# ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model?

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 October 9, 2003, 22:45 ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model? #1 David Guest   Posts: n/a I see some referrences give the inflow tubulence boundary condition as follows: k=const. epsilon=Cu^0.75*k(z)^1.5/l (this is the most common style gives epsilon) l£½(Cu^0.5*k(z))^0.5/d/dz (why???) so: epsilon=Cu^0.5*k(z)*d/dz My question is: where the expression of epsilon comes from? whether this turbulence condition means turbulence has been in equilibrium ? thanks.

 October 10, 2003, 08:04 Re: ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model #2 sylvain Guest   Posts: n/a In the log law region : k=uf^2/Cu^0.5 and k equilibrium means Production of k = Dissipation of k, so epsilon = Cu k^2/epsilon * (du/dy)^2 with du/dy = uf/Kappa.y = Cu^0.25 k^0.5 / Kappa.y it comes epsilon = Cu^0.75 k^1.5 / l , with l = Kappa.y This is a CONSEQUENCE of the k-eps model.

 October 10, 2003, 10:54 Re: ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model #3 David Guest   Posts: n/a Thank you,Sylvain.

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