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, Pm , available at. its shaft

T = Pmrot                              (14.22)

where ɷrot is the angular velocity of the rotor.

Since the angular velocity of the rotor is

ɷrot = 2πn/60

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

s = (nl - n)/ nl = (ɷf - ɷrot)/ɷf

or

ɷrot = ɷf (1 - s)

The angular velocity of the rotating magnetic field, ɷf , 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

ɷf = ɷ/p

where p is the number of pole pairs.

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

T = P/(1-s)*Pm                      (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 = m2prw2I22s (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 = E2,stI2 (m2P/ɷ) cos φ2                                               (14.26)

On recalling Eq. (14.12),

E2,st =4.44fw2kw2Φr = ɷw2kw2Φr/√2

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

T =(1/√2) m2Pw2kw2ΦrI2 cos φ2=const X ΦrI2 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.