Power Flow In Synchronous Motor

Net input to the synchronous motor is the three phase input to the stator.
.^..                      P_in = ?3 V_L I_L cos? W

where         V_L = Applied Line Voltage

I_L = Line current drawn by the motor

cos? = operating p.f. of synchronous motor

or                P_in  = 3 ([er phase power)

= 3 x V_ph I_aph cos? W

Now in stator, due to its resistance R_a per phase there are stator copper losses.

Total stator copper losses = 3 x (I_aph)² x R_a W
.^..     The remaining power is converted to the mechanical power, called gross mechanical power developed by the motor denoted as P_m.
.^..                P_m = P_in – Stator copper losses
Now    P = T x ?
.^..                P_m  = T_g  x (2?N_s/60)  as speed is always N_s

This is the gross mechanical torque developed. In d.c. motor, electrical equivalent of gross mechanical power developed is E_b x I_a, similar in synchronous motor the electrical equivalent of gross mechanical power developed is given by,

P_m = 3 E_bph x I_aph x cos (E_bph ^ I_aph)
i) For lagging p.f.,
E_bph ^ I_aph = ? – ?
ii) For leading p.f.,
E_bph ^ I_aph = ? + ?
iii) For unity p.f.,
E_bph ^ I_aph  = ?

Note : While calculating angle between E_bph and I_aph from phasor diagram, it is necessary to reverse E_bph phasor. After reversing E_bph, as it is in opposition to V_ph, angle between E_bph and I_aph 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.

.^..                      P_out = P_m – mechanical losses.

where         T_shaft = Shaft torque available to load.
P_out = Power available to load
.^..     Overall efficiency = P_out/P_in