ring design and voltage drop

[esc]

Hi guys. Ive generally no problem going through the deign process of cable selection for a circuit to be installed, (i try keep my hand in now and then rather than just looking in the OSG) but when i look at a RFC as in appendix 15, i get a little stuck. Ib = 32A, so In = 32A breaker, take it the cable is clipped direct to avoid any correction factors in this example, so far so good. So now i find a cable Iz, from the tables in appendix 4 to carry this 32A, im looking at table 4D5. Here, i feel sure we treat this ring as two radials supplying any particular SO on the ring. So each leg now needs to be able to carry 16A, and im assuming here, that we select 2.5mm T&E because of the time current characteristics of our 32A CB?, as both 1.0 and 1.5mm sq cable can carry this load, but could become a serious hazard under fault conditions, whereas 2.5 shouldn't?!
Also, do we calculate the voltage drop over the entire length of the cable, taking it that a SO could be midway or a few inches from the origin of the circuit? and the long leg would suffer far more volt drop?
Cheers

 

[Deleted member 26818]

The minimum cable rating for a standard RFC supplying BS1363 accessories is 20A, after correction factors have been applied.
There are a number of ways that you could use to work out VD, although it is generally accepted that if the total floor area covered by the RFC is no greater than 100m², it will be within limmits.

 

[Engineer54]

Hi guys. Ive generally no problem going through the deign process of cable selection for a circuit to be installed, (i try keep my hand in now and then rather than just looking in the OSG) but when i look at a RFC as in appendix 15, i get a little stuck. Ib = 32A, so In = 32A breaker, take it the cable is clipped direct to avoid any correction factors in this example, so far so good. So now i find a cable Iz, from the tables in appendix 4 to carry this 32A, im looking at table 4D5. Here, i feel sure we treat this ring as two radials supplying any particular SO on the ring. So each leg now needs to be able to carry 16A, and im assuming here, that we select 2.5mm T&E because of the time current characteristics of our 32A CB?, as both 1.0 and 1.5mm sq cable can carry this load, but could become a serious hazard under fault conditions, whereas 2.5 shouldn't?!
Also, do we calculate the voltage drop over the entire length of the cable, taking it that a SO could be midway or a few inches from the origin of the circuit? and the long leg would suffer far more volt drop?
Cheers

Ring final circuits should NEVER be designed/installed with a long leg as you put it. The load should always be evenly distributed (as far as humanly possible) throughout the overall rings length...

 

[ackbarthestar]

Interlacing Socket outlets through conduit is one way of achieving equally spaced loads around a RFC.

 

[Engineer54]

Interlacing Socket outlets through conduit is one way of achieving equally spaced loads around a RFC.

That's what i call, and was taught to call ...''A Staggered Ring Circuit''

 

[ackbarthestar]

That's what i call, and was taught to call ...''A Staggered Ring Circuit''

DO you mean, 'well I'm staggered I managed to pull all that cable though such a small pipe?:sifone:

 

[telectrix]

staggered home from the pub, more like. LOL.

 

[esc]

Thanks for that spinlondon. I'm ok working out VD on a radial cicuit, but in a RFC where theres no furthest outlet as such, i get a little confused. Cant quite get my head around the VD at each outlet of a RFC, wether it differs from outlet to outlet or is the same. If r1 is the same at each SO, instinct would have me believe VD is the same at each SO. Irrespective of that, instinct also tells me to calculate VD over the entire circuit length, CU back to CU. I appreciate its accepted to be within limits if the circuit is installed according to appendix 15 and would be adjusted for any correction factors. Im sure there are more important questions. I was just curious about the details.
Cheers

 

[esc]

Hi engineer54. I appreciate distributing the load in a RFC, my appollogies for exagerating a SO being so close to the CU. What i was trying to point out is that the SO's vary in distance from the CU and each other, is the VD at each SO the same, unlike a radial, A2 orA3 and would i calculate VD over the total length of cable used for the RFC, ie CU back to CU
Cheers

 

[i=p/u]

hmmm, if you do any sum on this , which i havent really a clue as never thought of it before, just the 100m area, the readings still come out more than 11.5v.........

 

[esc]

HI i=p/u. Thanks for letting me know the readings dont tally either. The only way ive looked at this is as follows. I know this is rubbish, it has to be, im completly missing something, not only for a RFC, but also for the A2 and A3 radials in appendix 15. From the OSG the resistance of 2.5 T&E with 1.5 CPC is 0.01951 ohms per meter (m). Our PD is rated at 32A on the RFC and A2 radial and our V (VD) is 11.5V, so using R = V / I we have

0.01952 * m = 11.5 / 32

m = 18.41 ????

with a PD of 20A for the A3 radial

m = 29.45

i know we can construct a RFC and a A3 radial out of that much cable, but i wouldn't be able to cover an area 100m sq and 50m sq respectivly

For an A2 radial 4.0mm sq cable with 2.5cpc the resistance is 0.01202 ohms per m, so again

0.01202 * m = 11.5 / 32

m = 29.89 ?????
its beyond me. Shall just take appendix 15 as it is i suppose!

Cheers

 

[i=p/u]

im reading the big green book, or i would attack this from all angles... but i never heard anyone talking about it... i say just take appendix 15 as it is

 

[i=p/u]

isit the further down the ring the less current you use, so the calcs are ajusted to load on ring

 

[i=p/u]

einstein has worked it out and says for us to leave it out to technical

 

[esc]

Well, whos gonnan argue with einstein! I was just reading the big red book i-p/u, im still saving up for a green one. The bottom of pg 257 and the top of pg 258 sum it up i suppose. Have a look when you have time, if you havent burnt your red one already. Ive just looked at the mV/A/m values in table 4D5 and got different answers to earlier!, but according to pg 257-258 these new figures are correct.
mmmmmm????
Just don't get how they introduce VD as a necessary design element and then show standard circuits for construction in appendix 15 that fall short of their own criteria

 

[JamesBrownLive]

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

 

[esc]

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

 

[JamesBrownLive]

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.

 

[Deleted member 26818]

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.

 

[JamesBrownLive]

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