2396 style question, 2400, short circuit, RCBO,

[Richard Burns]

Possibly it might be worth checking on your level of current training, you say you are studying for the 2396 and so have presumably got a good existing background in electrical compliance.
You started the thread with a fairly high level design query but are now getting confused about basics.

Now I do not profess to know lots about the subject but I tend to work from first principles and apply common sense, this is not always a good idea for the wiring regulations!
Always remember that I am posting my ideas in a similar manner to the way you are posting yours. Always check back with your notes and ensure I am not making errors.

If you have reference books that state that the fault current should be determined at the extremity of the circuit then they are more likely to be correct than I am. However do check the reasoning and intent behind the reason why they are determining the fault current.

In your last post you ask if you should use one of two equations but the equations are the same equation just mathematically rearranged.
S= √ (I² × t) ∕ k
multiply both sides by k
S * k = √ (I²t)
square both sides (or multiply by √I²t which you know equals Sk)
S²k² = I²t
(and if you divide both sides by I² you get t=k²S²/I²)
either way works.

My reasoning for using the PFC as fault current when determining energy withstand or cpc suitability, as opposed to ADS disconnection time, is hopefully explained by the attached diagram.

 

[Richard Burns]

OK so actually looking things up clarifies things.
Because of the nature of the tripping curve for fuses as the fault current decreases the time to disconnect increases.
For circuit breakers there is little difference in trip time once the fault current has reached the level of operating the magnetic trip.
Therefore when considering the energy withstand of the cable, even though the fault current is lower at the extremity of the circuit the increased time means that the cpc has to withstand a greater total energy so this is the worst case scenario.
I considered only circuit breakers where there is not much difference, but even then it is intended that the minimum fault current should be used. (not that I necessarily agree with this).
It would be necessary for checking the breaking capacity of the protective device to consider the maximum fault current at the origin if the circuit, but in other cases it is the minimum fault current (or maximum resistance).
A lack of attention to detail and just randomly answering lead me to the erroneous conclusion of thinking I²t included time. Sorry for causing confusion.