Consider the phasor diagram of a synchronous motor running on leading power factor cos? as shown in the Fig. 1.
The line CD is drawn at an angle ? to AB.
The lines AC and DE are perpendicular to CD and AE.
and angle between AB = Ebph and Iaph is also ?.
The mechanical per phase power developed is given by,
In triangle OBD,
BD = OB cos? = Ia Zs cos?
OD = OB sin ? = Ia Zs sin
Now BD = CD – BC = AE – BC
Substituting in (2),
Ia Zs cos? = Vph cos (?-?) – Eb cos?
All values are per phase values
Substituting (3) in (1),
This is the expression for the mechanical power developed interms of the load angle ? and the internal machine angle ?, for constant voltage Vph and constant Eph i.e. excitation.