Temperature Impact on LRC 1

[enirwin]

Is there a simple way to estimate the impact of winding temperature on the locked rotor current of an AC induction motor? I know the locked rotor current at room temperature and need to estimate it at colder temperatures. I don't need an exact prediction, just a ballpark estimate.

Thanks.

 

[dpc]

For locked rotor current, you can assume a power factor of about 0.2. From this you can calculate the approximate X and R values of the locked rotor impedance.

The only thing that will change with temperature (first-order effects) will be the resistance.

So based on aluminum or copper windings, you can use the appropriate temperature coefficient to estimate the decrease in resistance. Then calculate a new Z. I don't think it will have much effect.

 

[dpc]

And it's going to heat up real fast anyway.

 

[enirwin]

dpc,

It didn't occur to me that the resistive portion of the impedance would be small and therefore, even though the resistance might change quite a bit, the impact on the current would be small. Very helpful.

Thanks.

 

[7anoter4]

Hi enirwin
I agree with dpc that theoretically you may calculate the starting resistance R=U/Ist*cosFI.
and recalculate Rcold=Rst/(1+alpha*(T-To)).
And also I agree with dpc that this not an appropriate way to calculate cold R.
Rst depends on its own temperature not only on room temperature.
As for starting short time the heat dissipation from conductor to exterior will be very small,
almost all the heat produced in the conductors will increase the the resistance.
So Rst(1+alpha*(T-To)) formula will be more complicate as T will depends upon R*I^2 .
Also, both Resistance and X are depended on rotor current frequency as at the start will be 60 or 50 HZ and at synchronism will be 0 [theoretically at least].
For squirrel cage [usually made of deep bar of copper or aluminum] this change will be dramatically so R will decrease up to 1/3 or1/4 very fast.
and X =2*PI*freq.*L will not decrease as expected but as L[inductance] will increase more the result will be an increasing of X.
So I am not sure that your measurement will be so accurate.
Best Regards