Consider a two-pole, 60Hz, 13.8kV, 50MVA superconducting generator which achieve
ID: 2082516 • Letter: C
Question
Consider a two-pole, 60Hz, 13.8kV, 50MVA superconducting generator which achieves rated open-circuit armature voltage at a. field current of 1520A. It achieves rated armature current into a. three-phase terminal short circuit for a field current of 413A. (a) Calculate the synchronous reactance. Consider the situation in which this generator is connected to a 13.8kV distribution feeder of negligible impedance and operating at an output power of 43MW at 0.9 pf lagging. Calculate: (a) the field current in amperes, the reactive-power output in MVA. and the rotor angle for this operating condition. (b) the resultant rotor angle and reactive-power output in MVA if the field current is reduced to 1520A while the shaft-power supplied by the prime mover to the generator remains constant.Explanation / Answer
Synchronous reactance = 1/(short circuit ratio)
Short circuit ratio = (Ratio of field current req to build rated voltage at open circuit)/(Ratio of field current req to build rated current at short circuit)
So, short circuit ratio = 1520/413 = 3.68(data given in question)
Hence, Synchronous reactance = 1/scr = 1/3.68 = 0.27 p.u