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)
The main function of an interposing CT is to balance the currents supplied to the relay where there would be an imbalance due to the ratios of the main CTs. Interposing CTs are equipment with a ratio that can be selected by the user to achieve the balance required. An interposing CT is installed between the secondary winding of the main CT and relay.
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
EXAMIANS