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Brutal

I have installed Ring Final Circuit in 2.5 LOR 120m protected by 32A breaker

Volt drop calcs - 18mv/a/m x 32A x 120m / 1000 = 69.

So the volts drop is 69 volts? which is way over the stated 3 or 5 %.

Am I diong this corectly and if so what can I do about the volts drop?

Cheers
 
yep sounds better, 69 volts is a lot on 120 meters buddy.

It does seem a lot but the calcs give 69 volts.

If my calcs are wrong.... Where am I going wrong?

mv/a/m - 18

Ib - 32a

LOR 120

18 x 32 x 120 = 69120

divide by 1000 cos in milli volts gives me 69.1

The 120 meters is the whole circuit (as in ring)

surely I would just need to measure to the furthest point? say 60m

Also If it is a ring circuit and the CSA is doubled as two 2.5mm cables.... The total CSA is 5mm - does this have an impact?

I've been running my business for a while now so need to start getting this stuff right!
 
You may be right, as I said I didn't have my head on, but that sounds a hell of a lot of volt drop for that circuit, however If your sure?? then fair enough
 
You may be right, as I said I didn't have my head on, but that sounds a hell of a lot of volt drop for that circuit, however If your sure?? then fair enough

ok.... so what can I do about it? I have to provide calcs for the job, I can't show 69v volt drop
 
it's more like (18 x 20 x 120)/4000 = 10.8V.

20A because that's about the correct value of Ib. 4000 because it's a RFC. so you divide by 2 for the 2 cables, and 2 again because the cables are in parallel. same as R1+R2= ( r1+r20/4.
 
it's more like (18 x 20 x 120)/4000 = 10.8V.

20A because that's about the correct value of Ib. 4000 because it's a RFC. so you divide by 2 for the 2 cables, and 2 again because the cables are in parallel. same as R1+R2= ( r1+r20/4.

so I need to put circuits on 20a breakers?
 
no. it's a 32A RFC. you assume 1/2 the current goes down each leg, but allow a bit more ,20A instead of 16A, as at most points on the ring, 1 leg will be shorter than the other and so carry more of the load current (less resistance). it's only at the mid point that each leg will carry the same current.
 
well that's the way i've always calculated VD for a ring. chances are that most of the time the load will be well less anyway. most of the time, except for kitchens, the current will be < 5A.
 
Tel is just about there

You Divide the length by 4 so your calc will be 18 x 26 x 30/1000 = 14.04v then corrections for operating temperature and that would be 14.04 x 0.923 = 12.95 volts

This assumes 20 amp at the furthest point and 12 amp distributed evenly around the ring so, average = (32 + 20)/2 = 26

So I'm afraid your circuit fails. And you could check this as the OSG guide Table 7.1 (i) for a 2.5mm RFC is 106mts on RCD/MCB. It is slightly longer on a BS 3036 fuse becasue of the 0.75 rating factor
 
i was a bit more generous with current demand. apart from a kitchen RFC, it's extremely rare to see anywhere near 20+A on a RFC.
 
Agree 100% mate, and in the real world a 120m ring is not going to fail, unless it's going to be loaded up, but the OP said that he needed to show his calcs, and if he does it fails.

If he needs to show his clacs somewhere along the line there is going to be an ole f##t like me telling him sorry son that is not acceptable
 
malcolm. you're allowed to say "fart" as it's not in the banned list. it's also in the OED.
 

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