Boussinesq Assumption
Hi all,
I've read that the Boussinesq assumption linearized the relationship between the turbulent stresses and the mean strain rate, so: u_i*u_j=2/3*k*del_ij - 2*nu_t*S_ij I've also seen people say that the Boussinesq assumption makes the normal stresses isotropic, so: uu=vv=ww=2/3*k My question is where does the second term go? Does the Boussinesq assumption also assume that du/dx = dv/dy = dw/dz = 0? Thanks! |
When the normal stresses are made isotropic, the first term vanishes and you have the linearized equation for the turbulent shear stress (second term)
as: tau_xy = mu_t * du/dy where mu_t is the turbulent viscosity |
Thanks, but why does the first term vanish? It should vanish if the Kronecker delta is zero, which happens for the off-diagonal terms, not the diagonal terms.
I'm clearly still confused. I appreciate your help! |
Hi,
Expression for Tou ij =2*mue_t*Sij-2/3*k*rho*del_ij; Sij=2(dui/dxj+duj/dxi)----------Terms here are time averaged. When i=j; Sii=0 for incompressible flows. Recollect the time averaged continuity equation. |
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