IEC 60909 - Short Circuit Calculations

[Umtully1]

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

 

[mgtrp]

At the time of initial fault, there will be a decaying DC offset, which can result in peak currents that are higher than the symmetrical fault current. This is why Ip is much higher than what you calculated.

 

[Umtully1]

I know there's going to be a decaying DC offset. What I don't know is how ip relates to my equipment rating.

 

[juleselec]

These calculated values are necessary to size switchgear (among other things). Busbars are typically rated to the initial fault current Ik'' or the thermal fault current Ith (i.e. the Joule equivalent fault current) for some specified period of time such as 1s or 3s. The busbars and switchgear assembly are also rated to withstand some peak current Ip, which accounts for the potentially large mechanical forces that occur during a fault. The symmetrical breaking current Ib is the fault current that the circuit breakers should be able to clear, e.g. without arcing.

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.

 

[Umtully1]

That makes sense. I kind of figured out that I"k was related to switchgear and MCCs, but its good to have confirmation.

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

 

[jvrielink]

The transformer's impedance and X/R ratio largely determine your fault levels for a short-circuit on either side of it, you can't have 'too much' fault current through it. It'll only be damaged if you don't interrupt it in time.

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.

 

[7anoter4]

Based on IEC 60909 there are the following IEC standards. Here you can know what is Ith and where to employ Ip.
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.