Hey guys I've been wondering when your wiring in parrallel do you divide the R1+R2 values by 4 when calculating thermal constraints and checking that the Zs value is okay.
Also when wiring in parrallel if you only use one of the earth connections would you divide by 3.
Hope this makes sense.
It's perhaps not always recognised that the even the traditional adiabatic approach is dealing with energy let-through as well. k²S² ≥ I²t is just S ≥ √(I²t)/k rearranged. The only difference is rather than working out I (and hence I²) and t separately and multiplying them together, you get the final I²t number direct from manufacturers data or standards.
- Andy.
and using units of amps squared seconds rather hides the fact that we are really talking about joules (heating power) per ohm (of resistance in the fault loop) power being I squared R and all that.
When you know the power and the mass of the metal and its heat capacity tell you the temp rise during the zap. If you know this there is no need for a special formula to remember, you can hit the tarmac running as it were and re-derive it from 1st principles each time you need it, if that is the way you like to operate.
It also confuses folk when folk bandy about kA2s as to whether the k relates to the amps and should be squared as well (so 1kA2s = million times 1 amp amp second), or to the unit, when it should not. (really it should be the latter and 1kA2s = thousand times 1 amp amp second, but I have seen it wrong in a few places)
'Joules per Ohm' and 'kiloJoules per ohm' are clearer ,and explain why things get cooler for a given fault when the cable is fatter and the resistance is lower. (assuming of course that the fault current is set near constant by other impedances not under consideration and not changed by the fatter wire.. )
Mike.
The only difference is rather than working out I (and hence I²) and t separately and multiplying them together, you get the final I²t number direct from manufacturers data or standards.
Yes, and the important point about that, is the numbers are different from the separate I and t you get from the manufacturer's date on I vs t curves.
t also confuses folk when folk bandy about kA2s as to whether the k relates to the amps and should be squared as well (so 1kA2s = million times 1 amp amp second),
Yes, I understand the argument that doesn't really comply with SI units, but please take it from me, the kilo is not squared, '1 kA2s' means '1000 A2s'. However, this is a comment on product standards and data, rather than BS 7671.
The only difference is rather than working out I (and hence I²) and t separately and multiplying them together, you get the final I²t number direct from manufacturers data or standards.Yes, and the important point about that, is the numbers are different from the separate I and t you get from the manufacturer's date on I vs t
From the limited information I have available, the level 4 2396 Design exam requires candidates to work out the I2t value from circuit parameters. Older questions were based on fuses with fault currents being calculated between the 0.1 and 5s limits, but more recently mcbs are the OCPD in the question often with calculated fault currents greater than those required for 0.1s disconnection.
Despite advice to use manufacturer’s data for I2t when using mcbs in their projects, almost 100% of candidates choose to find the fault current, square it and multiply by 0.1 in their calculations for thermal compliance.
There is a problem with that approach.
The 'I' they are using is, I assume, the RMS value of a current that is in the shape of a sinewave. In fact the fault current may well be asymmetric, that is the sinewave climbs on top of the zero reference and then decays back to a normal sinewave over about 5 cycles (100ms) of the supply. This I²t will not be the same as one calculated assuming a sinewave RMS value.
Regards
Geoff Blackwell
There is a problem with that approach.
The 'I' they are using is, I assume, the RMS value of a current that is in the shape of a sinewave. In fact the fault current may well be asymmetric, that is the sinewave climbs on top of the zero reference and then decays back to a normal sinewave over about 5 cycles (100ms) of the supply. This I²t will not be the same as one calculated assuming a sinewave RMS value.
This is absolutely the reasoning, as the wording in earlier Editions of BS 7671 (and the Note to the Wiring Regs that preceded) attests. Assymetric fault current, which always occurs if the fault is less than 0.02 s, but often occurs if the fault is cleared in less than 0.1 s, as you get an exponentially rising fault current to a crest (of the fault current waveform, not mains sine wave) which then disconnects often part way through a half-cycle (but arcing current may flow until zero-crossing).
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