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 CFDtoy May 3, 2006 11:00

Derivation of k-e model from momentum equation

Hello:

Could someone direct me to some notes/ books on k-e 2 equation turbulence model derivation from momentum equation having a "source term".?

I am not sure if I add a source to momentum, I should be adding a source to k-e equations or is it taken into the governing equations by themselves?

CFDtoy

 Tian_FB May 3, 2006 23:12

Re: Derivation of k-e model from momentum equation

The book by Pope is Ok for u

 Mani May 4, 2006 13:02

Re: Derivation of k-e model from momentum equation

It depends on the nature of the source term, or rather: on the physical process your source term is supposed to describe. Is it an unsteady or steady term? If unsteady: are the associated time scales close to turbulent time scales or not? If you can consider your source term as a time average (in the sense of the RANS equations), then you may be able to just add it to the time-averaged momentum equation. However, if your source term is coupled to other terms, such that the Reynolds averaging process produces new terms, you'll have to go through the derivation to see which terms actually show up in the RA momentum and which (if any) would show up in the turbulence model.

 CFDtoy May 9, 2006 19:23

Re: Derivation of k-e model from momentum equation

I have some Scalar Eq:

d(alpha)/dt + Del. (alpha*U) = source(alpha)

Momentum Equation: incompressible

d(U)/dt+ Del. (UU) = -Del(p) + Source (say Alpha*U)

Now, I thought since k-epsilon are derived based on momentum eqn, if I update the momentum and hence U , k-epsilon would adjust itself to the new momentum source?

Do I need to add an explicit source term due to this (alpha) scalar ?

Regarding the time scales, I can assume alpha source acts at the same time scale as the turb kinetic energy etc...

Thanks,

CFDtoy

 Jeff Moder May 10, 2006 12:51

Re: Derivation of k-e model from momentum equation

Just to make things a little more exact, let me write your momentum equation with density included:

rho*[d(U)/dt+ Del. (UU)] = -Del(p) + vec_S

where vec_S = (Sx, Sy, Sz) = source term vector

If you are doing Reynolds averaging to derive k-e eqns (u = [u] + u' ; [u'] = 0) then you may derive the k-eqn by taking the Reynolds-average of the momentum equation dotted with the velocity vector vec_U = (u,v,w) and then subtract from this the Reynolds-averaged momentum equation dotted with the Reynolds-averaged velocity vector:

[momentum dot vec_U] - [momentum] dot [vec_U]

In this case, you would have the extra term

[Sx * u'] + [Sy * v'] + [Sz * w']

added to the k-eqn. If vec_S = rho*(gx,gy,gz) for example (g = gravity vector), then the extra term in the k-eqn is

[rho' u'] gx + [rho' v'] gy + [rho' w'] gz

If you are doing Favre-averaging

rho = [rho] + rho' ; u = {u} + u" ; {rho u"} = 0

then deriving the k-eqn is a little more involved. In Wilcox's "Turbulence Modeling for CFD" book, he dots the instantaneous momentum equation with (u", v", w") and Favre averages, which would give you the extra term

{Sx * u"} + {Sy * v"} + {Sz * w"}

If again we had vec_S = rho*(gx, gy, gz), then we would get

{Sx * u"} = {rho u" gx} = {rho u"} gx = 0

since {rho u"}=0 by definition of Favre averaging, and similarly for the v and w components. The effect of density fluctuations does "show up" in the k-eqn for the gravity body force (through the presure work term {u" dp/dx + v" dp/dy + w" dp/dz} since [rho] u" = - [rho' u'], etc...), but does not show explicity for Favre averaging, as it did for Reynolds averaging.

I am hoping the above examples help guide your derivation of your k equation.

 CFDtoy May 10, 2006 16:53

Re: Derivation of k-e model from momentum equation

Greetings Moder: That example is very helpful. As posted before my source term is a function of the scalar (which is function of the flow field). I would like to use Reynolds averaging and hence, I guess the sources, as you explain, should show up explicitly in the k-equation.

I am working on few more details on the source terms. shall get back to you soon in this concern.