Well, technically it is the area under the i2 curve seen during the fault, which would not be the rms fault current i2 times t, on a short duration fault.

Indeed, and that illustrates what I'm been saying about the potential confusion about the 'entity' which (confusingly, and usually strictly incorrectly) is usually denoted by "I²t".

As you say, what we really should be using is the area under the I²/t curve for the duration of the fault current - something which seems to be called 'let through', and it is in terms of that which a 'proper adiabatic equation' would be written. In other words, the equation should be written in terms of ...

... rather than I²t. The reason for our interest in this quantity is clear enough, since if one multiplies it by the total resistance of a conductor, it then gives the total amount of energy dissipated in that conductor during passage of the current and, more important to our context, if one multiplies it by, say, the resistance per metre of the conductor, one will get the total amount of energy peer metre dissipated in the conductor - which, under adiabatic conditions, will, in turn, be directly proportional to temperature rise in that conductor.

I²t is nearly always an approximation to the 'let through' (AUC) we really should be using. It is only exactly the same as 'let through' if 't' is an exact number of half cycles. If t is large in relation to a half cycle (i.e. 'many half cycles, probably plus a bit of a half-cycle'), then it will be approximately correct. You have suggested (as is often suggested) that that equation gives reasonably correct results down to t=0.01s. However, that is only one half-cycle, which I think is a bit iffy (unless it were

__exactly__ 0.01s). In other words, I think that It would give a pretty poor approximation to 'let through' (AUC) with, say, t=0.015s, or even 0.025s.

Having said all that, we are presumably, as always, expected to consider 'worst case' scenarios, and I think that makes things a lot simpler. With a very short disconnection time (only a small proportion of a half-cycle) that will presumably occur when the period of fault current is centred at the peak point of the voltage waveform. We are then back to the situation in which the true 'let through' (AUC) actually is (approximately) a sort-of I²t - the 'I' now actually being √2 times what would be the RMS current (calculated from V and Z, perhaps Uo and PEFC) if the fault current were sustained for many (or an exact number of) half-cycles. In other words, if we knew 't' (from adequate curves), we could undertake a worst-case adiabatic calculation. It's interesting that, for such a short disconnection time, the situation would actually be 'worse', by a factor of √2 than would be the case if one undertook ('naive'?) 'RMS calculations'.

Rapid yes and more rapid then the normal 50hz cycle, but not instant.

That's what I wrote, isn't it? I wrote

"... very fast rate of rise ... with high frequency components", not "infinitely fast rate of rise ... with infinitely high frequency components".

Kind Regards, John