MGVCL Exam Paper (30-07-2021 Shift 3)
A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.4 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)) = (1.1/0.6)*cosδ = (1.1/0.6)*0.5 = 0.91 M = (H*s)/(πf) = 4/(50π)
The tendency of a Reach of Distance Relay to operate at impedance larger than its setting value is known as overreach. Overreach is the presence of d.c. offset in the fault current wave, as the offset current has a higher peak value than that of a symmetrical wave for which the relay is set.
1.6 inches = 1.6 inches x 1,000 mils per inch = 1,600 mils 0.25 inch = 0.25 inch x 1,000 mils per inch = 250 mils Area = 1,600 x 250 = 400,000 square mils
EXAMIANS