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# Generator Short Circuit Current

Assumption: Generator Xd''=0.20, Xs=1.5 pu.
Now, Generally Speaking if we speak of bus fault at generator terminals we say fault current =1 (pu voltage)/0.20= 5 pu. Similarly fault current after 5 seconds or so =1/(1.5)=0.66 pu.

1) The concern is that 1 pu voltage assumed is generator open circuit terminal voltage. So internal voltage generated by machine is 1+(armature current(phasor)*reactance(phasor)). This value reaches up to 1.5-2 times the terminal voltage. All the reactances like subtransient, transient, synchronous are after this internal voltage of machine. So this wrong assumption of calculating fault current assuming 1 pu voltage is how much valid (certainly not accurate) as clearly internal voltage is appreciably higher than terminal voltage. For ex 1.5/0.20=7.5 pu is 30% higher than fault current calculated above.

2) Popular software package like Aspen does have a field called internal voltage in generator parameter dialog. But if I have only buses and lines and no loads modeled and hence load flow solution, am I compelled to use 1 pu voltage as I cannot find internal voltage of machine?

The analysis ignores the fact that a short circuit at the terminals of a machine results in a terminal voltage of zero, not the internal voltage plus the voltage drop across the internal impedance, which is closer to the analysis of a machine operating at full load feeding a fault through impedances such as transformers and transmission lines.

Another thing to consider is, if the machine is under manual voltage control (the usual assumption for the Thevenin equivalent model of a machine that are using), how can the internal voltage of the machine rise when the field voltage/current is fixed and doesn't change throughout the time of the fault.

Even if the AVR was in service, the Rotor Time Constant (tau) is sufficiently long (on the order of 1-10 seconds for larger machines) so that the change in field current (if any) will be negligible for the duration of time before the protective relaying clears the fault and/or trips the unit.

When using an internal voltage of 1 pu, the assumption of the generator is not loaded just before the short-circuit. As there will be no current flow and no voltage drop over the generator internal impedance, the internal voltage will equal the terminal voltage. The maximum voltage drop over the generator impedance will occur at full load, this voltage added to the terminal voltage gives you the internal voltage and the short-circuit at maximum rated excitation.

However, bear in mind that the maximum initial short-circuit is a function of the sub-transient impedance, which is dependent on the design of the generator damper windings. When the sub-transient has been subsided the excitation control is already winding back. A short-circuit near the generator has little real power and is highly reactive, both are reasons for the excitation system to reduce terminal voltage.

Generator