Boussinesq Assumption

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 October 19, 2009, 12:50 Boussinesq Assumption #1 Member   Join Date: Apr 2009 Posts: 78 Rep Power: 9 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!

 October 22, 2009, 08:57 #2 Member   JP Join Date: Mar 2009 Posts: 57 Rep Power: 10 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

 October 22, 2009, 10:12 #3 Member   Join Date: Apr 2009 Posts: 78 Rep Power: 9 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!

 October 26, 2009, 06:11 #4 Member   Krishna Join Date: Oct 2009 Posts: 34 Rep Power: 8 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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