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Average thermal conductivity (?) 
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#1
Jan913, 05:29 AM

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Hey all,
i'm trying to set up a thermal model of a permanent magnet synchronous electric motor. This by doing a lumped system analysis of the motor. In the thermal network that I've set up, I'm trying to introduce an equivalent thermal conductivity value (W/mK) for the copper conductors plus their insulation. This way I can introduce one thermal resistance in the network for the conductors in the stator. The thermal conductivity of the insulation is 0.22 W/mK and the typical thermal conductivity of copper is 390 W/mK. Is there a way to get an average thermal conductivity value from these? Can I simply add them up and devide by 2? please help, regards, Kyo. 


#2
Jan913, 08:12 AM

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PF Gold
P: 12,252

I don't understand the basis for your model. There is no thermal equivalent to Reactance, and this is quite relevant to the behaviour of a motor.
The only valid analog between thermal and electrical systems would, afaik, involve thermal and electrical resistances. 


#3
Jan913, 09:48 AM

P: 5

I do believe I mentioned the term Thermal Resistance.....
But yeah, I'm building a thermal resistance network to model the electric motor. The thermal resistance of a cylinder is given as: R_th= LN(R/r)/k*2∏*L i'm considering 1/18 of the stator, and therefore i have to introduce a correction value, which is known to me, the only thing i miss right now is a thermal conductivity that can be plugged in. So again my question is, can I determine a equivalent thermal conductivity value from the given two.... 


#4
Jan913, 10:07 AM

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PF Gold
P: 12,252

Average thermal conductivity (?)
Sorry  I may have got the wrong end of the stick here. You want to calculate how hot the motor will get. It's an electric model of a thermal system  not specifically an electric motor.
It looks like you have a power source (hot wires) and a number of series thermal resistive elements with a radiative element, finally. Thinking aloud, as it were . . . .That, at its simplest, would be a current source in series with two series electrical resistances. This would represent the thermal power dissipated internally (a fixed amount, hence the current source), a resistor for the conduction path and a resistor representing the radiative path  assuming a sink at 300K. At equilibrium, the coil temperature and the motor surface temperature would correspond to the voltages at the current source terminal and the junction of the two resistors. The 'radiation' resistor would be nonlinear and would follow the Stefan Boltzman law (4th power of the temperature difference). 


#5
Jan913, 10:56 AM

P: 5

We're almost on common ground now! I want to determine the thermal "bottlenecks" within the electric motor. This by doing a lumped system analysis of the motor, e.g. determining all the different thermal resistances of the different parts of the motor. I choose the copper winding cores as the primary heat source, so that the motor heats up from the windings. The heat flow is equivalent with the copper loss term. So that would mean P=I^2 * R, which is the current source in the model ( one of many, every loss term is equal to a current injection within the thermal resistance network).
So yes i do agree with your story. BUT, back to the question, to determine the thermal (equivalent) resistance of the windings, including their insulation within the stator slots, I'd like to introduce only ONE thermal resistance. This means I want to bypass the fact that there are 780 different conductors running through 1/18 th of the motor, with each of those 780 conductors/wires their own insulation. To bypass this, I need to know if I can determine a thermal conductivity term (k) which is a "educated guess" of the complete thermal conductivity value of copper+insulation. I just want to simplify the winding part of the thermal model, so all the other resistances, e.g. radiation and convection, plus conduction to the rotor part of the motor are all evaluated. 


#6
Jan1013, 03:52 AM

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PF Gold
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You have picked a difficult one here, I think!
Copper is a good conductor (thermal) and I don't reckon you'll get a big temperature gradient between inside and outside of the windings  at least it won't dominate. The iron armature and case will be relevant but the biggest unknown, I should have thought, will be thermal path across the air gap between armature and field magnet assembly. Bearing in mind that there will be air flow as the armature rotates, this must be a massive unknown in the equation. Motors are usually cooled by the fan, fixed to the shaft and there is significant heat carried away that way  as you can feel by the warm air blown out of a vacuum cleaner or heavy drill. All that will constitute a parallel path in an electrical analogue. I guess you would need to do a lot of actual measurements if you want realistic numbers for all the variables in this. Motor designers must have tables of data for this. 


#7
Jan1013, 04:36 AM

P: 5

About the current injections, I believe I should mention I'm studying the motor at a steady state situation. Take the heating of the bearings due to friction, or iron losses in the stator and rotor. These losses occur on different nodes within the thermal network, therefore I introduced different current injections within the system. Each of these current sources are grounded. And of course I agree that there are many different resistances in parallel.
About the airgap, with the use of the Nusselt and Taylor numbers, the evaluation of the airgap simplifies tremendously. A few different equations are known to evaluate this part of the model. I'll add the simpified network I'm using/adjusting at this very moment. But back to the question again, are you saying that I should only see the copper as a conductor without a thermal resistance within the evaluated network? If so, is this because of the small temperature gradient that occurs over the copper? Or is there some other reason that I'm missing? 


#8
Jan1013, 04:47 AM

P: 5

i forgot to add the image, so here it is.
This is not the exact used thermal network at this point, I'm not allowed to show all of my work because the company I'm working for at the moment ordered it to be classified. 


#9
Jan1013, 05:51 AM

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PF Gold
P: 12,252

I can see this is getting too complicated for me to be able to contribute much, I'm afraid. The principle is straightforward enough but there are so many details involved. I see that you want to involve those other heat sources, too.
I did have a root around for stuff on transformer design  relating to the temperature distribution in the copper windings. I found this link, which may be interesting or not. He is using a similar approach to yours, 


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