There are generally 3 power stages of an induction motor
- Stator iron loss which generally consists of hysteresis and eddy losses is dependent on the flux density and supply frequency in the core of iron which is also stated as practically constant.
- Since the frequency of rotor current is always smaller under the normal running conditions, it results into negligible rotor iron loss.
- the total rotor copper loss = 3I22R2
Considering I2 and R2 as respective current and resistance of the rotor
17.1. Rotor output
Stator input = stator output + losses
Rotor input = stator output
(Stator output will entirely inductively transfer to the rotor circuit)
Rotor gross output = rotor input – rotor copper losses.
The rotor output then gives rise to the gross Torque Tg by converting into the mechanical energy. Please note that some of the gross torque is also lost because of the windage and friction loss in the rotor. The rest of the gross torque appears as Tsh as the useful/shaft torque.
If Tg is in N-m and N is the actual speed of the rotor in r.p.m.
17.2. Synchronous watt
A synchronous wall is termed as the value of torque as the synchronous speed of motor. This new unit of torque develops a power of one watt. Let’s consider an example; if the motor is developing a torque of 750 synchronous watts, it reflects that the input of the rotor is 750 watts and the torque which is examined such that the power will be developed by 500 watts. It reflects that the rotor is developing the same torque which synchronized run.
It shows that the synchronous wattage is equal to the power transferred to the rotor across air-gap of an induction motor.
17.3. Factors determining torque
Let’s assume that a 3-phase induction motor is operated with slip‘s’ on a certain load, then
Induced electromotive force (e.m.f) per phase in rotor = sF2 volts
(Where E2 = standstill rotor induced e.m.f.)
Rotor impedance per phase, Z2= R2 + jsX2
(Where R2 and X2 are rotor resistance and standstill reactance per phase)
Example 1. The power input to a 3-phase induction motor is 60 k W. The stator-losses total 1.5 kW. If the motor is running with a slip of 5 per cent, find the total mechanical power developed,
Solution. Stator power input = 60 kW
Stator losses = 1.5 kW
Slip , = 4% (= 0.04)
Mechanical power developed:
Rotor input = stator output = stator input – stator losses
= 60 – 1.5 = 58.5 kW
Now mechanical power developed
= (1 – s) rotor input
= (1- 0.04) × 58.5
= 46.16kW. (Ans.)
Example 2.A 60 H.P., 6-pole, 60 Hz, slip-ring induction motor with a rotor current of 50 A is running on full Load at 960 r.p.m. Allowing 400 W for copper loss in the short-circuiting gear and 1500 W for mechanical losses, find the resistance R2 per phase of the 3-phase rotor winding.
Solution. Pout = 60 H.P. = 60 × 735.5 = 44130 W ;
p = 6 ;
f = 60 Hz ;
N = 980 r.p.m. ;
I2 = 50 A ;
copper loss in the short-circuiting gear = 400 W ; mechanical loss = 1500 W.
Rotor output = motor output + short-circuiting gear loss + mechanical losses
= 44130 + 400 + 1500 = 46030 W