MGVCL Exam Paper (30-07-2021 Shift 2)
In a power network, 400 kV is recorded at a 415 kV bus. A 50 MVAR, 415 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor?
The capacity of reactor = 50 MVAr when it is operated at 415 kV. Impedance of reactor = V2/MVAr = (415*415)2/50 = 3444.5 Ω For 400 kV, Reactive power observed = (400*400)/3444.5 = 46.45 MVAR
Or
VAr capacity of capacitor or reactor is proportional to V²
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
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