Correct Answer - Option 2 : 921.9 kW
Synchronizing power of a synchronous machine is given by,
\({P_{syn}} = 3\frac{{{E_f}{V_t}}}{{{X_s}}}\sin \delta \)
Where Ef is the internally generated emf
Vt is the terminal voltage
XS is the synchronous reactance
δ is the torque angle
Calculation:
Given that,
\({E_f} = {V_t} = \frac{{6600}}{{\sqrt 3 }}V\)
XS = 1.65 Ω
Mechanical angle (θm) = 1°
Number of poles (P) = 12
Electrical angle, \({\theta _e} = \frac{{{\theta _m}}}{2} \times P = \frac{1}{2} \times 12 = 6^\circ = \frac{\pi }{{30}}rad\)
\(P = 3 \times \frac{{\frac{{6600}}{{\sqrt 3 }} \times \frac{{6600}}{{\sqrt 3 }}\; \times \;\cos 0^\circ }}{{1.65}}\; \times \;\frac{\pi }{{30}}\)
= 2764.6 kW
= 921.53 kW per phase