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mf and mef huge difference

Posted 18 janv. 2013, 06:08 UTC−5 Low-Frequency Electromagnetics Version 4.3a 5 Replies

Julien Givernaud
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Hi,

I want to compare mf and mef physics in order to know which one I have to use for my study.
I work with a coil and a conductive core. I'm interesting in the electromagnetic power dissipated in conductive elements mf.Qh and mef.Qh respectively.
Under same excitation set up (current surface like in cold crucible tutorial) I obtain a huge difference between values of dissipated power in the core: mf.Qh=12.14 kW and mef.Qh=3.7 kW
Mesh is the same for the 2 physics and increasing mesh density has no significant impact on results.

Magnetic fields seems to be very different in the 2 cases, i join a picture of streamlines

I attach as well my simple model.

Can someone can explain this huge difference?

Thank you
5 Replies Last Post 18 janv. 2013, 10:57 UTC−5
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 18 janv. 2013, 08:37 UTC−5
Hi

check the dependent variables: in MF it's the magnetic filed via the magnetic vector potential A, MEF adds in also the Electric Potential = Voltage (hence more degrees of freedom and heavier to solve)

If you study the induced currents, you would probably need the voltage for solving correctly, but in MEF you can add an "Amper law's" alone, skipping the current conservation or Electric potential part for some domains, check the equations

--
Good luck
Ivar
Hi check the dependent variables: in MF it's the magnetic filed via the magnetic vector potential A, MEF adds in also the Electric Potential = Voltage (hence more degrees of freedom and heavier to solve) If you study the induced currents, you would probably need the voltage for solving correctly, but in MEF you can add an "Amper law's" alone, skipping the current conservation or Electric potential part for some domains, check the equations -- Good luck Ivar
Julien Givernaud
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Posted: 1 decade ago 18 janv. 2013, 09:30 UTC−5
Thank you Ivar for your answer,

In fact I don't understand why I have to add a new Ampères law because in mef there is already by default the Ampères law and current conservation in the first node.

Can you explain to me why I have to add the same equation in a new node?

I try what you say in adding a new Ampères law which overriddes the first on the core domain, it give some crazy results extremely sensitive to the mesh density (I use boundary layers), I think no realistic results.

Thank you
Thank you Ivar for your answer, In fact I don't understand why I have to add a new Ampères law because in mef there is already by default the Ampères law and current conservation in the first node. Can you explain to me why I have to add the same equation in a new node? I try what you say in adding a new Ampères law which overriddes the first on the core domain, it give some crazy results extremely sensitive to the mesh density (I use boundary layers), I think no realistic results. Thank you
Julien Givernaud
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Posted: 1 decade ago 18 janv. 2013, 10:01 UTC−5
I have tried a new excitation type;

-10 V terminal for mef
-10 V single turn coil domain for mf

It gives me the same results in power dissipation in the core for the 2 physics.
I don't understand what's happens really but it works,

thank you Ivar
I have tried a new excitation type; -10 V terminal for mef -10 V single turn coil domain for mf It gives me the same results in power dissipation in the core for the 2 physics. I don't understand what's happens really but it works, thank you Ivar
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
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Posted: 1 decade ago 18 janv. 2013, 10:11 UTC−5
Hi

you do not need to, but if you have i.e. some air without any conductivity, you cannot get a "current conservation" variable running correctly with 0 conductivity.
Then its better to add a simple "Ampere law" node and select only the air domain (and leave the rest in Ampere law with current conservation), and solve in MEF with V there where the current can flow and as MF where there is no current flowing

Check carefully the equation tab (not COMSOL equation node) for ampere law and Ampere law with current conservation

--
Good luck
Ivar
Hi you do not need to, but if you have i.e. some air without any conductivity, you cannot get a "current conservation" variable running correctly with 0 conductivity. Then its better to add a simple "Ampere law" node and select only the air domain (and leave the rest in Ampere law with current conservation), and solve in MEF with V there where the current can flow and as MF where there is no current flowing Check carefully the equation tab (not COMSOL equation node) for ampere law and Ampere law with current conservation -- Good luck Ivar
Julien Givernaud
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Posted: 1 decade ago 18 janv. 2013, 10:57 UTC−5
When I add an Ampère law on the gas domain results change
it gives more power dissipated in core with mef, around 50 % more, and about the same power dissipated in the coil
Thank you again
When I add an Ampère law on the gas domain results change it gives more power dissipated in core with mef, around 50 % more, and about the same power dissipated in the coil Thank you again

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