A 440 V, 50 Hz, \Delta-connected synchronous generator has a direct-axis reactance of 0.12 \Omega and a quadrature-axis reactance of 0.075 \Omega / \text { phase } ; the armature resistance being negligible. The generator is supplying 1000 A at 0.8 lagging pf. (a) Find the excitation emf neglecting saliency and assuming X_{s} = X_{d}.
(b) Find the excitation emf accounting for saliency.
Compare and comment on the results of parts (a) and (b).
On equivalent star basis
X_{d}=0.12 / 3=0.04 \Omega, \quad X_{q}=0.075 / 3=0.025 \Omega(a) Saliency ignores; X_{s}=X_{d}
\bar{E}_{f}=(440 / \sqrt{3}) \angle 0^{\circ}+j 0.04 \times 1000 \angle-36.9^{\circ}
=279.8 \angle 6.6^{\circ}
Or E_{f}=279.8 V or 484.6 V (line)
(b) I_{a}=1000 A , \phi=+36.90^{\circ} (lagging)
\tan \psi=\frac{V_{t} \sin \phi+I_{a} X_{q}}{V_{t} \cos \phi+I_{a} R_{a}}=\frac{254 \times 0.6+1000 \times 0.025}{254 \times 0.8} ; V_{t}=440 / \sqrt{3}=254 V
= 0.873
Or \psi=41.1^{\circ}
\delta=\psi-\phi=41.1^{\circ}-36.9^{\circ}=4.2^{\circ}
E_{f}=V_{t} \cos \delta+I_{d} X_{d} ; \quad I_{d}=I_{d} \sin \psi=1000 \sin 4.2^{\circ}
=254 \cos 4.2^{\circ}+73.2 \times 0.04
= 256.3 V or 444 V (line)