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)
Value of power factor is varies from 0 to 1. R→PF = 1 L→PF = 0 lag C→PF = 0 lead R-L→0 lag < PF < 1 R-C→0 lead < PF < 1 R-C-L→depends on value of R, L and C. PF = R/Z
Symmetrical breaking current is the moment when fault occurs maximum fault current is observed and decays with respect to time. Symmetrical breaking current = RMS value of ac component of short circuit current at the instant of separation of the contacts. Symmetrical breaking capacity is less than asymmetrical breaking capacity.