Circuit Wire For Long Run

[Cookie]

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?

 

[Cookie]

If it is 75-30 = 45C colder then it would be 17.7% lower for the resistive part.

For big cables with X dominating Z the overall difference is going to be less, as that won't change much with temperature.

So my ignorance will show, but if I type in 17.7% of 2.0 in Google I will have the 30*C Ac resistance of 12 gauge wire?

 

[pc1966]

So my ignorance will show, but if I type in 17.7% of 2.0 in Google I will have the 30*C Ac resistance of 12 gauge wire?

No, you need it reduced by 17.7% so multiply by 0.823

 

[Cookie]

No, you need it reduced by 17.7% so multiply by 0.823

Will this work with all wires sizes from 2.08mm2 to 1013mm2?

 

[pc1966]

Will this work with all wires sizes from 2.08mm2 to 1013mm2?

Yes, but only the the R term, the X term won't vary much with temperature and for large conductors that inductance starts to dominate the overall Z.

E.g. looking at post #40 you see X becomes comparable to R at cables around the 150mm or so size.

 

[Cookie]

Yes, but only the the R term, the X term won't vary much with temperature and for large conductors that inductance starts to dominate the overall Z.

E.g. looking at post #40 you see X becomes comparable to R at cables around the 150mm or so size.

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.

 

[pc1966]

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]

What about 150*C?

 

[Soi disant]

What about 150*C?

Ummm......

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]

Ummm......

I'm still getting the hang of this.

 

[Cookie]

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 ElectriciansForums

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...

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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.

There is more, but thats the general part of it.