Principle of Operation of 3ϕ Induction Motor

PRINCIPLE OF OPERATION OF 3Φ INDUCTION MOTOR

Three phase supply is given to the stator windings. Due to this, current flows through the stator winding. This current is called stator current. It produces a rotating magnetic field in the space between stator and rotor. Synchronous speed is produced by this field.

N_s = 120ƒ/P

The rotating magnetic field cutting the rotor conductors, an emf is induced in the rotor. If the rotor winding is shorted then the induced emf produces current. this current produces a rotor field.

The interaction of stator and rotor field develops torque. Then the rotor rotates in the same direction as the rotating magnetic field. When the rotor is standstill, the frequency rotor emf is equal to the supply frequency.

As the rotor speed increases, the frequency of rotor emf and the magnitude of rotor emf decreases.

The rotor tries to each up with the rotating magnetic field. But the rotor could not rotate at the synchronous speed. Therefore, the rotor runs at a speed slightly less than the synchronous speed.

The difference between synchronous speed and rotor speed is called the slip speed.

Slip speed = N_S - N

Slip (S) = N_S – N/ N_S

N = N_S (1-S)

% Slip = N_S – N/N_S × 100

At no load the difference between synchronous speed and rotor is only about 1%

Squirrel cage induction motor

Advantages

Cheaper, light weight, higer efficiency, less maintenance

Disadvantages

Moderate starting torque, there is no external resistance in the rotor circuit, so the starting torque cannot be controlled.

Applications

Lathes, drilling machines, fans, blowers, water pumps, grinders, printing machines etc

Slip ring induction motor

Advantages

Starting torque can be controlled by varying the rotor circuit resistance, speed is also controlled

Disadvantages

Heavier in size, high cost, high rotor inertia, high speed lamination, maintenance and reliability problems due to brushes and slip rings.

Applications

Lifts, hosts, cranes, elevators, compressors etc

Example: 21

A 6 pole 3ϕ, induction motor is connected to 50 Hz supply. If it is running at 960 rpm, find the slip.

P = 6, f = 50 Hz, N = 960 rpm

N_S = 120ƒ/P = 120 × 50/6 = 1000 rpm

Slip, S = N_S – N/ N_S × 100

=1000–960/1000 × 100

= 0.04 (or) 4%

Example: 22

A 2 pole, 3ϕ, 50 Hz induction motor is running on no load with a slip of 4% calculate (i) the synchronous speed (ii) speed of the motor.

P = 2, f = 50 Hz, S = 4% (or) 0.04

 (i) Synchronous speed

N_S = 120ƒ/P = 120 × 50/2 = 3000 rpm

(ii) Speed of the motor

N = N_S (1-S)

= 3000 (1 - 0.04)

N = 2880 rpm

Frequency of rotor current or emf

When the rotor is stationary, the relative speed between the rotor winding and the rotating magnetic field is N_S. Hence the frequency of emf induced and the resultant current is PN_S /120 that is same as the supply frequency (ƒ). As the rotor speed increases, the relative speed is (N_S - N) and the rotor frequency is

Slip S = N_S – N / N_S

SN_S = N_S - N = S(120ƒ/P)

ƒ_r = Sƒ

Example: 23

A 4 pole, 3ƒ induction motor operates from a supply whose frequency is 50 Hz. Calculate.

i) The whose speed at which the magnetic field of the stator is rotating

ii) The speed of the rotor when the slip is 0.04

iii) The frequency of the rotor currents when the slip is 0.03

iv) The frequency of the rotor currents at stand still.

P = 4, ƒ = 50 Hz

(i) Speed of the stator

N_S = 120ƒ/P = 120×50 / 4 = 1500 rpm

(ii) Rotor speed at slip 0.04

N = N_S (1-S)

= 1500 (1 - 0.04)

= 1440 rpm

(iii) Frequency of rotor current at slip = 0.03

ƒ_r = Sƒ

= 0.03 × 50

= 1.5

(iv) Frequency at standstill

S = 1

ƒ_r = Sƒ = 1 × 50