__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} = 2π*n*/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* = *m*_{2}*pr*_{w2}*I*^{2}_{2}/ɷ*s* (14.25)

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

*r*w_{2}*I*_{2} = *E*_{2} cos φ_{2}

and

*E*_{2}/*s* = *E*_{2},_{st}

it follows that

*T* = *E*_{2,st}*I*_{2} (*m*_{2}*P*/ɷ) cos φ_{2 }(14.26)

On recalling Eq. (14.12),

*E*_{2,st }=4.44*fw*_{2}*k*_{w2}Φ_{r} = ɷ*w*_{2}*k*_{w2}Φ_{r}/√2

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

*T *=(1/√2) *m*_{2}*Pw*_{2}*k*_{w2}Φ_{r}*I*_{2} cos φ_{2}=const X Φ_{r}*I*_{2} cos φ_{2 } (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.