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Appropriate solver for Nusselt Number in a Channel |
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October 24, 2017, 05:18 |
Appropriate solver for Nusselt Number in a Channel
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#1 |
Member
Join Date: Oct 2015
Location: Finland
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Hey,
I want to simulate a flow in a 5 mm width channel of 150 mm length with heated walls. I then want to calculate the Nusselt number from this simulation and compare this result with the given correlations for the channel flow. What is the appropriate solver to do this? I initially started off with a modified icoFoam with temperature equation added as passive scalar but was unable to use the wallHeatFlux utility with that. So now I am using rhoSimpleFoam for simulation but couldn't wrap my head around the flux I get when running the wallHeatFlux utility. Can somebody recommend a suitable solver and post-processing steps to obtain what I want? Best, Bulut |
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October 24, 2017, 13:48 |
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#2 |
Senior Member
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buoyantSimpleFoam/buoyantPimpleFoam would be a good place to start to obtain a steady temperature field or a time-averaged one with some post processing. What version of openfoam are you using? I just tested the wallHeatFlux utility in version 4.x and it returns an integrated heat flux in Watts for all wall patches. You could also calculate approximate gradients from a steady or time-averaged temperature field.
Caelan |
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October 24, 2017, 13:53 |
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#3 |
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Hey,
I got some meaningful results with rhoSimpleFoam but buoyant solvers seem more appropriate. I use both 4.0 and 2.4 so can switch to suitable one. I have a follow up on wallHeatFlux though. When I run the utility it gives me a wallHeatFlux with the unit m/s^3. Is it equivalent to Watts in SI units somehow or is it doing something wrong? Best, Bulut |
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October 24, 2017, 14:16 |
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#4 |
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If I were to guess, those units are some sort of power normalized by density and related to the dimension of your problem. I did run a rhoSimpleFoam tutorial in version 4.x and the wallHeatFlux returned units of Watts -- can you try with version 4.x?
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October 24, 2017, 14:25 |
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#5 |
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Hey,
I run squareBend tutorial of 4.x with wallHeatFlux utility run as post-processing. When I look at wallheatFlux files in time directories the unit given is: dimensions [1 0 -3 0 0 0 0]; Oh I get it now. Watt is kg-m^2/s^3 . This is kg/s^3, so its the flux. When I multiply this with the area I will get the Watts. Much clear now. |
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October 24, 2017, 14:36 |
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#6 |
Senior Member
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Glad you figured it out. To erase any further confusion, here the description from the source code:
"Calculates and writes the heat flux for all patches as the boundary field of a volScalarField and also prints the integrated flux for all wall patches." So you'll end up with printed values of integrated heat flux (in watts) for walls specifically, and regular heat flux for other patches written to each directory as you've found. Caelan |
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