Trying to understand short-circuit current rating and kA
Trying to understand short-circuit current rating and kA
(OP)
I am trying to understand short-circuit current rating ,
and how it is calculated. Would anyone have a fairly simple
explanation .
and how it is calculated. Would anyone have a fairly simple
explanation .






RE: Trying to understand short-circuit current rating and kA
What do you need exactly: understand or calculated something?
Actually. Short Circuit current calculated according to
Ohm law: I=U/R.
R it's a game.
Used Z1,Z2,ZO ( short circuit impedance components) or X"d subtransient reactance fot the generator, Uk for the transformer.
If you want understand it as well, please see attached FAQ
FAQ238-1287: What are good references for a Power Engineer?.
Regards.
Slava
RE: Trying to understand short-circuit current rating and kA
The actual short circuit current often contains an off-set or DC component. The actual peak current depends on:
The base KVA of the transformer bank.
The X/R ratio of the transformer bank.
The point in the sine wave cycle at which the short circuit is applied.
The off set or DC component decays to the "Available short circuit current" fairly quickly. The decay may last cycles or seconds depending on the properties of the circuit.
This value of short circuit current is of interest when designing protection and when calculating the maximum physical forces that will be exerted on components by the magnetic field of the short circuit current.
The "Available Fault Current" or the "Available Short Circuit Current" is calculated by dividing the full load current rating of the transformer by the Per Unit impedance of the transformer. This is the steady state short circuit current. It is used to select switch gear, fuses, breakers and other rated equipment that will withstand and or interrupt the maximum short circuit current in the event of a fault.
You should be seeing a discrepancy here!
The maximum short circuit current may be double the available short circuit current, and yet I have said that the available fault current is used to select equipment to withstand the associated mechanical forces.
The answer is found in the way that equipment is tested and rated for available fault current.
A switch or breaker or other equipment with an available short circuit rating will have been designed and tested to withstand the maximum current that may result from the rated level of available fault current.
Another way of stating this is to say that a piece of equipment rated for 10kA will safely withstand the currents and forces that may result from fault current supplied from a transformer that is capable of supplying a steady state fault current of 10kA.
Because of the general high levels of current involved the current is often listed as kA (1000 amps)rather than amps.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Trying to understand short-circuit current rating and kA
The short circuit rating is generally determined by test.
RE: Trying to understand short-circuit current rating and kA
RE: Trying to understand short-circuit current rating and kA
The IEEE Red Book has a very nice discussion on calculating fault currents.
RE: Trying to understand short-circuit current rating and kA