[tstorm]

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!

[srjp]

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

[tstorm]

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!

[t.krishnamohan]

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.