Question 4.18: A 1500 kVA, 3300V, 50Hz, three-phase, star-connected synchro...
A 1500 kVA, 3300V, 50Hz, three-phase, star-connected synchronous generator has an armature resistance of 0.2 Ω per phase. A field current of 50 A produces a short-circuit current of 262 A and an open-circuit EMF of 1200 V between the lines. Calculate voltage regulation of the generator on full load at 0.8 power factor lagging and at 0.8 power factor leading.
Total kVA = 1500
kVA per phase = \frac{1500}{3} = 500
Per-phase voltage = \frac{V_{L}}{\sqrt{3}} = \frac{3300}{\sqrt{3}} V = 1905 V
Per-phase current, I_{a} = \frac{500 × 1000}{3300 / \sqrt{3}} = 262 A
Given that at a field current of 50 A, short circuit I_{a} is 262 A and the open-circuit line voltage is 1200 V
Synchronous impedance/phase Z_{s} = \frac{open – circuit voltage per phase}{short – circuit current} = \frac{1200}{\sqrt{3} × 262} = 2.64 Ω
R_{a} per phase = 0.2 Ω
\cosΦ = 0.8, Φ = 37°, \sinφ = 0.6At lagging power factor load
E_{1} = \sqrt{(V \cos Φ + I_{a} R_{a})² + (V \sin Φ + I_{a} X_{s})²}and at leading power factor load
E_{2} = \sqrt{(V \cos Φ + I_{a} R_{a})² + (V \sin Φ – I_{a} X_{s})²}
= 2418 V
Percentage regulation at full load 0.8 power factor lagging
= \frac{(2418 – 1905) × 100}{1905} = 26.9 per cent
=1640 V
Percentage regulation at full load 0.8 power factor leading
\frac{ E_{2} – V}{1905} × 100 = \frac{ (1640 – 1905)}{1905} × 100 = – 13.9 per cent.
This shows that regulation is negative at 0.8 leading power factor load. This is because the full-load terminal voltage is more than the no-load voltage.