Skip to main content

Hunting; loaded operation of synchronous motor

Short answer

Hunting is rotor oscillation about synchronous position due to sudden load change, damped by amortisseur windings. Under load, rotor falls back by angle δ; back emf shifts, armature current increases, developing torque to match load.

What is hunting in synchronous motor? Explain the loaded operation of three phase synchronous motor.

Hunting

Hunting is the oscillation of the rotor of a synchronous machine about its steady-state synchronous position (the load angle δ\delta swings above and below its mean value) following a sudden change in load or supply.

Cause: when the load changes suddenly, the rotor does not take up its new load angle immediately. Because of its inertia it overshoots, and the restoring (synchronising) torque pulls it back, so it oscillates about the new position. The oscillations die out slowly.

Effects: variation in current, power and voltage, possible loss of synchronism, mechanical stress and noise.

Prevention: damper (amortisseur) windings on the pole faces damp the oscillations by developing a torque that opposes relative motion. A flywheel (to increase inertia) and proper design of the governor (prime mover control, for generators) also help.

Loaded operation

At no load, the rotor poles lie almost in line with the stator rotating-field poles. The back emf EbE_b is almost equal to VV and nearly in phase with it (δ≈0\delta \approx 0), so the net voltage Er=V−EbE_r = V - E_b across the armature is tiny and the current is small.

When mechanical load is applied:

  1. The rotor momentarily slows down and falls back by the load angle δ\delta; it then continues at synchronous speed, with only its position shifted.
  2. EbE_b shifts by angle δ\delta with respect to VV, so the resultant voltage Er=V−EbE_r = V - E_b increases.
  3. The armature current Ia=ErZsI_a = \frac{E_r}{Z_s} increases and lags ErE_r by θ=tan⁡−1(Xs/Ra)\theta = \tan^{-1}(X_s/R_a).
  4. The motor torque rises until it equals the load torque.
  5. The speed remains constant at NsN_s for all loads up to the pull-out torque.
P=3VEbXssin⁡δ,T=PωsP = \frac{3VE_b}{X_s}\sin\delta, \qquad T = \frac{P}{\omega_s}

Maximum power occurs at δ=90∘\delta = 90^\circ (pull-out). If the load exceeds it, the motor loses synchronism and stops.

 Phasors (motor):
 O------------------------> V
  `-._  delta             |
      `-._                |
          `-._   Eb       | Er = V - Eb
              `-._        |
                  `-._    |
                      `-._*
   Ia = Er/Zs lags Er by theta (close to 90 deg)

✓ Answer checked by Claude Opus 5.5 (high effort)

All Electrical Machines chapter 6 questions →