Net input to the synchronous motor is the three phase input to the stator.
Now in stator, due to its resistance Ra per phase there are stator copper losses.
Total stator copper losses = 3 x (Iaph)2 x Ra W
... Pm = Pin – Stator copper losses
Now P = T x ?
... Pm = Tg x (2?Ns/60) as speed is always Ns
Pm = 3 Ebph x Iaph x cos (Ebph ^ Iaph)
i) For lagging p.f.,
Ebph ^ Iaph = ? – ?
ii) For leading p.f.,
Ebph ^ Iaph = ? + ?
iii) For unity p.f.,
Ebph ^ Iaph = ?
Positive sign for leading p.f.
Neglecting sign for lagging p.f.
can be obtained by subtracting friction and windage i.e. mechanical
losses from gross mechanical power developed.
where Tshaft = Shaft torque available to load.
Pout = Power available to load
... Overall efficiency = Pout/Pin