Question

A 75 MVA, 25 Kv, three-phase, 60 Hz synchronous generator has a synchronous reactance of X= 9 Ω per phase and a negligible resistance. The generator is delivrimg rated power at a 0.8 power factor lagging at the rated terminal voltage to an infinitie bus, as shown in the following figure 1.

E/8 X 8 Ιφ V.0

Figure 1.The synchronous generator connected to infinite bus in single phase.

1)   The excitation voltage is E = 26.87 KV, The generator delivring 60 MW. Compute the powe angle δ , the armature current I ,the power factor PF  ? when the field current is adjusted for the following excitation:

*The excitation voltage is decreased to 89.8 percent (89.8%) of the value E ?

	A.	δ = 25.6°   I= 1832A   = -26. 52°   cos =   0.895 Lag
	B.	None of these
	C.	δ = 27.6°   I= 1732A   = -26. 52°   cos =   0.895 Lead
	D.	δ = 27.6°   I= 1732A   = -26. 52°   cos =   0.895 Lag

Answer 1

ELS Х Given 2000 ILO 25 x=92 given TS MVA , 85kV & t/ph © The excitation voltage E = 26.87 KV. Three phase active power = 60 MW. P = EV sins 60 Х 26.97X13) (25) sin 8 = 60 १ sins=0.464 => 8 = 27.650 8 27.6° 2 The rated armature current 1- 75 3 KA = 1732 Amps 3x25 Pf tos Excitation is reduced to 89.8%8 E.

P= constant EV sins X Esins = constant Esins, & sind sinsz = sins, Ez EI sin Ei 10.89851 sim (270) = 31-05° Ia Es-vé X 90 sins, = 0.515 62 s(26. /31-05 9190 la 0-898/ 26.87 37)KA 25/0° V3 01698/25870) 315" 14,433-75 Jo 다. 24,129.26/31.05 -14,433-75%0' 9190² 13921044/63.379 9490 = 1546-8-26-65 Paz 154608/-26-63 1546-8 (-26.52 Cos (26.63) = • new pf 0.895 lag option @ 8= 27-6°; 1 = 1732 A , Pfangle Pf=0.895 lag +26.52