## Power Transfer Stages of Induction Motor :

An induction motor converts electrical power supplied to it into mechanical power. The various stages in this conversion are called power transfer stages in an induction motor. The 3-phase power input to an induction motor i.e, stator input is,

*P*

_{in}= √3 V_{L}I_{L}Cos Î¦Where V_{L} and I_{L} are the line values of stator supply voltage and current and Cos Î¦ 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 rotor input P_{2}. Therefore,

*P*

_{2}= P_{in}- stator iron and copper lossesThe rotor losses consist of the majority of copper loss and very small rotor iron loss which are generally neglected. By subtracting 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 lossesA 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 is called Shaft Torque or Useful torque T_{sh},- T
_{sh} = Gross torque - T
_{g} = Torque lost due to the friction and windage losses.

_{2}= Rotor input_{c}= Rotor copper losses_{m}= Gross mechanical power developed or rotor output_{g}= Gross torque developed by the rotor in N-m._{sh}= Gross torque_{g}= Torque lost due to the friction and windage losses.Now input to the rotor is through the air gap with the help of a 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,

Now torque developed remains the same, but the rotor output which is gross mechanical power developed P_{m} is at a speed 'N' rpm. So from the output side, we can write,

#### We know that, rotor copper losses, Dividing equation 3 by 1, we get, So rotor copper losses are slip times the rotor input. Now gross mechanical power developed, So gross mechanical power developed is (1 - s) times the rotor input. Dividing equation 4 by 5, we get, From the above, it can be concluded that,

## Gross Torque and Shaft Torque :

The torque produced by the rotor is gross mechanical torque and due to mechanical losses entire cannot be available to drive the load. The load torque is net output torque called shaft torque or useful torque. Therefore, shaft torque is given as,

#### Gross Torque is given as,
where,- T
_{lost} = Torque lost due to mechanical losses - P
_{out} = Motor output - P
_{m} = Mechanical power developed - N = Motor speed.

_{lost}= Torque lost due to mechanical losses_{out}= Motor output_{m}= Mechanical power developed## 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.