IEC 60909 - Short Circuit Calculations
IEC 60909 - Short Circuit Calculations
(OP)
I've got a question regarding IEC 60909. Can someone explain how the different fault levels, ip, I"k, and Ik relate to different pieces of equipment. Specifically in relation to MCC, variable frequency drives and transformers?
I've gone through the standard and understand how those three numbers are calculated, but I don't recall ever once seeing how to apply them to equipment specifications.
For example, at a 2.5MVA transformer, according to etap we have I"k=33kA, IK=30kA, and iP=82kA. Now a hand calc, given the voltages and impedance of the transformer tells me that the maximum fault current that should be able to go through the transformer is about 35kA (690V, 6%Z). So how is it that iP = 82kA? The fault current through the transformer will automatically be limited by the impedance to 35kA right? There's no motor contributions on the secondary.
Am I missing something?
Scott
I've gone through the standard and understand how those three numbers are calculated, but I don't recall ever once seeing how to apply them to equipment specifications.
For example, at a 2.5MVA transformer, according to etap we have I"k=33kA, IK=30kA, and iP=82kA. Now a hand calc, given the voltages and impedance of the transformer tells me that the maximum fault current that should be able to go through the transformer is about 35kA (690V, 6%Z). So how is it that iP = 82kA? The fault current through the transformer will automatically be limited by the impedance to 35kA right? There's no motor contributions on the secondary.
Am I missing something?
Scott






RE: IEC 60909 - Short Circuit Calculations
RE: IEC 60909 - Short Circuit Calculations
RE: IEC 60909 - Short Circuit Calculations
When you did your hand calc, you probably calculated the steady-state value based on the short circuit impedance of the transformer, i.e. if the LV side was shorted, then you calculated the fault current based on Ohm's law. But a short circuit through a transformer is essentially like an RL switching transient - you generally can't go from normal load current to steady-state fault current instantaneously, and there needs to be a continuous transition path from load to fault. This is the short circuit transient waveform which has a decaying dc component (depending on the time of switching). It is basically the solution to the differential equation:
V = Ri + L di/dt
The transformer impedance will limit the fault current in the steady state, but there will be an exponentially decaying transient current that can have much higher currents. The decay time of the transient also depends on the X/R ratio.
RE: IEC 60909 - Short Circuit Calculations
And you're right about the transformer hand calc, i used:
Isc = KVA/(sqrt(3)*LV*%Z)
And that would make sense that its a steady state value and furthermore that the instantaneous current would not change from SS to peak value instantaneously.
So I guess my new questions are:
1. It looks like the Isc calculated above is the same as Ik, but I"k is larger than Ik (obviously) will this damage a transformer?
2. ip has no bearing on transformers? Only for equipment with busbars?
Thanks,
Scott
RE: IEC 60909 - Short Circuit Calculations
As for Ip and ratings on switchgear / circuit breakers, the ratio between symmetrical (I''k) and peak current (Ip) is standardized for LV equipment in IEC 60947-2: see this link for typical values.. If your X/R is unusually large you may need to size larger equipment to get an adequate dynamic withstand rating.
RE: IEC 60909 - Short Circuit Calculations
IEC 60865-1 /1993 Short-circuit currents-Calculation of effects Part 1 Method
IEC 60865-2 /1994 Short-circuit currents-Calculation of effects Part 2 Examples.