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Induction motor working and torque equation

Short answer

The stator's rotating magnetic field induces rotor currents; interaction produces torque T = K s E2^2 R2 / (R2^2 + (sX2)^2). Near synchronous speed T ∝ s (linear); at high slip T ∝ 1/s. The motor runs stably only in the linear low-slip region.

How does three phase induction motor work? Derive the torque equation and show the relationship between speed and torque.

A 3-phase induction motor works on electromagnetic induction: the stator's rotating field induces current in the rotor, and the interaction produces torque.

Working

  1. A balanced 3-phase supply to the 3-phase stator winding produces a rotating magnetic field of constant magnitude moving at Ns=120f/PN_s = 120f/P.
  2. The field cuts the rotor conductors and induces emf in them.
  3. The rotor circuit is closed, so rotor current flows.
  4. Current-carrying conductors in the field experience force; by Lenz's law the rotor turns in the direction of the field to reduce relative motion.
  5. The rotor settles at speed N<NsN < N_s; slip s=(Ns−N)/Nss = (N_s - N)/N_s. If N=NsN = N_s, no emf, no torque.

Torque equation

At slip ss: rotor emf =sE2= sE_2, rotor reactance =sX2= sX_2.

I2=sE2R22+(sX2)2,cos⁡ϕ2=R2R22+(sX2)2I_2 = \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}, \qquad \cos\phi_2 = \frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}}

Torque T∝ϕI2cos⁡ϕ2T \propto \phi I_2\cos\phi_2 and E2∝ϕE_2 \propto \phi:

T=k ϕ I2cos⁡ϕ2=k′E2⋅sE2R22+(sX2)2⋅R2R22+(sX2)2=K s E22R2R22+(sX2)2\begin{aligned} T &= k\,\phi\,I_2\cos\phi_2 \\ &= k' E_2 \cdot \frac{sE_2}{\sqrt{R_2^2 + (sX_2)^2}}\cdot\frac{R_2}{\sqrt{R_2^2 + (sX_2)^2}} \\ &= \frac{K\, s\, E_2^2 R_2}{R_2^2 + (sX_2)^2} \end{aligned}

Using air-gap power, T=Pagωs=3I22R2/s2πNs/60T = \dfrac{P_{ag}}{\omega_s} = \dfrac{3I_2^2R_2/s}{2\pi N_s/60}, so K=32πNs/60K = \dfrac{3}{2\pi N_s/60}.

Starting torque (s=1s = 1): Tst=KE22R2R22+X22T_{st} = \dfrac{K E_2^2 R_2}{R_2^2 + X_2^2}.

Maximum torque: sm=R2/X2s_m = R_2/X_2, Tmax=KE222X2T_{max} = \dfrac{KE_2^2}{2X_2}.

Relation between speed and torque

Put s=(Ns−N)/Nss = (N_s - N)/N_s:

  • Near synchronous speed (small ss): sX2≪R2sX_2 \ll R_2, so T≈KE22R2 s=KE22R2⋅Ns−NNsT \approx \dfrac{K E_2^2}{R_2}\, s = \dfrac{K E_2^2}{R_2}\cdot\dfrac{N_s - N}{N_s}. Torque rises linearly as speed falls — stable region.
  • Low speeds (large ss): T≈KE22R2sX22T \approx \dfrac{K E_2^2 R_2}{s X_2^2}, i.e. T∝1/sT \propto 1/s. Torque falls as speed falls — unstable region.
 N
 Ns|'-._
   |    '-.  normal running
   |       '.
 Nm|--------: Tmax (pull-out)
   |       .'
   |     .'   unstable
   |   .'
  0|_.'___________ T
     Tst

So the motor runs at almost constant speed from no load to full load, like a DC shunt motor.

✓ Answer checked by Claude Opus 5.5 (high effort)

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