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August 22, 2011, 04:17 
Calculating heat flux

#1 
New Member
Join Date: Jul 2011
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Hi,
I'm dealing with heat conduction in solids using laplacianFoam and want to calculate the heat flux through walls from Openfoam results. However, unexpected difficulties are arising which I want to explain with the following example: The analytical result for the heat flux through the walls of the annulus shown in the figure is q"=2460.01 W/m (related to length in z direction). 1. Using wallHeatFluxLaminar (from this thread: wallHeatFlux utility in OpenFoam1.6), the following values are calculated: Outside: q"_a=780.284 W/m^2 > multiplication by inner circumference gives q"=2451.3 W/m Inside: q"_i=1299.32 W/m^2 > q"=2449.2 W/m Compared to the analytical solution, these values are too small. 2. Using foamCalc magGrad T, the following values are calculated: Outside: magGrad T=78.0284 K/m > multiplication by k and the inner circumference gives q"=2451.3 W/m (same value as given by wallHeatFluxLaminar) Inside: magGrad T=129.932 K/m > q"=2449.2 W/m Thus, using magGrad T leads to the same heat flux as wallHeatFluxLaminar for the present example. However, this is not the case for all geometries I dealt with. 3. Calculating the gradient magnitude for myself from mag(grad T) = sqrt(gradTx^2 + gradTy^2), I obtain different heat fluxes: Outside: sqrt(gradTx^2 + gradTy^2)=78.307 K/m > q"=2460.09 W/m Inside: sqrt(gradTx^2 + gradTy^2) > q"=2460.07 W/m These heat fluxes are very close to the analytical value. Now, my questions are:  Why does the gradient magnitude calculated as sqrt(gradTx^2 + gradTy^2) differ from magGrad T calculated by foamCalc?  Why is the heat flux value calculated from sqrt(gradTx^2 + gradTy^2) apparantly "better" than this one calculated from WallHeatFluxLaminar or magGrad T? I really appreciate any comment on this problem. Regards Hugo 

August 22, 2011, 05:18 

#2 
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Join Date: Jul 2011
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I found the reason for the different results: I had writePrecision set to 6. When I increase it to 8, also wallHeatFluxLaminar and magGrad T give the correct values.
Regards Hugo 

May 7, 2015, 18:51 

#3 
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Juan David Rodriguez P
Join Date: Jan 2015
Posts: 18
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Hugo, thanks for your contribution.
I want to visualize heat flux lines (orthogonal to isotherms) in paraview. My case is a similar one to yours (heat conduction in 2D from a buried pipe to soil). Do know how to proceed? Best, Juan. 

May 27, 2015, 06:09 

#4 
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Nicole Andrew
Join Date: Sep 2014
Location: Pretoria, South Africa
Posts: 22
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Hi Juan,
Have you had any success yet? If you are used to modifying the solver, I would recommend just adding a new volScalarField, qFlux or something, calculated from fvc::grad(T) in the solver. I am not sure if this would be the best way. Apparently there are also calculators in paraview, perhaps one of these would be useful? Let me know if you want me to elaborate 

May 27, 2015, 12:47 

#5 
New Member
Juan David Rodriguez P
Join Date: Jan 2015
Posts: 18
Rep Power: 2 
Hi Nicole,
I finally used funkySetFields to calculate grad(T). I think it is ok and am still learning to modify code in OF, but I will try what you suggest. Thanks. Best, Juan. 

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