The Torque of an Induction Motor

The Torque of an Induction Motor

An expression for the torque of an induction motor may be deri­ved from the equation of the mechanical power, P_m , available at. its shaft

T = P_m/ɷ_rot                              (14.22)

where ɷ_rot is the angular velocity of the rotor.

Since the angular velocity of the rotor is

ɷ_rot = 2πn/60

it. follows that the angular velocity of the magnetic field of a motor, ɷ_f , and that of its rotor, ɷ_rot are connected by the slip

s = (n_l - n)/ n_l = (ɷ_f - ɷ_rot)/ɷ_f

or

ɷ_rot = ɷ_f (1 - s)

The angular velocity of the rotating magnetic field, ɷ_f , is equal to the angular frequency of the sinewave current in the stator phase windings, 00, for a two-pole motor (p = 1). In the general case of a multipole motor the angular velocity of the rotating magnetic field is

ɷ_f = ɷ/p

where p is the number of pole pairs.

On expressing ɷ_rot in terms of W in Eq. (14.22), we get

T = P/(1-s)*P_m                      (14.23)

From the equivalent circuit of a motor phase, Fig 18 , it fol­lows that the mechanical power at the. shaft is

Substituting Eq. (14.24) into Eq. (14.23) yields an expression for the torque. of an induction motor

T = m₂pr_w2I²₂/ɷs (14.25)

Since, on the basis of the phasor diagram in Fig 15

rw₂I₂ = E₂ cos φ₂

and

E₂/s = E₂,_st

it follows that

T = E_2,stI₂ (m₂P/ɷ) cos φ₂                                               (14.26)

On recalling Eq. (14.12),

E_2,st =4.44fw₂k_w2Φ_r = ɷw₂k_w2Φ_r/√2

Eq. (14.26) may be re-cast as follows

T =(1/√2) m₂Pw₂k_w2Φ_rI₂ cos φ₂=const X Φ_rI₂ cos φ₂                                  (14.27)

That is, the torque of an induction motor is proportional to the pro­duct of the rotating magnetic flux and the current in the rotor win­ding.

Labels: Electricity and Magnetism