Induction Motor Slip decrease with frequency decrease for same load.

AI Thread Summary
The slip in an induction motor decreases with frequency under the same load due to the relationship between frequency and rotor speed, where lower frequency results in lower slip. At higher frequencies, no-load losses such as iron losses, ventilation, and bearing friction are more pronounced, contributing to higher slip. Windage losses are influenced by the square of the speed, meaning that as frequency decreases, these losses also decrease. The no-load current, primarily reactive, is about 30% of the rated current, and copper losses can rise with increased no-load conditions. Overall, with less power required to maintain rotor rotation at lower frequencies, the slip decreases even if the torque remains constant.
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Hi,

Does anyone know or could help me understand the reasoning for the amount of slip within an induction motor decreasing with frequency under the same load conditions? I.e. For no external load, the slip is higher at 50Hz than at 20Hz etc? Thanks
 
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What about internal load, ie bearing friction and windage loss?

Is windage proportional to speed2 or to speed3 ?
 
“No-load” it does not mean “no load at all”. There are no-load losses –iron losses, ventilation and friction-bearing losses and copper losses. The no-load current-mainly reactive-is about 30% of rated and copper [winding] losses could be elevated. As jim hardy already said, ventilation and friction are proportional with square of rpm, so less frequency less rpm [rpm=60*freq/p where p= no.of pole pairs] less slip. The iron losses decrease with the frequency and so the no-load current and in turn no-load copper losses. The mechanical power requested in order to maintain the rotor in rotation will decrease even if the torque remains constant P=T*(2*PI()*Freq:
P=T\omega=2\pi_fT
Less power less slip.
 
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