ring design and voltage drop | Page 2 | on ElectriciansForums

Discuss ring design and voltage drop in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

I thought you calculate VD based on Ib not on In :thinking:


Edit:
Sorry, I see back in your OP that you are saying your Ib and In are the same. I just assumed that your Ib would have been lower, and that you were basing your example on a 32A CPD.

Cheers
 
Last edited by a moderator:
Hi jamesbrownlive. You're quite right that when calculating VD we use Ib, sorry if i used In above, but in this instance the(A1 A2 and A3 circuits) from appendix 15 have, Ib = In. Appendix 15 gives us the PD values for A1 A2 and A3 and we know we need Ib<=In, so the value of Ib can be equal to but no more than the PD ie 32/30A A1, 32/30A A2 and 20A A3
 
Having thought about it further, when calculating VD for an RFC you need to divide the VD by 4, because the length of the cable is halved, with twice the CSA. Hope that makes sense.
 
No the expected load at a BS1363 accessory is a maximum of 26A, though it is generally taken as being 20A.
You also appear to be forgeting to halve the figures for a RFC.
 
So if you had a 40m RFC (Just enough to cover the 100m2 floor area) with an Ib of 32A and using 2.5mm T&E, you would calcualte the VD as:

18 X 32 X 40

= 23,040

/1000

Gives you a total VD of 23.04V

Divided by 4 gives a working VD of 5.76V
 
Thanks for that jamesbrownlive. What you say is making sense to me and i understand the logic. Your answer reminds me about the fact that resistance is proportional to length and inversly proportional to CSA. Will ponder this and try get my head around it in relation to how a RFC is constructed.

With Thanks
 
Well, i wouldn't of thought so i = p/u, least i hope not. Diversity cannot be applied during the design process/cable selection process, we have to select cable size according to the full load and after considering correction factors. Diversity is not a part of cable selection
 
kk i remeberr somewhere someone did say divide by 4 now .....

what is the sum you were doing a above too work out how many meters you could get?? what did m stand for? say 11.5v divide by what = what...... i like to know things inside out.....
 
Ive used both ohms law R = V / I and used the values of mohms per meter out of table 4D5 in BS 7671.


From the OSG 2.5 T&E has a resistance of 0.01951 ohms per meter, so R = 0.01951 times m (meters) that is

R = 0.01951 * m

Ive taken V as volt drop at 11.5V and I as 32A, the rating of the PD, giving:


0.01951 * m = 11.5 / 32 ohms law


or m = (11.5 / 32)/0.0195 = 18.43m ??


Using the mV/A/m from table 4D5 of BS 7671, we have a value of 18mV, therefore

(18*32*m)/1000 = 11.5V


or m = 19.96m ??

Ive got this all wrong i know i = p/u. Think my perceptions of how a RFC handles voltage and current is off the scale!! But now ive had a thought from a post earlier.

Is from jamesbrownlive's post above, pg2, and how you both mention about dividing by 4. It would seem logical that when considering a RFC's construction, we have to consider that as a ring the CSA is doubled and the length effectivly halved, which would reduce the resistance to 1/4 of its standard value of 0.0195 ohms per m, in this case that would become 0.004875 ohms per m. Recalculating we have:

0.004875 * m = 11.5 / 32

m = 73.7m
which now seems ample length of cable to construct an A1 ring

What you think?
 
Last edited by a moderator:
Of course diversity is used during the design process/cable selection.
When else would you use it?
It would be a bit pointless applying it after you have designed and installed the circuit.
 
Widler has a good drawing for this. I think it’s in one of the sticky’s. It’s easy when it’s drawn out. (I’ve got to get this bloody PC sorted out so I can post drawings!)
 
Tried both, they both work until it gets to this site, and then I’m back to “mobile crap” screen. PLEASE someone HELP it doing my bloody head in ! ! ! ! !
 

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