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Old   September 3, 2003, 03:42
Default momentum equations and continuity???
  #1
Lucas Rainer
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Dear CFD friends

Can anyone explain to me in SIMPLE terms, no pun intended,why the continuity equation multiplied by the dependent variable should be subtracted from the momentum equations before solving the equations? I see references to this practice everywhere but no explanations to WHY exactly it is required.

Thank you Lucas.
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Old   September 3, 2003, 05:29
Default Re: momentum equations and continuity???
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jdc
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Hi Lucas,

It seems to me that you are refering to the form of the equation. There is two common forms: (1) conservative form, (2) convective form. Navier-Stokes equations are expressing conservation of mass, momentum and are naturally expressed using the conservative form. The convective form is obtained by substracting the continuity equation multiplied by the dependent variable from the momentum equations.

The conservative form will be used in finite volume code, while the convective form could be used for finite difference. I don't know about finite-element method.

For finite difference, the advantage of using the convective form could/might be: (1) less terms to discretize, (2) less sensible to the "non-conservative" nature of finite-difference. Note: there is some litterature on the discretisation of the convective terms for finite-difference (look for skew-symmetric).

I hoppe this will help and is not too complex. Sincerely, Julien
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Old   September 3, 2003, 07:05
Default Re: momentum equations and continuity???
  #3
mukhopadhyay
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i thought it is well explained in Patankar's book. Also see the book by Anderson,et.al
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Old   September 3, 2003, 09:17
Default Re: momentum equations and continuity???
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Abhijeet Mohan Vaidya
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your question is very very valid.

Actually Patankar started from the conservative form of the scalar transport equation. This is a good start as you know that descretizing conservative form is good for obvious reasons. But during discretization , he multiplied continuity equation by phi and substracted from the scalar equatioin. This will basically take the scalar equation back to the non-conservative form. You can do this in differential forms of the two equations for better understanding. He did that to fit the formula he derived in earlier section. which is (J_e-F_e.phi_p)=a_E(phi_E-phi_P) Substitution of above formula finally leads to the discretized equation.
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Old   September 3, 2003, 09:57
Default Re: momentum equations and continuity???
  #5
Lucas Rainer
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Thank you for the answers.

But if it is a good thing to use the conservative form of the equation - I agree on this, why would we want to transfer the equation to non-conservative form. I'm missing something here. Why not just discretise and solve the conservative form of the equation as is?

If we do the subtraction are we actually discretising the non-conservative equation? Why would we want to do that?

Lucas
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Old   September 3, 2003, 12:01
Default Re: momentum equations and continuity???
  #6
frederic felten
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Well one reason why you may want to use a form other than the conservative form has to do with what quantity you want to conserve implicitly.

For example, if you use the:

1)- Divergence form: d(Ui Uj)/dXj, you will conserve momentum automatically and kinetic energy only if continuity is verified.

2)- Advective form: Uj d(Ui)/dXj, you will conserve momentum and kinetic energy only if continuity is verified.

3)- Skew-symmetric form: 0.5*d(Uj Ui)/dXj + 0.5*Uj*d(Ui)/dXj, you will conserve kinetic automatically and conserve momentum only if continuity is verified.

For more insight and great discussion what some of these issues, please check the following paper:

Morinishi, Lund, Vasilyev and Moin, "fully conservative higher order finite-difference schemes for incompressible flow", Journal of Computational Physics, Vol 143, pp 90-124, 1998.

Sincerely,

Frederic Felten.
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Old   September 3, 2003, 12:05
Default Re: momentum equations and continuity???
  #7
AMV
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Please understand the section 5.2-7 Due to the form of equations 5.44,5.45 (and also refer 5.34), we have to substract continuity from momentum. If you want to just use simple 1st order up-wind for convection terms and central difference for the diffusion terms, you don't have to substract continuity from momentum but proceed the discretizing the eqn 5.50(page 97) and derive the complete formulation. But if you want to use power law then you will have to do it the way patankar did it. This is my understanding. Finally, the way Patankar derived his formulation as given by eq. 5.56 /page 99, it works excellent. I have solved and written atleast 25 cases/programs and the formulation was highly successful. I am using this method for IC engine CFD. Bye.
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Old   September 4, 2003, 06:58
Default Re: momentum equations and continuity???
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Lucas Rainer
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Thank you all for the help!

I spent a fruitful morning working through Patankar. By the way, AMV i particularly appreciated the explanation on p99. thanks for the reference

Lucas
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Old   September 15, 2003, 05:00
Default derivation of momentum equation
  #9
sunny
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hi guys

i have been trying to derive equation of momentum in cylindrical coordinates. but i couldn't do it and i was getting all wrong equations.

can anybody help me out by mailing me the derivation???????????!!!!!!!!

please...........
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