Synchronous motor adjusting to increasing load

A synchronous motor meets an increasing shaft load without changing speed. It does so by increasing its load (torque) angle $\delta$, the angle by which the rotor poles (and $E_f$) fall behind the stator field (and $V$).

Step-by-step action

1. At no load, the rotor poles are almost exactly under the stator poles; $\delta$ is very small and the motor draws only a small current for losses.
2. Load is increased. Load torque becomes greater than the developed torque, so the rotor decelerates momentarily.
3. Rotor falls back. Because the stator field keeps rotating at $N_s$, the rotor poles slip back by a larger angle $\delta$. The magnetic "spring" linking the poles is stretched more.
4. More current. $E_f$ keeps the same magnitude (field current unchanged) but swings further behind $V$. The resultant voltage $E_r = V - E_f$ increases, so $I_a = E_r/X_s$ increases.
5. More power. Developed power rises:

$$P = \frac{3VE_f}{X_s}\sin\delta$$

6. New balance. When the developed torque equals load torque plus losses, the rotor again runs at exactly $N_s$, but with a larger fixed $\delta$.

  V ------------------------------>
   \ \
    \  \  delta2 (more load)
     \   \
      \    '----> Ef2  (same length)
  delta1
       '---------> Ef1 (light load)

  Er2 > Er1  ->  Ia2 > Ia1

Limit of loading

• Power rises with $\delta$ only up to $\delta = 90^\circ$ (cylindrical rotor), where $P_{max} = 3VE_f/X_s$.
• If load is increased beyond this pull-out torque, the rotor cannot stay locked; it falls out of step, and the motor stalls with heavy current (the protection then trips).
• The pull-out limit can be raised by increasing excitation ($E_f$).

Other effects

• With fixed excitation, the power factor changes with load; for a given $E_f$ the pf moves towards lagging as load increases.
• A sudden load change makes the rotor overshoot and oscillate about the new $\delta$ (hunting); damper windings reduce this.