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Old   November 5, 2001, 15:03
Default k-e values
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Hi, there,

I am a new beginner working with k-e equations for turbulent flows. I am using low-reynolds version k-e equations with simple eddy viscosity method.

I have got a converged solution. But I found that the value of the dispation rate (e) is three order magnitude higher than the value of turbulent kinatic energy (k) generally. Is this reasonable result? I have no idea of this.

In general, is there any relationship between these two values?

I would be very grateful if any experienced one can tell me this. It is really hard to find such informaiton in the text book.

Thank you very much.

Best regards
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Old   November 7, 2001, 06:48
Default Re: k-e values
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for y+ (y+ = y*u_f/nu) in [30:300], you can make the assumption that :

k = (u_f)^2/sqrt(c_\nu)

eps = (u_f)^3/(Kappa*y)

so k/eps which has to do whith the characteristic turbulent time scale is propto (Kappa*y)/(sqrt(c_nu)*uf)

which could be very small.

hope that help,


PS : Kappa = 0.41 ; c_nu = 0.09 ; nu = 1.5e-5 for air in mks ; u_f is the skin friction velocity.

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