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¡¾Q¡¿inflow turbulence B.C. of k-epsilon model? |
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October 9, 2003, 23:45 |
¡¾Q¡¿inflow turbulence B.C. of k-epsilon model?
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#1 |
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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<u>/dz (why???) so: epsilon=Cu^0.5*k(z)*d<u>/dz My question is: where the expression of epsilon comes from? whether this turbulence condition means turbulence has been in equilibrium ? thanks. |
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October 10, 2003, 09:04 |
Re: ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model
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#2 |
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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. |
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October 10, 2003, 11:54 |
Re: ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model
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#3 |
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Thank you,Sylvain.
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