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 = Pm/ɷrot (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 = m2prw2I22/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 = 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.