frac{1}{3}co{s}^{-1}\left(\sqrt{6}W\right)$ , $W=\frac{{S}_{ij}{S}_{jk}{S}_{kj}}{\sqrt{{S}_{ij}{S}_{ij}}}$ ,

${C}_{1}=max\left[0.43,\frac{\eta}{\eta +5}\right]$

where $\eta =S\frac{k}{\epsilon}$ , $S=\sqrt{2{S}_{ij}{S}_{ij}}$ is the strain rate magnitude.