Circuit Wire For Long Run

Cookie · Apr 2, 2021

Ok, say you've got a lighting run with the last fixture being 1,280 meters away. 1,800 watts per phase 3 phase, 277/480Y volts though in this example we can use 230/400Y for the ease of discussion.

My question is- what size wire do I run for each phase and earth? What disconnection time do I use? Do I use a Type B, C or D breaker? What should my max R1+R2 come out to be?

[ElectriciansForums.net] Circuit Wire For Long Run

pc1966 · Apr 9, 2021

Cookie said:
Last question- I promise ?

Any idea what the AC R would be at 150*C and 250*C? I'd like to use that as a basis to calc final short circuit conditions.

The basic assumption is copper is fairly linear w.r.t. temperature up to around 1000C (so anything that might be usable for a cable) so you have:

R = R20 * (1 + alpha20 * (t - 20))

Where R20 is the temperature at 20C (e.g. from tables) and alhpa20 is the coefficient at 20C which is around 0.00393 (depending on copper purity and annealing, etc, but not far off this).

So if looking at 250C and the 20C tabulated values for UK cables, then the adjustment factor is:

1 + 0.00393 * (250 -20) = 1.9039

So basically almost double the cold resistance.

Cookie · Apr 9, 2021

What about 150*C?

Soi disant · Apr 9, 2021

Cookie said:
What about 150*C?

Ummm......

pc1966 said:
R = R20 * (1 + alpha20 * (t - 20))

Where R20 is the temperature at 20C (e.g. from tables) and alhpa20 is the coefficient at 20C which is around 0.00393 (depending on copper purity and annealing, etc, but not far off this).

So if looking at 250C and the 20C tabulated values for UK cables, then the adjustment factor is:

1 + 0.00393 * (250 -20) = 1.9039

Cookie · Apr 10, 2021

Soi disant said:
Ummm......

I'm still getting the hang of this.

Cookie · Apr 10, 2021

pc1966 said:
The basic assumption is copper is fairly linear w.r.t. temperature up to around 1000C (so anything that might be usable for a cable) so you have:

R = R20 * (1 + alpha20 * (t - 20))

Where R20 is the temperature at 20C (e.g. from tables) and alhpa20 is the coefficient at 20C which is around 0.00393 (depending on copper purity and annealing, etc, but not far off this).

So if looking at 250C and the 20C tabulated values for UK cables, then the adjustment factor is:

1 + 0.00393 * (250 -20) = 1.9039

So basically almost double the cold resistance.

If curious here is why I ask:

Parallel Arc Fault Resistance Found to be 0.03 Ohms | on Free Electrical Advice - Electricians Forums

Find Parallel Arc Fault Resistance Found to be 0.03 Ohms Advice and Help. How-to Parallel Arc Fault Resistance Found to be 0.03 Ohms in the UK Electrical Forum advice boards on ElectriciansForums.net | Free Electrical Advice - Electricians Forums: Electricians' Talk Forums. Parallel Arc Fault...

www.electriciansforums.net

UL and the NFPA are using 25*C for their loop impedance calcs and a Ze of 0.25 ohms.

It is the belief of many electricians that the fault current is universally much higher than 500 amps.

If this is the case, that arc fault coverage reaches much further into the circuit. If we can prove that to be the case, much of 210.12 is nullified.