Tabulated current carrying capacity

[schoe]

Hi all can some one help me out with this,

It = In divided ccxcixcgxca

I have a 32 amp mcb divided by cc 1 x cg 0.80 x ca 1.03 = 38.8 amps current carrying capacity.

What would the tabulated current carrying capacity be, as it should be higher than the current carrying capacity.

Hope that's not to confusing,

Thanks chris

 

[schoe]

Oh and I didnt have to use ci as it doesn't pass through any insulation

 

[Rampage]

Depends what cable type you've been specified. Find the table in the regs that corresponds to the type specified an find the cable with a higher rating (according to reference method) than the 38.8A.
I take it the Ca was 25c?

 

[schoe]

Depends what cable type you've been specified. Find the table in the regs that corresponds to the type specified an find the cable with a higher rating (according to reference method) than the 38.8A.
I take it the Ca was 25c?

Yep ca was 25 c. The cable size is 2.5mm twin and earth. When I looked in the on site guide it looked like the maximum tabulated current for

 

[schoe]

Sorry about that, when I looked in the onsite guide it looked like the maximum for clipped directed 2 single cables flat or touching is 27 amps which would make my calculation to high, is this correct? Thanks rampage.

 

[Rampage]

Yes, the maximum for 2.5 T+E is 27A which is clipped direct.
Is it a ring final? If so, then multiply by 0.67 (that's the way I've been taught, but I don't think it's in the regs?!).
If its a radial then 2.5 isn't big enough.

 

[schoe]

Yes, the maximum for 2.5 T+E is 27A which is clipped direct.
Is it a ring final? If so, then multiply by 0.67 (that's the way I've been taught, but I don't think it's in the regs?!).
If its a radial then 2.5 isn't big enough.

Yep it's a ring, which bit are you multiplying 0.67 by? Thanks again mate.

 

[telectrix]

for a RFC, you multiply the 27A for 2.5mm T/E by 1.67 to get the ccc.

 

[schoe]

for a RFC, you multiply the 27A for 2.5mm T/E by 1.67 to get the ccc.

Thanks mate where does that figure come from? Regs? Also what about cc,ci,ca are they dis regaured? So by multiplying 27a by 1.67 should the figure be 45.09 amps

 

[Rampage]

The factor of 0.67 is to allow for the use of 2 cables in a ring, its In x 0.67
I don't know where it comes from though, I can't find it in the regs. If your at college then go by what the regs say!
Look at appendix 15 (green book) ring and radial final circuit arrangements.
Also reg 433.1.103 (green book) or 433.1.5 (red book).

 

[Rampage]

Out of interest telectrix, is the It x 1.67 in the regs?

 

[MarkieSparkie]

Out of interest telectrix, is the It x 1.67 in the regs?

It's not in the regs explicitly, because it's the ratio of the line conductor to the CPC for standard 2.5mm[SUP]2[/SUP] T+E cable e.g. 2.5mm[SUP]2[/SUP]/1.5mm[SUP]2[/SUP] = 1.67 If the T+E cable is a different size, then the ratio will be different for that cable, so for standard 6mm[SUP]2 [/SUP]T+E the ratio is 6mm[SUP]2[/SUP]/2.5mm[SUP]2 [/SUP]= 2.4

 

[i=p/u]

would ring not just equal to 5mm wired in 2.5mm , say 27A clipped direct x 2=54

 

[johnboy6083]

No mate because the loading is shared equally throughout the ring (or should be). What your proposing is the practice of parallel conductors. They must both be the same length and be protected by the same protective device ( unless you use electrical/mechanical interlocks). Each conductor must feed the same load, so that the current is shared equally between the conductors.

 

[i=p/u]

i remember now. as not same length and load distributed around the load... so what would you put on this waterboyspark lol