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Jack November 5, 2001 15:03

k-e values
 
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

sylvain November 7, 2001 06:48

Re: k-e values
 
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,

Sylvain

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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