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MOTOR ABSORBED POWER 1
- Thread starter duki
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1
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The absorbed power of a motor ( electric input power) is higher than the load brake horsepower.
The excess of power requirements are the motor losses developed due to the currents circulating in the rotor and stator windings, The friction and windage, the magnetic losses in the laminations and structural frame.
The factor that measures the input power is the motor efficiency (EFF) normally in percent.
EFF = (HPshaft output / HPinput) x 100
The excess of power requirements are the motor losses developed due to the currents circulating in the rotor and stator windings, The friction and windage, the magnetic losses in the laminations and structural frame.
The factor that measures the input power is the motor efficiency (EFF) normally in percent.
EFF = (HPshaft output / HPinput) x 100
- Thread starter
- #4
duki, it's the first one, provided that the bkW of load = rated motor power (the power at which the efficiency is defined). The relationship is as given by aolalde's equation.
Note that the "rated motor kw" is always the shaft power (i.e. the motor output power, not the motor input power).
The motor losses are given by:
motor absorbed power - shaft power
= (1-eff) x motor absorbed power.
Note that the "rated motor kw" is always the shaft power (i.e. the motor output power, not the motor input power).
The motor losses are given by:
motor absorbed power - shaft power
= (1-eff) x motor absorbed power.
duki
Your second expression is not correct.
IEEE STD 112 defines efficiency:
Efficiency is the ratio of output power to total input power. Output power is equal to input power minus the losses. Therefore, if two of the three variables (output, input, losses) are known, the efficiency can be determined by one off the following equations:
EFF= output power/input power
EFF= (input power – losses)/input power
EFF= output power/ (output power + losses)
Use the formula that better fits your available data.
You have named the input power as “absorbed power” and the output power as “bKW of load”
Note that the rated or Nameplate kW or HP apply only when the load connected to the shaft matches exactly the name plate power. It is obvious that the units must be consistent, kW or HP.
Your second expression is not correct.
IEEE STD 112 defines efficiency:
Efficiency is the ratio of output power to total input power. Output power is equal to input power minus the losses. Therefore, if two of the three variables (output, input, losses) are known, the efficiency can be determined by one off the following equations:
EFF= output power/input power
EFF= (input power – losses)/input power
EFF= output power/ (output power + losses)
Use the formula that better fits your available data.
You have named the input power as “absorbed power” and the output power as “bKW of load”
Note that the rated or Nameplate kW or HP apply only when the load connected to the shaft matches exactly the name plate power. It is obvious that the units must be consistent, kW or HP.
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