# incompressible limit for compressible codes

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 November 1, 2013, 17:40 incompressible limit for compressible codes #1 New Member   Join Date: Nov 2013 Posts: 4 Rep Power: 4 People have always been talking about the incompressible limit for compressible codes. It seems to me that there are some issues there, but it is not very clear. What would happen if one runs a compressible code, using pressure and velocity as variables, at very low mach number? What would they encounter in numerics? Is there anything diverging or blowing up? If possible, is there any journal paper that discuss this issue?

November 1, 2013, 17:49
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Filippo Maria Denaro
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 Originally Posted by liuju1985 People have always been talking about the incompressible limit for compressible codes. It seems to me that there are some issues there, but it is not very clear. What would happen if one runs a compressible code, using pressure and velocity as variables, at very low mach number? What would they encounter in numerics? Is there anything diverging or blowing up? If possible, is there any journal paper that discuss this issue?

Think about dp/drho, for very low Mach number, even very small errors in the density solution will be amplified in the pressure field...
Generally, preconditioning can be used

November 1, 2013, 17:55
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 Originally Posted by FMDenaro Think about dp/drho, for very low Mach number, even very small errors in the density solution will be amplified in the pressure field... Generally, preconditioning can be used
Thanks. I understand dp/drho is ill-defined for incompressible flow.

So I proposed to solve, for example isothermal NS, in terms of pressure and velocity. Hence, we may avoid solving the density. Is there any fundamental trouble there?

November 1, 2013, 18:02
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 Originally Posted by liuju1985 Thanks. I understand dp/drho is ill-defined for incompressible flow. So I proposed to solve, for example isothermal NS, in terms of pressure and velocity. Hence, we may avoid solving the density. Is there any fundamental trouble there?
Sorry but I don't understand your question... if you assume T= constant the pressure and density are still related and the density equation should be solved.
To solve in terms of velocity and pressure alone you must enforce the constraint Div V = 0.

 November 1, 2013, 18:14 #5 New Member   Join Date: Nov 2013 Posts: 4 Rep Power: 4 It is not necessary to enforce div(u) = 0. Considering an isothermal flow, perfect gas law gives that pressure is a linear function of density. Hence, you may write density as a function of pressure: rho = rho(p). In doing so, all density terms could be replaced by pressure and its derivatives. For example, the mass balance equation: rho,t + \nabal(rho u) = 0 can be written as drho/dp p_,t + \nabla( rho(p) u) = 0. As incompressible limit, rho becomes a constant, hence the drho/dp becomes zero. It does not blow up, and is still well-defined. In my understanding, if there is any unstability, the source could only come from the locking phenomena, same as the Stokes problem.

November 1, 2013, 18:28
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 Originally Posted by liuju1985 It is not necessary to enforce div(u) = 0. Considering an isothermal flow, perfect gas law gives that pressure is a linear function of density. Hence, you may write density as a function of pressure: rho = rho(p). In doing so, all density terms could be replaced by pressure and its derivatives. For example, the mass balance equation: rho,t + \nabal(rho u) = 0 can be written as drho/dp p_,t + \nabla( rho(p) u) = 0. As incompressible limit, rho becomes a constant, hence the drho/dp becomes zero. It does not blow up, and is still well-defined. In my understanding, if there is any unstability, the source could only come from the locking phenomena, same as the Stokes problem.

Sorry but I still see some thing I am not convinced about... if you are simoultaneously using constant temperature and density assumptions and also the gas law relation, then pressure must be also constant.
The density equation can be rewritten in terms of the pressure field by using dp = a^2 drho but this is valid for isoentropic flows

November 1, 2013, 18:37
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 Originally Posted by FMDenaro Sorry but I still see some thing I am not convinced about... if you are simoultaneously using constant temperature and density assumptions and also the gas law relation, then pressure must be also constant. The density equation can be rewritten in terms of the pressure field by using dp = a^2 drho but this is valid for isoentropic flows
well, if the flow is incompressible, you may regard rho as a constant. Then the compressible NS degenerates to incompressible NS, where pressure is still an unknown.

OK. This is the part not clear to me as well. In textbooks, it says the pressure is no more thermodynamic pressure, but becomes pure mechanical. Some math book says here pressure becomes a Lagrangian multiplier. Maybe someone can help me understanding what is mechanical pressure.

In anyway, I think there is no problem in solving the comressible NS equation in terms of pressure.

November 1, 2013, 18:47
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 Originally Posted by liuju1985 well, if the flow is incompressible, you may regard rho as a constant. Then the compressible NS degenerates to incompressible NS, where pressure is still an unknown. OK. This is the part not clear to me as well. In textbooks, it says the pressure is no more thermodynamic pressure, but becomes pure mechanical. Some math book says here pressure becomes a Lagrangian multiplier. Maybe someone can help me understanding what is mechanical pressure. In anyway, I think there is no problem in solving the comressible NS equation in terms of pressure.

yes, in the incompressible (omo-thermal) limit, pressure is not thermodynamic exactly because p=rho*R*T would simply drive to a constant pressure. Of course that does not make sense in a pressure-driven flow.
The system degenerates in the momentum quantity where a gradient of a scalar function appears, supplied by the cinematic constrain Div V=0.
Therefore, any scalar function producing a gradient in the momentum such that the velocity field is divergence-free ensures a solution. You can simply see that there is no thermodynamic meaning of this function. It can be shown it is a Lagrangian multiplier, see for example the book of Peric and Ferziger.
Furthermore, you can solve the compressible form of the equations by using a suitable preconditioner, for example

http://www.grc.nasa.gov/WWW/5810/rvc...nditioning.pdf

 November 1, 2013, 23:58 #9 Super Moderator     Praveen. C Join Date: Mar 2009 Location: Bangalore Posts: 255 Blog Entries: 6 Rep Power: 10 In the limit of zero mach number, solutions of compressible equations converge to solution of incompressible equations. This is not necessarily true for numerical solutions. For an excellent discussion see http://citeseerx.ist.psu.edu/viewdoc...=rep1&type=pdf http://hal.archives-ouvertes.fr/docs...DF/RR-4189.pdf __________________ http://twitter.com/cfdlab

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