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#1 |
Super Moderator
Tobias Holzmann
Join Date: Oct 2010
Location: Bad Wörishofen
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Hi all,
actually I don´t have a question but it would be nice to have someone who proofs my statement. Right now I am doing a rhoSimpleFoam calculation in an exhaust gas system. During the calculation, we solve for p, U, h. By knowing the fluxes, I am moving on with an own solver. There, I am solving the temperature equation as: Code:
cp *( ddt(rho, T) + div(phi, T)) ) == laplacian(lambdaEff, T); ![]() The laminar one was calculated using the VDI Wärmeatlas. So actually it is dry air with 10 % of water and is implemented as a polynomial function of temperature. However, the turbulent thermal conductivity is not known out of the box. In the thermodynamic libraries of FOAM, I saw that kappaEff (which is the effective termal conductivity) is calculated as: Code:
kappaEff = kappa + cp*alphat
Additionally, I made a test where I compared the chtMultiRegionFoam (with frozen flow - energy equation is solved) to my own solver (temperature equation is solved). In both equations the effective field in the laplacian term was used. At the end, both gave the same results (with some small errors). However, doing so, the effective conductivity of the air is between 1e-3 W/m/K to 200 W/m/K which is very high and I have no experience if this value is a physical range. Note, my fluid flow is high turbulent. I have around 50 m/s inside the pipes. My Question Can anyone confirm that I did the calculation of the turbulent thermal conductivity correct?
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Keep foaming, Tobias Holzmann |
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#3 |
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Santiago Lopez Castano
Join Date: Nov 2012
Posts: 354
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are you doing LES or RANS?
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#4 |
Super Moderator
Tobias Holzmann
Join Date: Oct 2010
Location: Bad Wörishofen
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I am doing RANS but it does not matter as the only difference is related to the mut calculation. Thus, alphaEff will change and that´s it.
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#5 |
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Santiago Lopez Castano
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#6 |
Senior Member
Ruiyan Chen
Join Date: Jul 2016
Location: Hangzhou, China
Posts: 162
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Tobi,
I'm interested in this topic, when you mentioned confirming your steps, are you comparing your results to an experimental or a numerical case? I'm kind of surprised that no Prandtl number appears in your implementation, or maybe it's already included in the calculation of alphat by OpenFOAM? Sometimes I see people use both a laminar Prandtl number and a turbulent Prandtl number for the laminar part and turbulent part of the heat diffusion coefficients, respectively. Thanks, Ruiyan |
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#7 |
New Member
Adam
Join Date: Jan 2019
Posts: 21
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Do you use any wallfunction for alphat? I just noticed that my turbulent Prandtl number is always 1 if I dont use any. Because of that I implemented a function for the turbulent Prandlt number.
So do you keep the turbulent Prandtl number constant? |
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