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February 24, 2015, 12:07 |
Symmetry and total surface heat flux
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#1 |
New Member
Ontario
Join Date: Feb 2015
Posts: 8
Rep Power: 11 |
Hi,
I am trying to determine the heat transfer coefficient from forced convection for a cylinder in cross flow from a jet. There is a symmetry condition to split the cylinder and jet in half. I am using a k-epsilon turbulence model. When I look at the total surface heat flux and try to calculate nusselt number from that the results are almost double what they should be. Is the total surface heat flux that fluent calculates calculated from the area modeled (which is half the actual length of the cylinder) or does it take into account that there is a symmetry face and double the overall area of the modeled cylinder? |
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February 26, 2015, 17:58 |
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#2 |
Senior Member
JSM
Join Date: Mar 2009
Location: India
Posts: 192
Rep Power: 20 |
Fluent gives total surface heat flux from the area modeled. You need to double the heat flux if you are using symmetry boundary condition.
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With regards, JSM |
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November 8, 2017, 05:47 |
heat flux
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#3 |
New Member
hamid reza
Join Date: Jun 2017
Location: iran
Posts: 1
Rep Power: 0 |
i have heat 5W,and my total area is 60*42 mm^2 and use 1 symmetry boundary condition
so should i consider heat flux=5/(60*42*10^-6)=1984 w/m^2 or 5/(60*21*10^-6)=3968 w/m^2 for heat flux BC? |
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Tags |
fluent, symmetry, total surface heat flux |
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