A 1500 kVA, star-connected, 6.6-kV salient-pole synchronous motor has X_{d} = 23.2 \Omega and X_{q}=14.5 \Omega / \text { phase } respectively; armature resistance being negligible.
Calculate the excitation emf when the motor is supplying rated load at 0.8 leading pf.
What maximum load the motor can supply without loss of synchronism, if the excitation is cut off? What will be the value of torque angle under this condition.
I_{a}=\frac{1500}{\sqrt{3} \times 6.6}=131 A ; \quad V_{t}=\frac{6.6}{\sqrt{3}}=3.81 kV
\tan \psi=\frac{V_{t} \sin \phi-I_{a} X_{q}}{V_{t} \cos \phi+I_{a} R_{a}}
=\frac{-3.81 \times 0.6-0.131 \times 14.5}{3.81 \times 0.8+0}=-1.373
Or \psi=-53.9^{\circ}
\delta=\phi-\psi=-36.9^{\circ}-\left(-53.9^{\circ}\right)=17^{\circ}
E_{f}=V_{t} \cos \delta-I_{d} X_{d} ; \quad I_{d}=I_{a} \sin \psi=0.131 \times \sin -53.9^{\circ}=-0.106
=3.81 \cos 17^{\circ}+0.106 \times 23.2
= 6.1 kV or 10.57 kV (line)
With excitation cut off, output is only reluctance power
P_{e}=V_{t}^{2}\left(\frac{X_{d}-X_{q}}{2 X_{d} X_{q}}\right) \sin 2 \delta=(6.6)^{2} \times \frac{23.2-14.5}{2 \times 23.2 \times 14.5}=563 kW