Confused over electrical cable length and size

[JohnW2]

Namely that most of it didn't really need saying in relation to the thread; would be of no help to DIYers and you didn't explain the reason for most of your statements.

Fair enough. I won't bother responding to your comments, then - since responses to comments about something that didn't need saying would be a bit pointless.

Kind Regards, John

 

[JohnW2]

I see not 'grave error'. As I said (and was quoted), I was talking about a circuit which was "fairly 'linear' in physical arrangement", and one with multiple branches would not qualify as that.

 

[ban-all-sheds]

Sockets around the walls of houses when internal walls are included are not "fairly 'linear' in physical arrangement". A "star topology" to connect every one could easily be the most economical use of cable.

 

[sparkwright]

I understood that the maximum length of a cable on a 2.5mm2 32amp ring was 73m, not 106m.

I would certainly go for two rings rather than one.

Also, traditionally at least, the floor area served by a ring should not exceed 100m2 - though this does seem pointless now bearing in mind things like electric heaters are not used in the way they were decades ago.

 

[JohnW2]

I understood that the maximum length of a cable on a 2.5mm2 32amp ring was 73m, not 106m.

I have not actually noticed it being mentioned in this thread, but the 106m figure is the one that ericmark is always quoting - and, as I understand it, it derives from the fact that he was once told at a meeting that one can/should work out the 'maximum VD' in a 32A ring circuit on the assumption that there is a 20A load at the centre of the ring with the other 12A being evening distributed down the two halves of the ring.

Although it doesn't sound particularly unreasonable an assumption to use, I've never seen it written, other than by eric.

Kind Regards, John
Edit: important missing word added!

 

[EFLImpudence]

Doesn't that, though, work out to 98m - using 18mV/A/m?

I can't get anything to come to 106m. - even using old figures of 240V or 6%.

 

[ban-all-sheds]

I have actually noticed it being mentioned in this thread, but the 106m figure is the one that ericmark is always quoting - and, as I understand it, it derives from the fact that he was once told at a meeting that one can/should work out the 'maximum VD' in a 32A ring circuit on the assumption that there is a 20A load at the centre of the ring with the other 12A being evening distributed down the two halves of the ring.

Although it doesn't sound particularly unreasonable an assumption to use, I've never seen it written, other than by eric.

If you want to see it written by someone other than Eric, look at Table 7.1 in the Red OSG. I'm sure it's in later versions, but I don't have any, so cannot provide a reference.

 

[JohnW2]

Doesn't that, though, work out to 98m - using 18mV/A/m? ... I can't get anything to come to 106m. - even using old figures of 240V or 6%.

Unless I'm missing something (a common occurrence ), it's far from a straightforward calculation. I need to scribble - watch this space.

Kind Regards, John

 

[JohnW2]

Doesn't that, though, work out to 98m - using 18mV/A/m? .... I can't get anything to come to 106m. - even using old figures of 240V or 6%.

Unless I'm missing something (a common occurrence ), it's far from a straightforward calculation. I need to scribble - watch this space.

Right - scribbling done. I actually make it about 104.3 metres. As I said, it's not a trivial calculation - so are you sitting comfortably .....

I will assume that the "12A evenly distributed along the two legs" is equivalent to 6A at the middle of each leg (i.e. a quarter of the total ring length from the CU).

Of that 6A, 4.5A will travel by the 'short route' and 1.5A by the 'long route'. The two 1.5A 'long route' currents (one from each side of the ring) travelling in the two 'middle quarters' of the ring will cancel, since they are going in opposite directions, and therefore will have no effect on VD.

Considering just 'one side' of the ring, the current in the half of that leg (i.e. quarter of the ring) more remote from the CU will just carry 10A (half of the 20A load at the centre of ring) (the two 1.5A 'long route' currents having cancelled). The current in the half of the leg closer to the CU will carry 14.5A (10A from central load, plus 4.5A of the 6A load on this leg).

Hence, if the total length of the ring in L, then the VD in the half of the leg remote from the CU will be:

10 x {L/4} x 0.018 volts

... and the VD in the half of the leg closer to the CU will be:

14.5 x {L/4} x 0.018 volts

Hence the total VD is:

(10 x {L/4} x 0.018) + (14.5 x {L/4} x 0.018)
= 24.5 x {L/4} x 0.018

Hence, for the guideline maximum VD of 11.5 V

24.5 x {L/4} x 0.018 = 11.5

Whence: L = (11.5 x 4) / (24.5 x 0.018) = 104.308 metres

So, if I've got it right, although all these figures are all pretty close, I'm not sure how eric's 106 metres or your 98 metres are derived.

Kind Regards, John

 

[ban-all-sheds]

I can't get anything to come to 106m. - even using old figures of 240V or 6%.

• No drawing the load at multiple points, 24A at the midpoint.

• The ring topology means halve the current and halve the length - easiest way to do that is current/4, so 6A

• 6 x 18 =108mV/m

• 11.5/.108 = 106.5

Interestingly, the figure in the 16th OSG is 84m - that one makes no sense.

