Saturday, February 2, 2013

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

The Torque of an Induction Motor

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.