MGVCL Exam Paper (30-07-2021 Shift 2)
A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given.
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (11/05)*cosδ = (11/0.5)*0.5 M = (H*s)/(πf) = 4/(50π)
Where, Pmp is maximum power in watt Vmp maximum voltage at maximum power in volt Imp is maximum current at maximum power in ampere Voc open circuit voltage in volt Isc is short circuit current in ampere
Fill Factor = Pmp/(Vosc*Isc) = (Vmp*Imp)/(Voc*Isc) = (2.65*0.5)/(0.7*3) = 0.63
Equivalent resistance of the circuit, Req = (12||18) + 8 Req = 7.2 + 8 Req = 15.2 ohm Source current = current through 8 ohm resistor, I = Vr/R8 I = 20/8 I = 2.5 A Supply votage V = I*Req V = 2.5*15.2 V = 38 V
Slip, s = (Ns - Nr)/Ns Ns = 120f/P 0.05 = (1000 - Nr)/1000 Nr = 950 rpm Speed of rotor with respect to stator magnetic field = 950 - 1000 = -50 rpm
Speed of the stator magnetic field with respect to stator core = 0 Speed of the rotor magnetic field with respect to rotor = sNs Speed of the stator magnetic field with respect to rotor magnetic field = 0 Speed of the rotor magnetic field with respect to stator core = Ns
EXAMIANS