Efficiency of induction motorlosses of induction motor
Losses of induction motor:
To talk about losses of induction motor we talk about constant(fixed) losses and variable losses:

Constant (fixed) losses:
From the name, fixed losses remain constant over the normal working range of the motor; We can determine these losses by the noload test of the induction motor; Fixed losses also subdivided into:
1 Iron (core) losses: core losses of the stator consist of eddy current losses and hysteresis losses that dependent on frequency and voltage of the supply. The supply frequency is constant so the stator core losses are constant.
We can minimize the eddy current losses by using lamination on the core which decreases the area so the resistance increases and the eddy losses decrease; We can also minimize hysteresis losses by using highgrade silicon steel
2 Mechanical losses: we can consider mechanical losses constant losses for a giving running condition; These losses include friction and windage losses.
3 Brush friction losses: which in general are proportional to the rotor speed.

Variable losses:
Variable losses or copper losses occurred due to the current flowing in stator and rotor windings; When the load changes the current flowing in the stator and the rotor changes so the losses changes. We can determine these losses by the blocked rotor test of the induction motor.
Power flow diagram of induction motor:
That’s good, but it would be better to deep in details of losses and that happens with the help of the power flow diagram;
The power flow illustrates the input power to the stator:
Pis= √3 .VL.IL.cos Φi =3.VSP.ISP.cos Φi.
While:
VL: The line voltage supplied to the stator of the motor.
IL: The line current.
Cos Φi: input power factor.
And the losses in the stator:
 Stator copper losses (losses in the stator winding resistance): I^2.R.
PSCL = 3.Isp^2*Rsp.
 Stator core losses (hysteresis and eddy current losses in the stator core).
Ps (h+e)
We have an output power of the stator:
Pos=PisPscPs (h+e)
The air gap between the stator and the rotor transfers the output power of the stator to the rotor. Hence we have air gap power (Pg).
So:
The power output of the stator= air gap power= input power to the rotor.
Pos= Pg= Pir.
When we enter the rotor there are:
 Rotor copper losses (losses in the rotor resistance): I^2.R.
Prc= 3.I2^2*R2.
 Rotor core losses (hysteresis and eddy current losses in the rotor core):
Pr (h+e)
 Friction and windage losses (Pfw).
 Stary load losses (Pmisc): which contain any other losses as losses due to the harmonic field.
 Developed mechanical power (Pmd).
Developed mechanical power= rotor input rotor copper loss.
Pmd=Pir– Prc = Pg– 3I2^2*R2.
And the output power of the motor will be:
Po= Pmd– Pfw– Pmisc.
This output power is also known as useful power or shaft power.
 Rotational losses: they are additional losses produced at starting and during acceleration; They are the sum of the friction, core, and windage losses and we can say they are constant losses with the change of speed.
Prot= Pfw+ P(h+e)+ Pmis.
So the final output power of the motor will be:
Po= Pmd– Prot = Pmd– Pfw– P(h+e)– Pmis.
Efficiency of induction motor:
We care about losses of induction motor because they are our entrance to the efficiency of induction motor; The net efficiency of any motor is the ratio between the input power and the output power;
Ƞ = output power/ input power.
= output power/ (output power+losses).
Which means that the losses play a vital role in estimating efficiency; The efficiency of induction motor varies between 85% to 97% as the losses are:
 Fiction and windage losses 5_15%.
 Iron (core) losses 15_25%.
 Stator losses 25_40%.
 Rotor losses 15_25%.
 And stray losses 10_20%.
Take care, the efficiency also varies with the motor size; The efficiency increases when the motor gets bigger; Small motors don’t have high efficiency; Motors in many kilowatts have an efficiency above 90%; To have a high efficiency we need high cost.
To achieve the maximum efficiency of induction motor the rotor should almost run as fast as the magnetic field in the stator to have a slip close to zero.
I think it would be better to have a short example:
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