Power Flow In Synchronous Motor
Net input to the synchronous motor is the three phase input to the stator.
... Pin = ?3 VL IL cos? W
where VL = Applied Line Voltage
IL = Line current drawn by the motor
cos? = operating p.f. of synchronous motor
or Pin = 3 ([er phase power)
= 3 x Vph Iaph cos? W
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
... The remaining power is converted to the mechanical power, called gross mechanical power developed by the motor denoted as Pm.
... Pm = Pin – Stator copper losses
Now P = T x ?
... Pm = Tg x (2?Ns/60) as speed is always Ns
This is the gross mechanical torque developed. In d.c. motor, electrical equivalent of gross mechanical power developed is Eb x Ia, similar in synchronous motor the electrical equivalent of gross mechanical power developed is given by,
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 = ?
Note : While calculating angle between Ebph and Iaph from phasor diagram, it is necessary to reverse Ebph phasor. After reversing Ebph, as it is in opposition to Vph, angle between Ebph and Iaph must be determined.
In general,
Positive sign for leading p.f.
Neglecting sign for lagging p.f.
Net output of the motor then can be obtained by subtracting friction and windage i.e. mechanical losses from gross mechanical power developed.
... Pout = Pm – mechanical losses.
where Tshaft = Shaft torque available to load.
Pout = Power available to load
... Overall efficiency = Pout/Pin
