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 ! ! ! ! !

 

[Richard Burns]

Sounds like your computer thinks it is a mobile phone, so the forum is sending you to the mobile site or vice versa.
Unfortunately I do not know how the forum does that (client or server) so can't help.
Usual things of clearing caches, directly typing the forum address in the address bar not using links/favorites. but you have probably tried all that. Make sure there are no spurious m's in the web address you go to.
Good luck

 

[Chr!s]

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?

I think your not far off, you need to make a further consideration to load distribution and conductor temperature and your there.

Regards Chris

 

[Guest123]

I use an (Ib) of 26A when calculating for an A1 RFC which would gives a maximum length of 98M.

It can be engineered further but I prefer to err on the side of caution with 26A and (Ct) of 1.

 

[i=p/u]

can you show us your calculation in full lenny..... inspre me baby

 

[esc]

Hi spinlondon, i never been taught to use diversity with cable selection. That doing so is is a definite no no. Diversity is applied to maximum demand to give a practical/realistic expectation of the current being drawn off he supply and hence through the service fuse at any one time.

 

[esc]

Cheers chris. Im trying to get my head around load distribution and the calculations.
Get what you say about conductor temperature.
I've done the calculation without any correction and method C to keep things as simple as possible

 

[esc]

Hi lenny, im with i = p/u. Can we see you full calc
Cheers

 

[Risteard]

Hi lenny, im with i = p/u. Can we see you full calc
Cheers

It assumes 20A at one end the of ring with a further 12A evenly distributed around the ring. Therefore it is 20 + (12/2) = 26A.

 

[Deleted member 26818]

Hi spinlondon, i never been taught to use diversity with cable selection. That doing so is is a definite no no. Diversity is applied to maximum demand to give a practical/realistic expectation of the current being drawn off he supply and hence through the service fuse at any one time.

Consider then a RFC with 10 socket-outlets.
What CSA of cable would your method indicate, and what rating of CPD would be required?

 

[malcolmsanford]

That is how I work out the length and volt drop in a Ring.

Length

2.5mm T&E ring.
1 conductor 2.5mm CSA @ 7.41mohm/m
1 conductor 1.5mm CSA @ 12.10mohm/m
{from page 166 OSG}

so if you have a measured value of say

r1 @ 0.58/7.41 x 1000 = 78.27m
rn @ 0.58/7.41 x 1000 = 78.27m
r2 @ 0.96/12.10 x 1000 = 79.33m

Volt Drop

Divide the length by 4 = 19.57 x 18 x 26/ 1000 = 9.16v

Bs 7671 pg 258, 6.1 correction for operating temp.

This calculation give you a ct of .923, x 9.16 = 8.45v

or

4 x 8.45v x 1000/26/18 = 72.22 / .923 = 78.24m

It assumes 20 amp at the furthest point and 12 amp distributed evenly around the ring so,

average = (32 + 20)/2 = 26 which is the value that Lenny posted.