CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Main CFD Forum

Derivation of k-e model from momentum equation

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Display Modes
Old   May 3, 2006, 11:00
Default Derivation of k-e model from momentum equation
  #1
CFDtoy
Guest
 
Posts: n/a
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
  Reply With Quote

Old   May 3, 2006, 23:12
Default Re: Derivation of k-e model from momentum equation
  #2
Tian_FB
Guest
 
Posts: n/a
The book by Pope is Ok for u
  Reply With Quote

Old   May 4, 2006, 13:02
Default Re: Derivation of k-e model from momentum equation
  #3
Mani
Guest
 
Posts: n/a
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.
  Reply With Quote

Old   May 9, 2006, 19:23
Default Re: Derivation of k-e model from momentum equation
  #4
CFDtoy
Guest
 
Posts: n/a
Hello Mani: Thanks for your reply. I am giving a brief description here of the problem. Please let me know your thoughts.

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
  Reply With Quote

Old   May 10, 2006, 12:51
Default Re: Derivation of k-e model from momentum equation
  #5
Jeff Moder
Guest
 
Posts: n/a
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.

  Reply With Quote

Old   May 10, 2006, 16:53
Default Re: Derivation of k-e model from momentum equation
  #6
CFDtoy
Guest
 
Posts: n/a
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.

Thanks for your help.

btw, is there any online PDF version derivation of k-e models that can be accessed? If so, please direct me to the same.

Thanks,

CFDtoy

  Reply With Quote

Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Discretization of momentum equation query siw CFX 0 June 20, 2011 08:38
Derivation of Momentum Equation in Integral Form Demonwolf Main CFD Forum 2 October 29, 2009 20:53
Momentum equation of Darcy's law sambatra OpenFOAM Running, Solving & CFD 2 April 17, 2009 03:27
Coupling between momentum and energy equation Mattia FLUENT 0 October 26, 2007 08:02
momentum equation (Normal to wall) kk Main CFD Forum 0 July 2, 2006 23:13


All times are GMT -4. The time now is 21:39.