**.**P

^{.}._{in }= √3 V

_{L }I

_{L }cosΦ W

_{ }

where V

_{L }= Applied Line Voltage I

_{L }= Line current drawn by the motor cosΦ = operating p.f. of synchronous motor

or P

_{in }= 3 ([er phase power) = 3 x V

_{ph }I_{aph }cosΦ W Now in stator, due to its resistance R

Total stator copper losses = 3 x (I

Now P = T x ω

P

i) For lagging p.f.,

E

ii) For leading p.f.,

E

iii) For unity p.f.,

E

In general,

Positive sign for leading p.f.

Neglecting sign for lagging p.f.

where T

P

_{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 lossesNow P = T x ω

**.**P^{.}._{m }= T_{g }x (2πN_{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,

_{m }= 3 E_{bph }x I_{aph }x cos (E_{bph }^ I_{aph})i) For lagging p.f.,

E

_{bph }^ I_{aph }= Φ - δii) For leading p.f.,

E

_{bph }^ I_{aph }= Φ + δiii) For unity p.f.,

E

_{bph }^ I_{aph }= δ**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.

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.

_{shaft }= Shaft torque available to load.P

_{out }= Power available to load**.**Overall efficiency = P^{.}._{out}/P_{in }
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