RANS vs LES

RANS&LES

RANS

雷诺应力

τij=ρˉuiuj~\tau_{ij}=-\bar\rho\widetilde{u_i'u_j'}

湍动能

k=12uiui~=12τii/ρˉk=\frac{1}{2}\widetilde{u_i'u_i'}=-\frac{1}{2}\tau_{ii}/\bar\rho

Boussinesq 假设

τij13τkkδij=2μt(Sij~13Skk~δij)\tau_{ij}-\frac{1}{3}\tau_{kk}\delta_{ij}=2\mu_t(\widetilde{S_{ij}}-\frac{1}{3}\widetilde{S_{kk}}\delta_{ij})

LES

亚格子应力

τij=ρˉ(uiuj~ui~uj~)\tau_{ij}=\bar\rho(\widetilde{u_iu_j}-\widetilde{u_i}\widetilde{u_j})

亚格子湍动能

k=12(uiuj~ui~uj~)=12τiiρˉk=\frac{1}{2}(\widetilde{u_iu_j}-\widetilde{u_i}\widetilde{u_j})=\frac{1}{2}\frac{\tau_{ii}}{\bar\rho}

Boussinesq 假设

τij13τkkδij=2μt(Sij~13Skk~δij)\tau_{ij}-\frac{1}{3}\tau_{kk}\delta_{ij}=-2\mu_t(\widetilde{S_{ij}}-\frac{1}{3}\widetilde{S_{kk}}\delta_{ij})

RANS 和 LES 的联系:

首先有:脉动的平均 =0 ???

并且有:

uiuj~=(ui~+ui)(uj~+uj)~=ui~uj~+ui~uj+uiuj~+uiuj~=ui~uj~+uiuj~\widetilde{u_iu_j}=\widetilde{(\widetilde{u_i}+u_i')(\widetilde{u_j}+u_j')}= \widetilde{ \widetilde{u_i}\widetilde{u_j}+\widetilde{u_i}u_j'+u_i'\widetilde{u_j}+u_i'u_j' }= \widetilde{u_i}\widetilde{u_j}+\widetilde{u_i'u_j'}

于是得到:

τij=ρˉuiuj~=ρˉ(uiuj~ui~uj~)=τsgs\tau_{ij}=-\bar\rho\widetilde{u_i'u_j'}=-\bar\rho(\widetilde{u_iu_j}-\widetilde{u_i}\widetilde{u_j})=-\tau_{sgs}

Author: Yan Zhang
Link: https://openfoam.top/RANS_LES/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.
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