The general approach with induction machines is a locked rotor test, wherein you ramp up the voltage at some frequency (e.g. 25% of rated frequency) until the current reaches the rated current.
This is an induction motor running as an induction generator above its sync. speed. to generate active power.
Therefore, for a 3-ph short circuit at 11kV bus the MAXIMUM contribution from the IG is its LRC = 1720.6 per the given data sheet.
But for a single line to earth fault at 11kV bus it will be little lower and one has to perform an EMT study
(required parameters are given in the last page of the data sheet) to get the exact value or contact the manufacturer.
Thank you Kirbanda, I assume that the LRC of 1720.6 A is synchronous short-circuit current.
Short-circuit current of 1720.6 A divided by rated current of 380 A is 4.51 FLA, or an IG impedance of 1 / 4.51 = 0.22 per-unit.
Although a typical induction motor locked rotor current is approx. 6 x FLA, the subtransient current is higher than 6 x FLA, typically 12 x FLA. x”d = 1/12 = 0.083
However an IG has no external excitation supply, therefore the short-circuit current would decrement faster.
1) For an IM typical LRC is taken as 6xFLA. But the new high efficient machines goes upto 7xLRC. It is mentioned in NEC/ NEMA MG1.
2) Since you have manufacturer's data sheet you must not use the typical value but use 1720.6A which is 4.51xFLA.
3) This is sym current. But for protection coordination you should take 1.25xLRC which is the peak current.
4) Yes you are correct. Since this sc current is generated by the trapped flux, it dies down fast because there is no source to maintain it similar to a sync. machine.
Thank you Kiribanda
Assuming:
1) Symmetrical sub-transient short-circuit current (I"k) is 1.25 * symmetrical synchronous short-circuit current (Ik).
2) IG X/R ratio of 30 ( IEEE Std 399-1997 )
PrG = 6300000 W
VrG = 11000 V
pf running = 0.87 pu
FLC = 382 A
Symmetrical Synchronous short-circuit current = 1720.6 A
Symmetrical Sub-transient short-circuit current = 1.25 * 1720.6 = 2151 A
Sub-transient impedance = 382 / 2151 = 0.1776 pu
Peak short-circuit current to IEC 60909-0 Equations 54 & 55 = sqrt(2)*I"k ( 1.02 + 0.98e^-3R/X ) = 5800 A
The sub-transient short-circuit current (I"k) should be considered if current limiting circuit-breakers or fuses are involved.
The mechanical forces are proportional to the square of the peak short-circuit current.
