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How to set the thermal conductivity?

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Old   July 22, 2020, 14:30
Default How to set the thermal conductivity?
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Alex
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I need to know what is the thermal conductivity value of my simulation. I am simulating a natural convection problem using buoyantPimpleFoam, but I do not know where to read k (thermal conductivity).

Is it somewhere in the thermophysical properties?

Code:
/*--------------------------------*- C++ -*----------------------------------*\
  =========                 |
  \\      /  F ield         | OpenFOAM: The Open Source CFD Toolbox
   \\    /   O peration     | Website:  https://openfoam.org
    \\  /    A nd           | Version:  6
     \\/     M anipulation  |
\*---------------------------------------------------------------------------*/
FoamFile
{
    version     2.0;
    format      ascii;
    class       dictionary;
    location    "constant";
    object      thermophysicalProperties;
}
// * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //

thermoType
{
    type            heRhoThermo;
    mixture         pureMixture;
    transport       const;
    thermo          hConst;
    equationOfState perfectGas;
    specie          specie;
    energy          sensibleEnthalpy;
}

pRef            101325;

mixture
{
    specie
    {
	nMoles          1;
        molWeight       28.9;
    }
    thermodynamics
    {
        Cp              1006.591;
        Hf              0;
    }
    transport
    {
        mu              1.85e-5;
        Pr              0.7;
    }
}


// ******************************
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Old   July 22, 2020, 23:27
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Fumiya Nozaki
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Hi,

You might want to check the definition of the Prandtl number (Pr):
https://en.wikipedia.org/wiki/Prandtl_number

The thermal conductivity k is uniquely identified because the values of mu, Cp and Pr are specified.

Hope this helps,
Fumiya
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Old   July 23, 2020, 07:05
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Quote:
Originally Posted by fumiya View Post
Hi,

You might want to check the definition of the Prandtl number (Pr):
https://en.wikipedia.org/wiki/Prandtl_number

The thermal conductivity k is uniquely identified because the values of mu, Cp and Pr are specified.

Hope this helps,
Fumiya
Hi Fumiya,

That was so helpful! I understood that the value of k that I was using was wrong because I changed mu in my simulation.

Then I have another question, what about the Thermal Expansion (Beta)? what are the parameter which affect the value?

I ask that because I need it to calculate the Grashof number
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Old   July 23, 2020, 10:04
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Fumiya Nozaki
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Hi Sedullo,

For a perfect gas, the thermal expansion is a inverse of the absolute temperature:
https://en.m.wikipedia.org/wiki/Thermal_expansion

Hope this helps,
Fumiya
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Old   August 28, 2020, 05:27
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Quote:
Originally Posted by fumiya View Post
Hi,

You might want to check the definition of the Prandtl number (Pr):
https://en.wikipedia.org/wiki/Prandtl_number

The thermal conductivity k is uniquely identified because the values of mu, Cp and Pr are specified.

Hope this helps,
Fumiya
Hi Fumiya,

If the vicous is sutherland law, does it has relation between k, As, Ts?
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Old   August 12, 2024, 17:53
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John Mary
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Hello,

I am newbie to OpenFOAM and as part of my course I am simulating a gas - solid fluidised bed.

I copied the fluidised bed tutorial of the reactingTwoPhaseEulerFoam and modified the geometry to add a horizontal cylindrical heating element close to the inlet.

I added the heater patch name to all files in the 0 repository and the simulation is running well. Right now my aim is to modifiy the conductivity model in the energy equation.

From all I have read, the thermal conductivity is implicitly defined in the Prandtl number but the problem with this approach is that it sets a uniform thermal conductivity in the bed whereas the conductivity in the bed is not uniform and highly dependent on the volume fraction of the solid particles.

I have found some suitable models of effective thermal conductivity but I find it difficult to integrate them in the solver, as OpenFOAM does not perform calculations in the transportProperties file.

That is why I have reached out to find if you can help.

Kind regards,
John
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