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Qs on symmetry of flow field, conducting solid |
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November 11, 2002, 12:06 |
Qs on symmetry of flow field, conducting solid
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#1 |
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hello there,
I'm modelling a gas combustion furnace, a horizontal cylinder with a burner in one end in the center, and outlet in another end. And I'm using CFX 5.5.1. Q1, I found the flow field is not symmetric. Is there anyone else experienced the similar phenomena? What can I do to improve/correct this? (I'm not going to use a 2D simulation.) Q2, If I consider a conducting solid (Fireclay) around the furnace, heat transfer coefficient BC. With a small timescale for "energy" equations, it would converge with a wrong T field. While if I apply a larger timescale (e.g. 10s), the convergence is very slow. I played around with "Solid time scale factor" etc., there was little help. Could anybody give me some suggestions on this? Thanks in advance. bo |
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November 11, 2002, 18:08 |
Re: Qs on symmetry of flow field, conducting solid
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#2 |
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Q1. When you unstructured grid is not symmetric, your solution will not be completely symmetric.
Q2. Have you tried to run with only the energy equation while using a huge timestep? Astrid |
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November 12, 2002, 08:05 |
Re: Qs on symmetry of flow field, conducting solid
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#3 |
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Thank you.
I thought solving a heat conduction problem would be fast. The calculation is running now with 60s for energy equations and 1s for the others. The RMS for energy in solid is about 2.3*10^-4 and converging very slow (rate=0.99 or even 1.00). Energy imbalance is about 50% in the solid area but decreasing slowly. Any more suggestions? I found out that you also attended the conference in Strasbourg, in an earlier post. Do you have any news of the CD? Best regards bo |
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November 12, 2002, 15:17 |
Re: Qs on symmetry of flow field, conducting solid
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#4 |
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Run with an even higher timestep on the energy equation, like 1.0E20. If the heat transfer is diffusion dominated it may take quite some time to converge with a small timestep.
Also, what is the Prantdl number for your fluid? If the thermal boundary layer is much thicker than the hydrodynamic (i.e. Pr>>1) then you will have to use quite large energy timesteps. If the fluids equations are solved well enough then you might also try turning them off with the "solve fluids =f" (maybe also solve tke eps=f) expert parameter and then just solve the energy equation with that large timestep. As far as the symmetry of the flow field goes, how were you ensuring this? Did you use symmetry boundary conditions? If not, then symmetry is not guaranteed. Neale |
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November 12, 2002, 18:31 |
Re: Qs on symmetry of flow field, conducting solid
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#5 |
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No news of the CD. Ask the helpdesk.
Astrid |
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