[dappe]

Hello everybody.
I'm working on an impinging jet and I managed to run my simulation with all the settings of a previous Fluent case. My problem is that I don't know how to add the wall thickness option present in the thermal dialog of the heated wall in Fluent. With this option, Fluent solves a 1D steady heat conduction through the wall. In order to do that, we need to specify the type of material, the thickness of the wall and the heat generation rate in the wall (set to zero in my case). With this thin wall calculation, it is only possible to set a constant value for k, otherwise we should use the shell conduction option.
I want to add this option because when I compare the results, I have different values for the wall heat flux and the heat transfer coefficient, and this is probably due to a different Twall used in the equations: Fluent uses the actual Twall calculated, while OpenFOAM uses the fixed one I set in the boundary condition (please correct me if I'm wrong).
I'm running rhoSimpleFoam and my turbulence model is kOmegaSSt. If you need other infos, just ask.
What do you suggest in order to do add the same feature in my OpenFOAM case?
Thanks in advance.
Alex

[Nicole]

Hi Alex,

Have you looked at externalWallHeatFluxTemperature boundary condition? I think it does what you want.

You can get a bit of info in the header (.H) file. In my OpenFOAM 2.4.0 installation it is located in:
src/turbulanceModels/compressible/turbulenceModel/derivedFvPatchFields/externalWallHeatFluxIncompressible

If you can't find it, just grep for it in the src folder. I hope this helps!

[dappe]

Thank you Nicole.
I had a quick view of the code but if I'm not wrong I can use this b.c. only if I have a fixed Heat Flux or a fixed Heat Transfer Coefficient.
In my case I have a fixed temperature at the wall.

Do you have other suggestions?

Alex

[Nicole]

Hi Alex,

Could you not hack it by making the heat transfer coefficient very large and then Tinf = Twall?

I'm sure there is a more elegant solution but hopefully someone smarter can help you with that

[dappe]

Sorry Nicole but I can't understand your suggestion.
If I make the heat transfer coeff very large then Tinf=Twall but I will not know neither the actual local temperature at the wall surface nor the right local heat transfer coefficient.
Maybe I misunderstood your solution, can you be a little bit clearer?
Thank you
Alex

[Nicole]

Hi Alex,

Which "local heat transfer coefficient" are you talking about? The one from inside your domain to the inside of the wall, or the one on the outside of your wall to the environment (this section is outside of your mesh in my understanding?)?

My understanding was that you had a fixed temperature on the outside of the wall surface, so which "actual local temperature at the wall surface" do you want to know?

I'm starting to think that I misunderstand your problem. (I know nothing about Fluent so I battle to understand your introduction) I am picturing this in 1D:

... meshed domain.....|........wall........|....outside....
.............................. Twall?..............Twall?.............Tinf
..................................|..conduction... |...convection...

What exactly would you like to specify in your boundary condition? And what are you interested in looking at after solving?

[dappe]

hi Nicole
Well I'm not sure about this because I don't know completely how this option works (they gave me the Fluent case already solved), but my case should be:
Twall(fixed 644K)-->0.002m wall thickness-->T at the wall surface (to be calculated)-->fluid domain with T field solved with rhoSimpleFoam (321K at the inlet of the jet).
The fluid is air while the wall is made of steel.
What I want to know is the local heat flux along this heated wall.
Maybe it could be done also with a groovyBC but I'm not sure about that.
Sorry if my post was a little bit confusing and thank you for your help.
Alex

[Nicole]

Hi Alex,
No problem. It's been fun.

(Note: I am assuming that you want the heat flux THROUGH the wall, and not "along the wall", but rather along the pathway that you have indicated with arrows in your picture, which I assume is perpendicular to the wall surface.)

If you want a quick solution you can still do what I said earlier, use externalWallHeatFluxTemperature and set h to a large value (you will have to see how big you can get it, I'm not sure if it will affect convergence) and Tinf = Twall = 664K. You will also have to set the wall thickness and conductivity. Then you can check the actual Twall calculated to see if it is suitably close enough for your desired accuracy. This will not fix the heat flux, it simply says that the convective heat loss is so good that the external wall temperature is effectively the same as the environment temperature, which means that the heat transfer "resistance" will be dominated by either the steel shell, or the heat transfer in the fluid domain.

Then you can use a post processing utility such as wallHeatFlux (just type this in the command line after you have run your case) to calculate the average heat flux over all of your boundary patches. I think it also makes a field that you can visualise in paraview to see the heat flux distribution over the face. wallHeatFlux calculates the heat flux from the centre of the last cell in the fluid domain to the "T at the wall surface (to be calculated)", which, of course, is equal to the conduction through the wall and the convection away from the wall assuming that the heat transfer is largely one dimensional in this region. If it is not then I think you will actually have to mesh the wall.

Please not that this is a "hack" and not a technically correct solution. I think ultimately you would need to code your own boundary condition but I have been trying to do that all day with no success so you will have to find someone else to help you with that

Nicole

[dappe]

ok I got it, I will try.
The fact is that in Fluent you don't have to mesh the wall in order to do that (maybe It does the mesh automatically but I don't think so). Using wallHeatFlux I had different results because that flux is calculated using the fixed T at the wall (644K), which is a non completely physical solution (the wall T should be lower next to the jet).
I think that Fluent does a rough "mesh" in the solid only one cell high and with the same number of cells of the fluid domain along the wall. Then it calculates:

q=Ksteel*(Twall-TrealSurface)/(wall thickness)

with q equal to the value of the heat flux from the surface to the fluid (Tfluid= T in the closest cell to the wall).
But this is only my advice and I'm not sure about it.
In fact with the no slip condition I don't know if the q from the surface to the fluid should be expressed in the conductive or in the convective form.
I mean, the q written above is equal to:

q=hAir(TrealSurface-Tfluid)
or
q=kAir*(dTfluid/dn) with n normal direction to the wall and so z in my case
?

Thank you for your support.
Alex