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

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Old   October 9, 2003, 22:45
Default ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model?
  #1
David
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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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Old   October 10, 2003, 08:04
Default Re: ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model
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sylvain
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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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Old   October 10, 2003, 10:54
Default Re: ¡¾Q¡¿inflow turbulence B.C. of k-epsilon model
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David
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Thank you,Sylvain.
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