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Net heat evaluation on the closed surface of the compressor |
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February 13, 2021, 19:51 |
Net heat evaluation on the closed surface of the compressor
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#1 |
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Hi all,
I am studying how non adiabatic wall conditions (constant temperature on walls) could affect the conditions of variables like pressure ratio, power, etc. I have modeled the machine. Right now I am interested in evaluating how much net energy is in the control volume of the diffuser (in both cases: adiabatic and nonadiabatic). So once I had the results of solver I went to CFD-Post. I define the enclosed volume of diffuser and then I try to determine the net-energy with: areaInt(Wall Heat Flux)@diffuser surface where difuser surface is: But I'm not sure if it is the correct way to evaluate what I want. Could anyone give me any suggestions? Thank you very much for the help. Something that I can't solve yet is that Heat Wall Flux is applied to walls but the inlet to the diffuser is not, nor the outlet, only the shroud and the hub. When I use the Heat Flux function I get strange results. When I use the Heat Flux (areaInt(Heat Flux)@diffuser surface) function I get out of range values, because the reference temperature that CFX-Post uses in the Heat Flux function is the reference temperature of the fluid. I am using the SST turbulence model, with yplus approximately 1. The wall conditions are total pressure inlet, mass flow outlet, walls with constant temperature, and periodic conditions on both sides. Any suggestion or help would be very useful |
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February 13, 2021, 20:36 |
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#2 |
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Glenn Horrocks
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Another way of looking at it is to look at the net heat gain of the fluid passing through. So calculate the heat flux at the inlet and the heat flux at the outlet and you have the heat gain. If your fluid is compressible you are going to have to account for those effects, and if you have viscous heat generation or some other heat generation or loss mechanism you will have to account for that.
But your initial statement says you want to see the effect of adiabatic vs heat transfer on pressure ratio and power. For these parameters you don't need to do anything like this - just run with and without heat transfer and compare the parameters.
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February 13, 2021, 21:44 |
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#3 | |
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Quote:
To calculate the total heat gained/lost from inlet to outlet, would it be enough to use the function Heat Flux (CFD-Post)? areaInt(Heat Flux@)Inlet + areaInt(Heat Flux@)Outlet I've been thinking about it but I don't really understand how to use this function since it uses a reference temperature of the fluid (in my case 20º). Finally, I have already calculated the pressure ratio, etc., but I want to justify those trends with something more than a result, a number. Thanks a lot Last edited by jmenendez; February 14, 2021 at 08:27. |
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February 14, 2021, 04:52 |
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#4 |
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Glenn Horrocks
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No, I am suggesting doing the integral of the heat in and the heat out. So that is something like areaInt(T)@Inlet and areaInt(T)@Outlet. You can get the total heat input to the fluid from this. Other functions like massFlow()@Outlet might help as well.
Note this assumes incompressible, constant properties fluids. If either of those assumptions is incorrect I trust I can leave it up to you to correct it.
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February 14, 2021, 08:38 |
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#5 | |
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Quote:
I'm below Mach 0.4 (except maybe a very small area near the blade), so I don't consider compressibility effects. I have some doubt because CFX-Post has two functions: Wall Heat Flux and Heat Flux. As far as I have been able to find out, the first one is only for walls, since in my inlet it gives 0W while the Heat Flux function 200W (both integrated to the inlet area). Thank you very much for your help, I really appreciate it. |
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February 14, 2021, 17:54 |
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#6 | |
Super Moderator
Glenn Horrocks
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Quote:
You should check this comment out, but I think Wall Heat Flux is heat flux through a wall (obviously), but Heat Flux is heat flux into a control volume through any control volume face. This means a simple flow through the control volume will result in a heat flux in and out due to the convection of the fluid and the heat it contains. Often they will cancel out, but not always.
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February 14, 2021, 18:14 |
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#7 | |
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Quote:
It is curious because the difference between Heat Flux in the inlet-outlet has the same order of magnitude as the energy transformed by the compressor in pressure and speed. I do not know how to calculate then the energy that the compressor has with respect to the adiabatic case. I'm trying several alternatives. One of them is to subtract the compressor power from the Heat FLux between inlet-outlet. These are more rational results But I still don't know how to calculate it. Thank you so much for your help. |
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Tags |
cfx post 19.1, heat balance |
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