# 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}