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Algorithm for compressible Low-Re flow

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Old   July 7, 2004, 04:34
Default Algorithm for compressible Low-Re flow
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Frederik
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Hello,

I am working on Low-Re flows, but with intermediate Mach -Numbers (Re<10000, Ma<0.3) using the Low-Re k-eps or v2f in v3.2 with ideal gas. This case leads to unsatisfactory convergence at very low relaxation factors. Is there any combination of solver algorithm, differencing scheme etc. which is predestined for such a case?

Thank you for advising - Frederik.
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Old   July 7, 2004, 06:56
Default Re: Algorithm for compressible Low-Re flow
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4xF
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a) For this Mach number, the flow can be considered as incompressible, so there is no point of using the ideal gas law. b) Re < 10,000 for a flat plate is below the the transition point from laminar to turbulent, so why use a turbulence model?
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Old   July 8, 2004, 07:52
Default Re: Algorithm for compressible Low-Re flow
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Frederik
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Ok. I didn't mention: It's an internal flow in a duct with a gap < 1mm. Pressure loss is quite high, say p_in = 3 bar, p_out= 2.3 bar, so the effect of expansion must be considered. The flow is no longer laminar at Re = 3000 and essentially turbulent at Re = 10000. Having said this, are there any answers to the initial question?
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Old   July 8, 2004, 14:16
Default Re: Algorithm for compressible Low-Re flow
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4xF
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a) the effects of thermal expansion may be considered, but this does not mean that the flow is compressible. A fluid can be compressible (density varies with temperature, for example) but *the flow* can still remain of incompressible nature (i.e. there are no variations of density or pressure because of the flow kinetic energy). I think that you want to model the fluid as being barotropic. b) Try with a Low-Re model. Since it is a flow in a duct and you know the Reynolds number, why don't you use a Low-Re model and try first with a combination inlet+outlet. Then switch to periodic boundary conditions to simulate an indefinite channel, so that you know how much pressure loss you will get per unit length. You should find more or less the same value for the pressure loss as you expect.

Hope this helps.
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