ring design and voltage drop

[esc]

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

 

[i=p/u]

so id spinlondon saying you can use diversity in rfc when working out volt drop

 

[esc]

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

 

[i=p/u]

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.....

 

[esc]

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?

 

[Deleted member 26818]

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.

 

[Knobhead]

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!)

 

[i=p/u]

open the window tony and launch it... im sure the forum could sponsor you new 1 as you a classy/smart contribuiting member.. tho grumpy

 

[Richard Burns]

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!)

Install chrome or Firefox

 

[Knobhead]

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 ! ! ! ! !