Induction Motor Slip decrease with frequency decrease for same load.

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SUMMARY

The discussion centers on the relationship between induction motor slip and frequency under constant load conditions. It is established that slip decreases as frequency decreases, with specific reference to the effects of no-load losses, including iron losses, ventilation, and bearing friction. The slip is higher at 50Hz compared to 20Hz due to the proportional relationship of ventilation and friction losses to the square of RPM. Additionally, the mechanical power required to maintain rotor rotation decreases with lower frequency, resulting in reduced slip.

PREREQUISITES
  • Understanding of induction motor principles
  • Knowledge of slip and its calculation
  • Familiarity with no-load losses in electric motors
  • Basic grasp of power equations in rotational systems
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  • Research the impact of frequency on induction motor performance
  • Study the relationship between slip and load conditions in induction motors
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  • Explore the effects of bearing friction and windage on motor efficiency
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Electrical engineers, motor control specialists, and anyone involved in the design or optimization of induction motors will benefit from this discussion.

Physicist3
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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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