Question

A 2000 V single-phase synchronous generator is being used as a backup
generator outside a diagnostic center. When the generator
terminals are short-circuited and a field excitation current of 2.5 A is passed, the
ammeter between the terminals read 100 A of full load current. If the terminals
of the same generator are opened and the voltage across them is measured using
a voltmeter, we get the value of 500 V with the same excitation current. The
armature resistance is 0.8 ohms.
Determine the voltage regulation of this generator if it is delivering 100 A
during normal operation with unity power factor, 100 A with 0.8 leading PF,
and with 0.71 lagging PF.
(Hint: Synchronous Impedance = (0.C Voltage) / (S.C Current))

Answer 1

Given rated voltage of synchronous generator - V = 2000 V Voca 500 Isc = 100 A synchronous Impedance Voc Isc SVO loo x = 52 R = 0.82 AL > cod=1 4 Sing=0 unity power Factor voltage regulation 2 TRcosø +3x9in )x100 VRE x 10$ 100 X 0.8x1 2000 VRE 4-1 At 0.8 leading PL - COSø= 0.8 >> sind :0.6 UR = two [6,8x0.8 - 5x0.6] 2606 x 10 € VR: -11.84 At 0.7 lag PE 1 coso: 0.75 sing = 0.704 VR - 100[ 0.8x0.71 + 6x0.704) x100 2000 VR= 20.444