Induction Motors: Power Consumed vs Power Delivered

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SUMMARY

The power consumption of a three-phase induction motor is calculated using the formula P = sqrt(3) * V * I * PF. For a motor operating at 460 volts and 30 amps with a power factor (PF) of 0.82, the power consumption is 19.6 kW. However, this value represents the electrical power consumed, not the mechanical power delivered to the motor's shaft. The difference arises due to losses such as iron losses, copper losses, and mechanical inefficiencies, which means the mechanical output power will always be less than the electrical input power.

PREREQUISITES
  • Understanding of three-phase induction motor operation
  • Familiarity with electrical power calculations
  • Knowledge of power factor (PF) and its significance
  • Basic concepts of motor efficiency and losses
NEXT STEPS
  • Research the efficiency ratings of different induction motors
  • Learn about the types of losses in induction motors, including iron and copper losses
  • Explore the relationship between mechanical power output and electrical power input
  • Study how to improve the power factor in industrial applications
USEFUL FOR

Electrical engineers, motor design specialists, and anyone involved in the operation and optimization of induction motors will benefit from this discussion.

Bergenheimer
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Hello

The equation for power consumption of a three phase induction motor is given by the equation:

P = sqrt(3)*V*I*(PF)

Does this differ from the power that is delivered to the motor. For example, if a motor is running at 460 volts, 30 amps with a PF of 0.82, then the power consumption is

P = sqrt(3)*460*30*.82 = 19.6 kW

What is the corresponding power being delivered to the motor? is it the same value? Just want to know if the power being delivered to the motor is different than the power the motor actually consumes.



Bergenheimer
 
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Hi
One basic question, how it would be possible, that power delivered wouldn't be consumed ?

Equations you are writing here are describing power delivered = consumed (consumed in many ways - iron losses, copper losses, friction, ventilation, mechanical power output).

There is one more thing: difference between power output (mechanical power on shaft) and power input (your equations). Maybe you are referring to this ?
 
Hello Berg:

Your OP - refers to "power consumed" and "Delivered to", these are the same thing and referring to the electrical circuit - a matter of perspective, from the sources perspective or from the motors perspective. It is the Mechanical Power (shaft) output of the motor that will not equal these values due to losses in the motor - both electrically and mechanically. The better the motor - the better the efficiency etc.
 

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