 

[flameport]

17th edition yellow OSG:

 

[ericmark]

I understood that the maximum length of a cable on a 2.5mm2 32amp ring was 73m, not 106m.

I would certainly go for two rings rather than one.

Also, traditionally at least, the floor area served by a ring should not exceed 100m2 - though this does seem pointless now bearing in mind things like electric heaters are not used in the way they were decades ago.

I was given the 106 meter length at an IET lecture, I also could not get the results, so next lecture I asked the question, and I was told for volt drop on a ring final we assume 20 amp drawn from centre point and the rest even distributed, so average for rest is 6A on a 32A MCB so we calculate design current as 26 amp when working out volt drop. With that then you do get 106 meters.

The problem is with radial we use same idea, 20A at furthest point with rest even distributed, which since likely the MCB is 20A means 32 meters on a radial that goes to 42 meters with a 16A radial.

Now next question is with two rings how to split? Splitting side to side of house reduces cable used compared with up/down so on a technical footing it is better, also since upstairs likely less current is used than down stairs likely load more equal between the two rings. However traditionally we go up/down so that is what people expect and to move from it could mean wrong MCB is switched off.

However a MCB does not isolate it does not switch neutral, and one should test for dead, so not realising the socket is not dead is the fault of person working on it.

I found calculating volt drop with Correction factor Ct is not easy, so years ago I wrote a Excel program, however when my phone stopped doing free Excel I moved to Java Script. I still use it today.

My thoughts were if I do a EICR and miss a volt drop issue, could it come back and bite me latter? One can calculate volt drop from the loop impedance, so some one can with relative ease check. However once you factor in reading errors then not so easy to show some one got it wrong.

So 0.35 loop impedance at consumer unit and 0.94 (line - neutral) at furtherest point and that shows 106 meters, but 0.34 and 0.95 and you have exceeded volt drop by 0.5 volt, reverse 0.36 and 0.93 and your under by 0.5 volt. So at best readings are +/- 0.5 volt. Now add to this the designed does not need to use 20 amp centre and 12 even spread, he could decide on 16 amp centre and 16 even spread so could work it out as 24 amp design current, that gives him an extra volt to play with.

So in real terms one has to allow for 2 volt over before one could be sure there was a design fault. So the installer would need to go over by 30 meters before one could be 100% there was an error and you would need to measure cold. So by time you take into account the circuit may not be cold before you could say without fear of being wrong that too much cable has been used, it would have to calculate as an extra 50 meters.

So over 1 ohm line - neutral impedance I would be making a comment, under that would not say anything unless asked. Hind sight should have used PSCC not impedance, but only use the program for comments of forums now so not worth rewriting. And when I look at the forms, it does not ask for line - neutral PSCC or loop impedance to be entered measured at furthest point. So one could hardly prove it was incorrect last time a EICR was done. It is possible some one has extended the circuit.

So does excessive volt drop matter? Well for refrigeration equipment yes, and for TV's Radio's and like before the switch mode power supply. But volt drop is becoming less and less important. I have seen on caravan sites where battery chargers have failed due to volt drop, and I well remember a shrink wrap machine failing due to volt drop.

Fluorescent fittings have failed after 20 years of use due to volt drop, however that is because the DNO have reduced the voltage to cater for solar panels, not house internal wiring, in March 1993 we had an extra 9 volt volt drop added to UK supply, there were auto transformers sold which would auto change tapping to maintain a steady voltage, however they stopped working when there was high demand, so most regarded them as snake oil.

It would be interesting to hear how many actually measure the volt drop with an EICR, I would guess around 2%?

 

[ericmark]

• No drawing the load at multiple points, 24A at the midpoint.

• The ring topology means halve the current and halve the length - easiest way to do that is current/4, so 6A

• 6 x 18 =108mV/m

• 11.5/.108 = 106.5

Interestingly, the figure in the 16th OSG is 84m - that one makes no sense.

I make it 85.7 meters so 84 meters is not far out. I make it 107 meters at 5% volt drop so the figures for 4% volt drops are near enough.

 

[ericmark]

Doesn't that, though, work out to 98m - using 18mV/A/m?

I can't get anything to come to 106m. - even using old figures of 240V or 6%.

Correction factor Ct = 0.917762660619803 on my software so using 16.5197278911565 mV/A/m not 18 mV/A/m it is such a pain working out correction factor I wrote a Java Script program to do it for me.

 

[ericmark]

Whence: L = (11.5 x 4) / (24.5 x 0.018) = 104.308 metres

So, if I've got it right, although all these figures are all pretty close, I'm not sure how eric's 106 metres or your 98 metres are derived.

Kind Regards, John

That is not the corrected figure, that's why you don't get same as me. In fact I don't get 106 I get 107 I will guess it is rounding down or up which accounts for difference, I would guess they use 16.52 not 16.5197278911565 as my software uses, it would all depend on calculator used but 4 s.f. is reasonable and will account for my calculations being 1.098127765025 meters out.