In article ,
"Will Dean" writes:

Hi,

This is an abstract question - but I'm just interested to see what I've
misunderstood here...

Imagine a domestic installation, with a PSC of 6kA at the incoming supply,
fused at 100A. Right next to the incoming supply, there's a consumer unit,
which contains one 20A MCB. The CU is connected to the incoming supplies
with very short 25mmsq tails (I've ignored their contribution to lowering
the fault level.)

This 20A MCB supplies a short (let's say 0.5m) piece of 2.5mmsq T&E, which
supplies a single socket next to the consumer unit.

My calculations suggest that the L-N fault current at the socket is going to
be about 5kA, and for adiabatic I^2t or (kS)^2 compliance, we would need to
break a short circuit in around 3ms max to avoid damaging the 2.5mm cable.

Really you want Andy for this, but the 5kA sounds wrong to me.

Supply impedance for 6kA PSC is 240 / 6000 = 0.04 ohms.
Add to that the 0.5m of 2.5mmsq T&E, which will have a
resistance of 0.09 ohms, and we have 0.13 ohms.
240V across 0.13 ohms is only 1.846kA.

Being a physicist rather than an electronic engineer, I'll
work out the temperature rise from first principles rather
than using the method in BS7671.

Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec.
500cm of 2.5mm^2 cable during a fault current of 1.846kA for 0.01 sec.
Mass of copper is 8.96g/cm^3 * 0.025cm^2 * 500cm * 2 = 224g,
Specific heat capacity of copper = 0.385J/g/K...

Power dissipated per 500cm:

P = I^2 * R = 1846^2 * 0.09 = 306kW

Energy dissipated per 500cm during 0.1 sec:

306kW * 0.01 = 3060J.

Temp rise = 3060 / 0.385 / 224 = 35.5 Centigrade

Thus a fault current of 1.846kA for 0.1 sec will result in a temperature
rise of about 35.5 Centigrade in the copper conductors. If cable was
previously within its correct operating temperature range (i.e. = 70 C),
it will survive this without damage (i.e. not exceed 160 C for PVC).

--
Andrew Gabriel