Revision history [back]

Agree with what perolofl has mentioned, also take into account that the plotted bus frequency is the time derivative of the generator internal rotor angle. If you have applied a solid three-phase fault, the active power output will be 0, while the mechanical power (or Torque in pu) will continue to be more or less constant during the first instants after the fault. This creates a Torque surplus and according to the generator power swing equation the rotor will start accelerating, giving rise to Δω as well as Δδ. As long as Δω increases,i.e. as long as the governor continues feeding the Torque surplus, you will be noticing a frequency rise.

Right after the fault clearance, and since Δδ has increased during the fault application, the generator (if still interconnected) will experience a wild active power output change, since the much larger δ (being integrated during the whole fault application acceleration interval) will impose a much larger active power output (much greater than what the governor applied Torque was) giving rise to a Torque deficit and subsequent deceleration interval, which creates a transient power swing.

If you perform LLG, or LG, not all voltages go down to zero and thus the generator continues to supply active power. The active power supplied during the fault might even grow, if the voltage reduction was less than the current increase, although the current increase will mostly be reactive, but still the generator will continue supplying active power giving rise to lower frequency increase.