Torque and Power Factor of an induction motor

TORQUE AND POWER FACTOR

Torque of an induction motor is proportional to the product of flux per stator pole, rotor current and the power factor of the rotor.

Τ α ϕ I₂ cosϕ₂

T = K ϕ I₂ cosϕ₂

I₂ - Rotor current at standstill

ϕ₂- Angle between rotor emf and rotor current

K - Constant

Ε₂ α ϕ

T α E₂ I₂ cos ϕ₂

T = K₁ E₂ I₂ cos ϕ₂

Starting Torque

The torque developed by the motor at the instant of starting is called starting torque.It is greater than the normal running torque.

Let

E₂ - Rotor emf/phase at standstill

R₂ - Rotor resistance/phase

X₂ - Rotor reactance/phase at standstill

Z₂ = √ R₂² + X₂² - rotor impedance/phase at standstill.

Starting torque, T_st = K₁ E₂ I₂ cos ϕ₂

If supply voltage V and ϕ is constant

E₂ is constant

K₂ is another constant

K₁ = 3/2πN_S

N_S → Synchronous speed in rps

Condition for maximum starting torque

If rotor resistance equals rotor reactance the starting torque is maximum.

Effect of change in supply voltage on stating torque.

E₂ α V

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K₃ is another constant

T_st α V²

Rotor emf under running conditions

Let

E₂ - Standstill rotor emf/phase

X₂ - standstill rotor reactance/phase

ƒ₂ - rotor current frequency at standstill

When rotor is stationary, S = 1, the frequency of rotor emf is same that of the stator supply frequency. At standstill emf induced is maximum.

When rotor starts running the relative speed and the stator flux is decreased.

Under running condition E_r = SE₂

Frequency of the emf, ƒ_r = Sƒ₂

Rotor reactance, X_r = SX₂

Torque under running condition

T α E_r I_r cos ϕ₂

T α ϕ I_r cos ϕ₂

E_r Rotor emf/phase under running condition

I_r - Rotor current/phase under running condition

E_r = SE₂

Consdition for maximum torque under running conditions

Torque under running condition is maximum at that value of slip S, makes the rotor reactance per phase equal to rotor resistance per phase

R₂ = SX₂

Example : 24

A 3ƒ slip ring induction motor with star connected rotor has an induced emf of 120 V between slip rings at standstill with normal voltage applied to the stator. The rotor winding has a resistance of 0.3W** and standstill leakage calculate

 (i) Rotor current/phase when running short circuited with 4% slip.

(ii) The slip and rotor current per phase when the rotor is developing maximum torque.

i) Induced emf/phase, E_r = SE₂

= 0.04 (120/√3) = 2.77 V

Rotor reactance/phase, X_r = SX₂

= 0.4 × 1.5 = 0.06 Ω

Rotor impedance/phase = √(0.3)² + (0.06)²

= 0.306 Ω.

Rotor current/phase = 2.77/0.306 = 9A

ii) For developing maximum torque

R₂ = SX₂

S = R₂/X₂ = 0.3/1.5 = 0.2

X_r = 0.2 x 1.5 = 0.3 Ω

Z_r = √(0.3)² + (0.3)²

= 0.42 Ω

E_r = SE₂ = 0.2 × 120/√3 = 13.86 V

Rotor current/phase = 13.86/0.42 = 33A.

Torque speed curve

The torque developed by a conventional 30 motor depends on its speed. Fig.3.93 shows the torque speed curve.

From the fig.3.93 the stating torque (at N = 0) is 1.5 T and the maximum torque also called breakdown torque. If the load torque exceeds 2.5T motor will suddenly stop