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#1 |
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Senior Member
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Hi all,
I am calculating a heated pipe (air = 800K) and do not want to simulate the air around that pipe but want to set a heat-transfer BC for that. There for I thought its possible to work with the BC Code:
wallHeatTransfer So the question to you all: is there a other BC for that or some examples with groovy? Thanks in advance. Tobi |
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#2 |
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New Member
Daniel Pielmeier
Join Date: Apr 2012
Posts: 26
Rep Power: 3 ![]() |
Just that I got you right. The air with 800K is flowing around the pipe and heats it up. The heat conducts through the wall and in turn heats up the fluid within the pipe.
Instead of solving the air around the pipe you can either apply a heat flux or a heat transfer coefficient and the ambient temperature at the wall using the externalWallHeatFluxTemperature boundary condition. |
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#3 | |
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Senior Member
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Quote:
Hi, in the attachement there you find a picture. I think that tells everyone more then words. - in the pipe is air with T=800K and flows inside through the pipe. - heat-transfer between fluid and solid due to chtMutli calculations - at least I want to make the case more realistic. for that I want to create/calculate the heat transfer from the solid to the ambience air. So my options are: - building the air-domain and create a new mappedWall for pipe->new_air_domain - or doing it with a BC like groovy? Hope everything is clear now
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#4 |
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New Member
Daniel Pielmeier
Join Date: Apr 2012
Posts: 26
Rep Power: 3 ![]() |
Okay so the air inside the pipe heats up the pipe walls and you want to add conduction on the outer wall of the pipe. I think the boundary condition I mentioned above should do the trick. Just use a negative heat flux or heat flux coefficient.
If you want somewhat correct solid temperatures you have to do this anyway else you will end up with 800K for the solid wall. |
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#5 |
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Senior Member
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Hi billie,
thanks for your replay. The problem for the "heat flux" is that its not possible to calculate that. The air = 800K changes in 1sec to 293K (thermal shock) and then the heat flux is completly different ... thats my problem. |
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#6 | |
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New Member
Daniel Pielmeier
Join Date: Apr 2012
Posts: 26
Rep Power: 3 ![]() |
Quote:
If you don't know the heat flux you can use the heat flux coefficient in combination with the ambient temperature.The convective heat transfer coefficient should be around 3-5W/m^2/K. You can also estimate the heat flux coefficient using the formulas for natural convection of external flow around a cylinder from [1]. [1] http://en.wikipedia.org/wiki/Heat_tr...ral_convection |
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#7 |
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Senior Member
Pei-Ying Hsieh
Join Date: Mar 2009
Posts: 249
Rep Power: 7 ![]() |
Hi, Tobi,
To clarify what Daniel suggested. Use the following for the external wall of the pipe: myPatch { type externalWallHeatFluxTemperature; kappa solidThermo; // solidThermo or lookup // q uniform 1000; // heat flux / [W/m2] Ta uniform 300.0; // ambient temperature /[K] h uniform 10.0; // heat transfer coeff /[W/Km2] value uniform 300.0; // initial temperature / [K] kappaName none; } ----------------- Pei-Ying |
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#8 |
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Senior Member
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Hi,.
thanks for your replay ![]() But I figured it out by my self (-> code) But thanks a lot Tobi |
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