DIYbanter - View Single Post - Adiabatic short-circuit compliance on very short short-circuits

In article ,
(Andrew Gabriel) writes:
These two combined make the result way off. I'll do the
calculation and post again.

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 0.009 ohms, and we have 0.049 ohms.
240V across 0.049 ohms is 4.878kA.

Let's assume fastest MCB trip time possible is half a cycle, 0.01 sec.
Specific heat capacity of copper = 0.385J/g/K
Density of copper = 8.96g/cm^3
So, mass of copper in 50cm of 2.5mm^2 cable is
8.96g/cm^3 * 0.025cm^2 * 50cm * 2 = 22.4g

Power dissipated per 50cm:

P = I^2 * R = 4878^2 * 0.009 = 214kW

Energy dissipated per 50cm during 0.01 sec:

214kW * 0.01 = 2140J.

Temp rise = 2140 / 0.385 / 22.4 = 248 Centigrade

Thus a fault current of 4878A for 0.01 sec will result in a temperature
rise of about 248 Centigrade in the copper conductors, which will
damage or wreck the cable.

So let's work backwards to identify the maximum fault current 2.5mm^2
T&E can handle...

Max fault current temperature rise allowed is 90C.
Mass of copper per metre = 44.8g
Max energy per metre = 90 * 44.8 * 0.385 = 1552 Joules.
Max power per metre for a 0.01 second fault = 1552 / 0.01 = 155200 Watts.
Max current = sqrt(P/R) = sqrt(155200 / 0.018) = 2936A

So you are always going to get some length of 2.5mm^2 T&E which
is inadiquately protected if the PSC at supply end is greater
than 2936A. In the case of your 6kA PSC supply and 2.5mm^2 T&E,
that's going to be the first 2.3 metres of cable (if I did the
calculation right;-)

--
Andrew Gabriel