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Old   January 18, 2009, 14:32
Default Incompressible vs compressible solvers
  #1
James
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Hi,

I have to model an internal flow situation where the Mach number varies from 0.03 to 1.2. I have a compressible flow solver at hand. I'm wondering whether a compressible solver can solve the flow even if I have very low Mach numbers?

I was told that I could re-scale my geometry so that I can have a higher Mach number and the re-scaling would a lower Re.

Cheers
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Old   January 18, 2009, 17:00
Default Re: Incompressible vs compressible solvers
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otd
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Not sure I understand:

If you change the geometry in the compressible solver (CS) in such a way that you change the velocity field, that would change the Mach number field. The Reynolds number has velocity and a geometry dimension in it - so Re might go up or down depending on what was changed in what direction.

It might be more useful to set up the same geometry in both compressible and incompressible codes - if both codes are well-understood and checked out.

Some of the older Los Alamos codes (sola-ice comes to mind) would work at low Mach numbers.

Hope this is of some use to you.
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Old   January 18, 2009, 20:39
Default Re: Incompressible vs compressible solvers
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Harish
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The compressible flow solver would work fine at the lower mach number range if you have preconditioning implemented in the compressible solver. You rescale the wave speeds to achieve this.

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Old   January 19, 2009, 00:32
Default Re: Incompressible vs compressible solvers
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momentum_waves
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What about viscous stresses? If the solver is based on Euler equations & not Navier-Stokes, you're going to be solving for the wrong physics. Surely?

mw...

www.adthermtech.com/wordpress3

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Old   January 19, 2009, 06:07
Default Re: Incompressible vs compressible solvers
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James
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Sorry I meant rescaling the geometry and using a higher Mach numbers so that the Reynolds number still remains constant (if that makes sense). Will this be valid?
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Old   January 19, 2009, 06:15
Default Re: Incompressible vs compressible solvers
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Timon
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No. For flows to be dynamically equivalent, both Reynolds and Mach number need to be the same!
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