Synchronous Motors

11.8 Motor Phasor Diagram
Consider an under-excited ^tar-connected synchronous motor (E b < V) supplied
with fixed excitation i.e., back e.m.f. E b is constant-
Let V = supply voltage/phase
E b = back e.m.f./phase
Z s = synchronous impedance/phase
(i) Motor on no load
When the motor is on no load, the torque angle α is small as shown in Fig. (11.7
(i)). Consequently, back e.m.f. E b lags behind the supply voltage V by a small
angle δ as shown in the phasor diagram in Fig. (11.7 (iii)). The net voltage/phase
in the stator winding, is E r .
Armature current/phase, I a = E r /Z s
The armature current I a lags behind E r by θ = tan -1 X s /R a . Since X s >> R a , I a lags
E r by nearly 90°. The phase angle between V and I a is φ so that motor power
factor is cos φ.
Input power/phase = V I a cos φ
Fig.(11.7)
Thus at no load, the motor takes a small power VI a cos φ/phase from the supply
to meet the no-load losses while it continues to run at synchronous speed.
(ii) Motor on load
When load is applied to the motor, the torque angle a increases as shown in Fig.
(11.8 (i)). This causes E b (its magnitude is constant as excitation is fixed) to lag
behind V by a greater angle as shown in the phasor diagram in Fig. (11.8 (ii)).
The net voltage/phase E r in the stator winding increases. Consequently, the
motor draws more armature current I a (=E r /Z s ) to meet the applied load.
Again I a lags E r by about 90° since X s >> R a . The power factor of the motor is
cos φ.
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