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Natural heat transfer in He tank with a heated rod inside |
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January 16, 2015, 10:12 |
Natural heat transfer in He tank with a heated rod inside
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#1 |
New Member
Claudio Torregrosa
Join Date: Sep 2013
Posts: 3
Rep Power: 13 |
Hello!
I am running CFX transient simulations of a small dimensions tank (0.6 x 0.23 m) filled with He at RT and atmospheric pressure. Inside the tank there is a small rod 8 mm diameter which is uniformly heated up to 600 C by an internal source in just 1 s, as well as a serpentine at constant temperature of 30 C. The goal of my simulation is to check how long it takes for the heat to be transferred from the rod to outside the tank and serpentine heat sink, as well as the max increase of He temperature during this transient. In the outer walls of the tank there are convective BC. I prepared a "fake" 2-D model of a cross section of 0.6 mm of the tank and rods. Symmetry boundary conditions in both extruded planes. I run simulations assuming buoyancy and He as a Ideal gas. I did not included radiation yet. My problem is the huge variation of the heat transfer coefficient between the rod and the Helium depending on the turbulence model in the fluid. I know that this quite obvious.., but still the variation is huge, and the models which, according to physics, are more suitable for my case give very high heat transfer coefficients. I am always running with quite small time steps 0.05 s and I don't have troubles of convergence in any of the simulations. -In the first trial I ran using k-e model, which I think it was not very suitable as I got error: 'The Reynolds number is outside of the range expected based on the Option selected for the TURBULENCE MODE". The simulation still ran anyway. I got a h in rod surface varing from h = 0.2 W/m2K to h =4 W/m2K depending on the time and the development of the He flow around the rod. The order of velocities in the fluid was 0.02 m/s -Then I used laminar flow -non turbulence-, given the low velocities of the He (0.35 m/s). Then the simulation runs perfectly, very good convergence. But, the averaged h in the rod surface is up to 3000 W/m2K ! -Finally, I tried SST turbulence model. The results are consistent with the ones using laminar flow. velocities up to 0.3 m/s and h in the rod surface around 2000 W/m2K I am really surprised concerning these high values of the heat transfer coefficient in the rods surfaces, taking into account that is natural convection with He My question is also if I am using the proper turbulent model or if I am missing something important. Or the reason why I get this high convective heat transfer coefficient... Regarding the mesh. I re-meshes several times the boundaries close to the rod surfaces, as well as run simulations with smaller time step (0.01s). I get values of convective heat transfer coefficient even higher. Any suggestions? Thank you in advance. |
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January 16, 2015, 12:24 |
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#2 |
Senior Member
Join Date: Jun 2009
Posts: 1,880
Rep Power: 33 |
How are you computing your heat transfer coefficient ? Recall heat transfer coefficient is a function of the reference temperatures involved.
A value of 3000 [W/mK] means nothing to me until the whole context of its definition is defined. For example, Q = HTC * A * (T_wall - T_???) you need to clearly define T_???. Let us say T_??? is the temperature of the first node near the wall, then the HTC will change as you refine the mesh, correct ? But the value of Q will always be correct. You may want to look at the output file for the HTC warning, and read the documentation. Hope the above helps, |
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January 16, 2015, 15:12 |
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#3 |
New Member
Claudio Torregrosa
Join Date: Sep 2013
Posts: 3
Rep Power: 13 |
Thanks very much for you answer!
The value of HTC is the one I get from CFD-post. So I guess is taking the T_?? as the adjacent cell of the wall. That could be the reason then for this high values..as I have a very fine mesh and the temperature in the adjacent nodes is very high.. Concerning the turbulent model; which do you think is more accurate for this application? STT? or just non turbulence model (laminar). Since convergence is working well with both of them. Thanks! |
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January 16, 2015, 18:59 |
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#4 |
Senior Member
Erik
Join Date: Feb 2011
Location: Earth (Land portion)
Posts: 1,188
Rep Power: 23 |
Calculate your Rayleigh number, that should tell you laminar or turbulent.
Your case should surely be turbulent though. |
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Tags |
heat transfer coefficient, helium, natural heat transfer |
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