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psi in compressibleInterFoam (OpenFoam)

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Old   April 4, 2013, 11:36
Default psi in compressibleInterFoam (OpenFoam)
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Clement Olivier
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Hi everyone,

I am currently using the solver compressibleInterFoam (of OpenFoam). I am trying to change the equation of state perfectFluid and therefore, I need to understand what is psi. It seems to be the compressibility defined by
\psi = \left( \frac{\partial \rho}{\partial p} \right)_T
But I am not sure at all.
Moreover, I do not understand the file pEqn.H, and thus it do not help me to understand how psi is used.

If someone could give me some help, I would be very thankful

Best regards

Clément
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Old   May 28, 2013, 06:44
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Hi Clement,

Have you managed to figure this out? I am also looking to implement an equation of state and am not sure what to make of the psi term.

Referring to the documentation for totalPressure boundary condition (which is more in line with my own interests), psi is defined as compressibility and has units m^2/s^2. Beyond that though, I have not seen any documentation about what it actually is ...

Edit:
Continuing with the dimensional line of thought,
\psi = \frac{P}{\rho} = \frac{kg/ms^2}{kg/m^2}=\frac{m^2}{s^2}
so the units work out in that regard.

Regards,
Aaron

Last edited by dalaron; May 28, 2013 at 06:57. Reason: Forgot to add something
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Old   April 24, 2014, 09:08
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Hi

I think the compressibility is actually the reverse that is being defined above. It is

psi = rho/p which makes it a isothermal compressibility


Thanks
RR
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Old   February 9, 2016, 06:52
Default totalPressure definition
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Hi,


In totalPressureFvPatchScalarField.H we can read that for compressible supersonic flow the total pressure is defined as :


\f[ p_p = \frac{p_0}{(1 + 0.5 \psi G |U|^2)^{\frac{1}{G}}}
\f]
where
\vartable
p_p | pressure at patch [Pa]
p_0 | total pressure [Pa]
\gamma | ratio of specific heats (Cp/Cv)
\psi | compressibility [m2/s2]
G | coefficient given by \f$\frac{\gamma}{1-\gamma}\f$
\endvartable



What compressibility is ? is its dimension really m2/s2 ? and really not s2/m2 ??.
I would expect s2/m2 in order to obtain psi*U^2 as an unidimensional number ..

I am also surprised by the G definition. It is really \frac{\gamma}{1-\gamma} ?
I would expect the inverse frac{1-\gamma}{\gamma} ?

Does someone has an opinion on it?

Regards
Fanny
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Old   February 12, 2016, 03:51
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Hi

I see it is s2/m2 because in the above equation it is multiplied by U2 to get a scalar value.

Best Regards,
RRK
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Old   November 21, 2016, 09:17
Default from the Programmer's Guide v 3.0.1
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Hi, everyone,

If it still can help anybody, in studying the Programmer's Guide OF v. 3.0.1, I,ve found, in section 3.4, the definition of psi (\psi):
\psi = \frac{\partial \rho}{\partial p},
which is called the barotropic relationship (compressibility), and its unit is s²/m² (at least for sonicLiquidFoam).
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Old   February 8, 2018, 12:31
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Alexander
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Quote:
Originally Posted by acs View Post
Hi, everyone,

If it still can help anybody, in studying the Programmer's Guide OF v. 3.0.1, I,ve found, in section 3.4, the definition of psi (\psi):
\psi = \frac{\partial \rho}{\partial p},
which is called the barotropic relationship (compressibility), and its unit is s²/m² (at least for sonicLiquidFoam).
The problem is that this definition of psi is unambiguous.
It can be
\psi = \left ( \frac{\partial \rho}{\partial p} \right )_T = \frac{1}{RT}
in isothermal case, and
\psi = \left ( \frac{\partial \rho}{\partial p} \right )_S = \frac{1}{\gamma RT}
in adiabatic case. Can we be sure that psi is determined by the first formula in all cases?
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Old   February 9, 2018, 01:32
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Quote:
Originally Posted by sahas View Post
The problem is that this definition of psi is unambiguous.
It can be
\psi = \left ( \frac{\partial \rho}{\partial p} \right )_T = \frac{1}{RT}
in isothermal case, and
\psi = \left ( \frac{\partial \rho}{\partial p} \right )_S = \frac{1}{\gamma RT}
in adiabatic case. Can we be sure that psi is determined by the first formula in all cases?
I checked the source code for compressibleInterFoam and based on the thermo library used, I can say that psi is surely 1/RT (at least for a perfect fluid). It may vary according to the the equation of state chosen at runtime.
Please correct me if I am wrong!
ES7
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Old   March 21, 2018, 17:39
Default OpenFOAM docs
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Arpit
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This might be useful. You can look up different versions (v3 and above).
https://cfd.direct/openfoam/user-gui...hermophysical/
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Old   May 11, 2019, 04:33
Default Add a non-linear equation of state to foam
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Hi foamers,

I want to add a non-linear equation of state (JWL) to foam to use in sonicFoam.
In this equation, I need to calculate rho, Cp, h, s, psi, Z, and cpMcv and some functions such as:


correct (which is used in EEqn.H: thermo.correct() )
pThermo(which is used in createFields.H: thermo.pThermo() )
p (which is used in createFields.H: thermo.p() )
rho (which is used in createFields.H: thermo.rho() )
he (which is used in createFieldRefs.H: thermo.he() )
psi (which is used in createFieldRefs.H: thermo.psi() )

But JWL is non-linear and calculation of parameters are not as easy as perfectGas! However, I have calculated rho using Newton method in rho() function. But I don't know how to calculate the others!

Thanks
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Old   May 12, 2019, 09:37
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Dear foamers,
my last problem is not solved yet!
In addition I want to know:
1- How can I calculate "h", "Cp" for my new equation of state? why Cp and h are zero for perfectGas model in openFoam? and how did they calculate "s=-R*log(p/Pstd)" ?

Thanks
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Old   October 30, 2019, 11:32
Default JWL implementation that works with OpenFOAM
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Hi,

Yes, we have just released a new solver (blastFoam) which includes the JWL equation of state (an extensions to model afterburn), as well as several other useful and interoperable equations of state (e.g. ideal gas, stiffened gas, tait, cochran-chan, van der waals, JWL), flux schemes (HLLC, AUSM+, Kurganov/Tadmor), and multiple examples/tutorial cases.


Source, user guide and examples are available here: https://github.com/synthetik-technologies/blastfoam

Hope this helps!
Peter Vonk
Synthetik Applied Technologies
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