## Power Transfer Stages of Induction Motor :

#### Induction motor converts an electrical power supplied to it into mechanical power. The various stages in this conversion is called as power transfer stages in an induction motor.

#### The 3-phase power input to an induction motor i.e,

#### Stator input,

####
**P**_{in} = √3 V_{L} I_{L} Cos Ï†

**P**

_{in}= √3 V_{L}I_{L}Cos Ï†####
Where V_{L} and I_{L} are the line values of stator supply voltage and current and Cos Ï† is the power factor of the motor.

####

A part of this power is consumed in stator iron and copper losses. The remaining power is transferred inductively to the rotor through the air-gap. This is called as Rotor Input, P_{2}

A part of this power is consumed in stator iron and copper losses. The remaining power is transferred inductively to the rotor through the air-gap. This is called as Rotor Input, P

_{2}#### So,

####
P_{2} = P_{in }-
Stator iron and Copper losses

####
The rotor losses consists of majority of copper losses and a very small rotor iron losses which are generally neglected.

The rotor losses consists of majority of copper losses and a very small rotor iron losses which are generally neglected.

####
By substracting the rotor copper losses from ' P_{2} ', we get the gross mechanical power developed by the motor, P_{m }

####
P_{m} = P_{2 }-
Rotor Copper losses

####
A part of ' P_{m} ' is consumed as mechanical losses and the remaining is the power available to the load at the shaft. This is called as Net Output Power of the Motor, P_{out}

#### The above stages can be shown diagrammatically called as Power Flow Diagram of Induction Motor.

## Relation Between Rotor Input, Rotor Copper Losses and Rotor Output :

#### Let,

####
P_{2} - Rotor input

####
P_{c} - Rotor copper losses

####
P_{m } - Gross mechanical power developed or rotor output

####
T_{g} - Gross torque developed by the rotor in N-m

####
The actual torque available at the shaft called as Shaft Torque or Useful torque, T_{sh}

####
T_{sh} - Gross torque,

####
T_{g} - Torque lost due to the friction and windage losses

####
Now input to the rotor is through the air-gap with the help of rotating magnetic field which is rotating at a speed of ' N_{s} ' rpm.

####

The rotor input can be expressed in terms of gross torque T_{g} and speed as,

####
P_{2} = 2Ï€ N_{s} T_{g}
/ 60 . . . ( 1 )

####
Now torque developed remains same, but the rotor output which is gross mechanical power developed,P_{m } is at a speed ' N ' rpm.

So from output side, we can write

Now torque developed remains same, but the rotor output which is gross mechanical power developed,P

_{m }is at a speed ' N ' rpm.####
**P**_{m} = 2Ï€** N T**_{g} / 60 watt

_{m}= 2Ï€

_{g}/ 60

#### We know that,

#### Rotor copper losses,

####
P_{c} = P_{2} - P_{m}

####
= ( 2Ï€ N
T_{g} / 60 ) ( N_{s} - N ) . . . ( 2 )

#### Dividing equation ( 2 ) by ( 1 ), we get

#### Therefore,

##
P_{c} = sP_{2} . . . ( 3 )

#### So rotor copper losses are slip times the rotor input.

#### Now gross mechanical power developed,

##
P_{m} = P_{2} - P_{c}

##
= P_{2}
- sP_{2}

_{}

#### Therefore,

####
P_{m} = ( 1 -
s ) P_{2} . . . ( 4 )

#### So gross mechanical power developed is ( 1 - s ) times the rotor input,

####

Dividing equation ( 3 ) by ( 4 ), we get

####
**P**_{c}
/ P_{m} = s / 1
- s

_{c}/ P

_{m}= s / 1 - s

#### From the above, it can be concluded that

##
P_{2}
: P_{c} : P_{m } = 1 : s
: 1 - s

## Gross Torque and Shaft Torque :

#### The torque produced by rotor is gross mechanical torque and due to mechanical losses entire cannot be available to drive load.

####

The load torque is net output torque called shaft torque or useful toque.

####

Therefore,

#### Shaft torque

#### Gross torque,

#### where,

####
T_{lost} = Torque lost due to mechanical losses

####
P_{out} = Motor output

####
P_{m} = Mechanical power developed

#### N = Motor speed

## Efficiency of an Induction Motor :

####
The ratio of net power available at the shaft ( P_{out} ) and the net electrical power input ( P_{in} ) to the motor is called as overall Efficiency of an Induction Motor.

####
**% Efficiency = ( P**_{out}
/ P_{in} ) × 100

_{out}/ P

_{in}) × 100