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August 6, 2014, 13:13 |
Energy balance
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#1 |
New Member
Join Date: Jul 2014
Posts: 21
Rep Power: 11 |
Hi all,
Can anyone advise me please how to check the energy balance in a simple straight tube? I have tube with one inlet and one outlet and one wall. At the inlet I impose constant temperature and at the wall a heat flux. When I use wallHeatFlux utility to calculate the power it shows at the inlet and outlet values in order lower than the heat loss at the wall. However, if I roughly estimate the power from the mass flow rate and the temperature difference between inlet and outlet (cp*m*dt) it looks similar to the heat loss trough the wall. But the problem is that there is already temperature profile at the inlet and I cannot set the outlet temperature exactly. The solver is buoyantSimpleFoam. Thanks Atlan |
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August 7, 2014, 04:59 |
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#2 |
Senior Member
Philipp
Join Date: Jun 2011
Location: Germany
Posts: 1,297
Rep Power: 26 |
I have no experience with "wallHeatFlux" utility, but just from the wording: Maybe it just calculates the heat flux as a spatial derivative of temperature times the heat conductivity, because that is how you would calculate the heat flux at a wall:
q=lambda*dT/dy. At the inlet and outlet you don't have a wall but the moving fluid. Thus, "q" gets an additional component by the heat that is transported by the moving fluid rho * u * E0. This could be the reason the balance doesn't work how you do it, because heat conduction at inlet and outlet ist just way less than convective heat transport.
__________________
The skeleton ran out of shampoo in the shower. |
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August 7, 2014, 09:54 |
Energy balance
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#3 |
New Member
Join Date: Jul 2014
Posts: 21
Rep Power: 11 |
You are right. I have realized, it is not possible to use the wallHeatFlux utility for inlet and outlet. It is applicable only for walls.
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