Cross sectional area current capacity

Joined
26 Apr 2015
Messages
51
Reaction score
0
Country
United Kingdom
How would one determine the current carrying capacity of 2.08mm2 conductor?
 
Sponsored Links
I dont think those are accurate, or would comply with the regs. For example 2.08 seems to be listed at 15 amps, but that is obviously based on method 101 as 1.5 is already 15 amps for installation method A. Dont the regs base the values off a certain equation? Either that use the current value inbetween 1.5mm and 2.5mm2.
 
Sponsored Links
I presume they are accurate for AWG and the US - AWG 12 (2.05mm²) 20A.

The maximum for 1.5mm² T&E is 20A but it is a different cable and conductor.
 
I presume they are accurate for AWG and the US - AWG 12 (2.05mm²) 20A.

The maximum for 1.5mm² T&E is 20A but it is a different cable and conductor.

I dont think so. I did some digging and found this. 12 gauge is 3.31 mm2. This wire here is 2.08mm2.
 

Attachments

  • awgtomm2.jpg
    awgtomm2.jpg
    145.2 KB · Views: 305
Found this, but I can't say how it relates to UK CCC's though. Realistically I'd treat it more like 1.5mm than 2.5mm

Is this wiring existing? Do you HAVE to use it or can you replace it?

DC_wire_selection_chartlg.jpg
 
Dont the regs base the values off a certain equation? Either that use the current value inbetween 1.5mm and 2.5mm2.
No published equation that we are aware of. However, if you look at the second post from me on this page of a recent thread, you will see that the BS7671 (Table 4D5) Method C CCC figures can be closely modelled by:

CCC = 15.774 x CSA^0.6077

... for a 2.08mm² conductor, that would give a (Method C) CCC of about 24.6A. Mind you, I don't know that an inspector or person doing an EICR would accept my modelling, so it might be safer to assume, conservatively, that it's 1.5mm² !!

Kind Regards, John
 
I presume they are accurate for AWG and the US - AWG 12 (2.05mm²) 20A.
The maximum for 1.5mm² T&E is 20A but it is a different cable and conductor.
I dont think so. I did some digging and found this. 12 gauge is 3.31 mm2. This wire here is 2.08mm2.
Perhaps they are inaccurate. I was only going by the first link.
 
Dont the regs base the values off a certain equation? Either that use the current value inbetween 1.5mm and 2.5mm2.
No published equation that we are aware of. However, if you look at the second post from me on this page of a recent thread, you will see that the BS7671 (Table 4D5) Method C CCC figures can be closely modelled by:

CCC = 15.774 x CSA^0.6077

... for a 2.08mm² conductor, that would give a (Method C) CCC of about 24.6A. Mind you, I don't know that an inspector or person doing an EICR would accept my modelling, so it might be safer to assume, conservatively, that it's 1.5mm² !!

Kind Regards, John


Thanks! That is what I was looking for. :) I think it does hold some value in my opinion.


However this equation only works for clipped direct I take it? What would it be reduced to when using other methods like method A? And the power curve is based on a 70*C conductor operating temperature?
 
Thanks! That is what I was looking for. :) I think it does hold some value in my opinion.
You're welcome - but, as I said, I'm not sure whether 'third parties' would necessary accept my figures :)
However this equation only works for clipped direct I take it? What would it be reduced to when using other methods like method A? And the power curve is based on a 70*C conductor operating temperature?
Yes, as I said, it's a modelling of (Method C) of Table 4D5 - so, yes, 70°C conductor temperature. If I had the time, I could try modelling Method A. However, looking at Table 4D5, the difference between Method C and Method A is always a factor of between 1.35 and 1.49 (1.38 for 1.5mm² and 1.35 for 2.5mm²), so if you divided the Method C figure by 1.49 (to be safe/conservative) or by 1.38 (based on the 1.5/2.5mm² figures), you wouldn't be far off - so, for 2.08mm² Method A (70°C), that would be about 16.5A (or 17.8A).

Kind Regards, John
 
Thanks! That is what I was looking for. :) I think it does hold some value in my opinion.
You're welcome - but, as I said, I'm not sure whether 'third parties' would necessary accept my figures :)
However this equation only works for clipped direct I take it? What would it be reduced to when using other methods like method A? And the power curve is based on a 70*C conductor operating temperature?
Yes, as I said, it's a modelling of (Method C) of Table 4D5 - so, yes, 70°C conductor temperature. If I had the time, I could try modelling Method A. However, looking at Table 4D5, the difference between Method C and Method A is always a factor of between 1.35 and 1.49 (1.38 for 1.5mm² and 1.35 for 2.5mm²), so if you divided the Method C figure by 1.49 (to be safe/conservative) or by 1.38 (based on the 1.5/2.5mm² figures), you wouldn't be far off - so, for 2.08mm² Method A (70°C), that would be about 16.5A (or 17.8A).

Kind Regards, John


That makes sense. Now my understanding is the factor is less restrictive for 1.5mm2 because smaller conductors can dissipate heat better? Or is it the other way around? Correct to think that heat dissipation of 2.08mm2 is in between 1.5mm2 and 2.5mm2?

Edit: just to confirm, by 1.35 you mean 1.38? My apologies.
 
That makes sense. Now my understanding is the factor is less restrictive for 1.5mm2 because smaller conductors can dissipate heat better? Or is it the other way around? Correct to think that heat dissipation of 2.08mm2 is in between 1.5mm2 and 2.5mm2?
Yes, I suppose that's got to be true to some extent, but there's hardly any difference between 1.35 and 1.38, so one might as well 'be conservative'! However, I also suspect that some of the small variation in the de-rating factors (from 1.39 for 1mm² {yes, fractionally higher than 1.5mm² and 2.5mm²} to 1.49 at 16mm²) may be as much to do with rounding as anything else.

Kind Regards, John
 

DIYnot Local

Staff member

If you need to find a tradesperson to get your job done, please try our local search below, or if you are doing it yourself you can find suppliers local to you.

Select the supplier or trade you require, enter your location to begin your search.


Are you a trade or supplier? You can create your listing free at DIYnot Local

 
Sponsored Links
Back
Top