The various losses in the stator of three phase induction motor are stator copper losses and stator iron losses. The stator iron losses consist of eddy current loss and hysteresis loss.

In rotor the losses are rotor iron loss and rotor copper loss. The other losses on the rotor side are friction losses at bearing and slip rings etc and the windage losses.

The stator iron losses depend upon the two factors namely flux density and the supply frequency. The rotor iron loss depends on the frequency of the rotor current. The frequency of the rotor current depends on the slip or speed of the rotor (f

^{’}=sf). The iron losses of the rotor are negligible, since the frequency of the rotor current is very small.**Construction of Three Phase Induction Motor**

**Power stages and various losses in an Induction Motor: **

### Power flow chart of an Induction Motor:

i. Rotor copper loss is determined by using the formula 3I

^{2}_{2}R_{2}.ii. Stator copper loss is calculate by using the formula 3I

^{2}_{1}R_{1}iii. Input to the motor on load is equal to √3V

_{L}I_{L}cosΦ watts.**Slip Torque characteristics of an induction Motor**

### Three Phase Induction Motor Rotor output:

In an induction motor is supplied to the stator winding. The power to the rotor circuit is transferred entirely by induction. The output available at the rotor shaft is mechanical energy.

During the power conversion, certain losses take place at various stages. The rotor output of an induction motor is given as follows:

Stator output = Stator input – stator losses

Stator output = Rotor Input

Rotor Output = Rotor Input – Rotor copper losses

Rotor output = Mechanical Energy

Mechanical Energy ----> Gross Torque T

_{g}_{}

Gross Torque T

_{g}= (useful torque or shaft torque T_{sh}) + (Winding and friction losses in the motor)Rotor output = 2πN

_{r}T_{g}_{}

Where, N

_{r}= actual speed of the motor in r.p.m T

_{g}= Gross torque in NMFrom this T

_{g}= Rotor output in watts / 2πN_{r }--------> 1_{}

For an ideal case, assuming no copper losses in the rotor circuit.

Rotor Output = rotor Input

For this condition, the motor should run at synchronous speed.

Then the torque,

T

_{g}= Rotor Input / 2πN_{s }--------> 2Using the eqn 1 & 2

Rotor output = 2πN

_{r}T_{g }--------> 3Rotor input 2πN

_{s}T_{g }--------> 4_{}

But in actual case

(Rotor input – Rotor output) = Rotor copper losses.

using the eqn 3 & 4

(2πN

_{s}T_{g}- 2πN_{r}T_{g}) = Rotor copper losses.That is Rotor copper losses = T

_{g}. 2π(N_{s}– N_{r}) --------> 5Using the eqn 4&5

Rotor copper losses/Rotor input = T

_{g}2π(N_{s}-N_{r}) / 2πN_{r}sT_{g} = (N

_{s}-N_{r}) / N_{s} = Slip S --------> 6

Thus , from the equation 6, Rotor copper losses equal to

s x Rotor input --------> 7

But Rotor Gross Output = Rotor input – Rotor copper losses

Using equation 7

Rotor gross output = (Rotor input – s) x Rotor input

= (1-s) Rotor input

Or

Rotor gross output/Rotor input = (1-s) --------> 8

Using s, (N

_{s}– N_{r}) /N_{s}Rotor gross output/Rotor input = {1- (N

_{s}– N_{r}) /N_{s}} = (N

_{s}-N_{s}+N_{r)}/ N_{s} =N

_{r}/N_{s}That is,

Rotor gross output/Rotor input = Actual speed of the motor / Synchronous speed

Hence this can be expressed as

Rotor efficiency = N

_{r}/N_{s}_{}

Using the expression (6 and 8), (6+8) gives

Rotor copper losses/Rotor Gross input = s/(1-s)

### Efficiency of an Induction Motor:

The induction motor receives power from the supply mains through the stator windings. The input power is electrical power. The power from the stator winding is transferred to rotor, via air gap by induction principle. The output power is a mechanical power. The mechanical power is obtained at the shaft of the motor to perform mechanical work.

During this power transfer, some losses such as copper losses in stator winding, copper losses in rotor winding, core losses in stator winding, frictional losses and windage losses take place.

Then the efficiency is determined by knowing the input power and the output power.

Efficiency of any machine is the ratio of output power and input power. The efficiency of an induction motor is obtained as

Efficiency η = Output power / input power

= Output power / (Output power + losses)