IEC60909 Query

Here is one link which may help : IEC60909FaultCalculations.pdf

This pdf file has a graph of the SC current at initiation...

As you can see from the graph, I_p is higher than √2*I"_k due to the presence of the DC component. The wave is quite asymmetrical at the start. IEC 60947 specifies I_p = 2.2 * I_rms when I_rms > 50kA.

I think you have the wrong idea about what Ik" represents, the Ik" you're looking at which is lower than Ip is for a symmetrical fault.. Ip is denoted for an assymetrical fault as well on the far left of the graph.

i had no idea at all (not even a wrong idea) about I_k" till you mentioned it here It seems to me that I_k" is a hypothetical value in any case, since all faults i have seen have a DC component, and are asymmetrical in nature.

Your original query :

_{In IEC60909: IK" is denoted as the initial symmetrical short circuit current (RMS) Ip = peak short-circuit current is it true then, that Ip = sqrt(2)*IK" ?}

Obviously from the graph, I_p = 2√2 * I_k". So your assumption is half of reality .... am i missing something here ?

No, Ip looks to be sqrt(2)*Ik", since 2*Sqrt(2)*Ik" is the peak to peak amplitude of Ik".. Ip looks to be the value 0 to peak of Ik"... The reason I ask this is because I am using SKM Powertools to do some modelling, and in my model it looks as if Ik" is ~2.4*Ik" when I perform the IEC60909 short circuit analysis..

Oh, i see that your knowledge level is much higher than mine, so you are probably right.

However, at the risk of appearing a bit thick, i believe that I_p is the first actual peak seen by the switchgear which switches on to the fault, this I_p having a theoretical maximum value of 2.828*I_rms. Reality, as shown by the graph, is somewhat less, and IEC 60947 and parts specify that as 2.1x @ ≤50kA and 2.2x @ >50kA.

A purely symmetrical fault (if such a thing exists) will appear like the upper graph of this image....in such a case, your figure of I_p = √2*I_k" will be correct.

ahh you're quite right, very good. Ip is the first peak seen by the switch gear, that makes sense. thanks

Very heartening to think that i haven't lived under a wrong assumption all my life, thank you.

Just to be sure,

1. the whole wave is shifted up wrt to the zero axis, so a peak of 2.2 * I_rms is realistic,

2. and the system, especially the switchgear, will suffer the enormous dynamic effect of this very high current. (Due to the short time involved, we can ignore the thermal effect)

1. yes correct, you can't read the 2.2 factor directly off the graphs you posted, but for an assymetrical fault the switchgear will always see close to 0A for at the first instant of the fault, with a peak value shortly after.. 2. this will be the highest current incurred by the switchgear for a very short period, however when you specify your switchgear you typically use the RMS fault current Ik".

True! But that is only for the Breaking capacity of the switchgear. For the making capacity, however, you must also specify the peak current, duly considering the decaying DC Component into account. Also, eletro-dynamic forces generated in the switchgear as well as elsewhere in the installation during a short circuit is proportional to square of the peak current, again, with the DC component given the due consideration. In any case, one cannot neglect the peak value of the short circuit current during the first half cycle.