Confused over electrical cable length and size

[JohnW2]

1 ... 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

Although that very simple approach gives a pretty similar answer to the one I got (104.3 metres) using a more exhaustive 'brute force from first intervals' approach (see post #24 above), I haven't yet managed to work out the basis of the "24A at the midpoint" assumption above.

Is there some flaw in the more extensive calculation I presented in post #24 (the assumption there being that the situation is the equivalent of 6A load at the centre of each leg of the ring)?

As can be see, the only bottom-line arithmetical difference between my approach and the one above, is that mine has a figure of 24.5 (arrived at by the reasoning described in my previous post), rather than 24.

Kind Regards, John

 

[JohnW2]

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.

If I used your corrected figure of 16.52 mV/A/m, my calculation would give an answer of 113.65 metres.

Kind Regards, John

 

[ban-all-sheds]

Is there some flaw in the more extensive calculation I presented in post #24 (the assumption there being that the situation is the equivalent of 6A load at the centre of each leg of the ring)?

Only that it is not needed to explain where the figure of 106m came from.

 

[securespark]

A radial circuit is not, not, NOT a single cable running from the CU through every socket.

It could be.

 

[ban-all-sheds]

Indeed it could, but an assumption that it naturally has to be can lead to false conclusions about how much cable it needs.

 

[JohnW2]

Only that it is not needed to explain where the figure of 106m came from.

Do you mean that it is your "24", rather than my "24.5" which is 'flawed', and that 'they' perpetrated the same 'flaw' as you? As I said, I would be interested to understand how you arrived at the 24A figure.

 

[ban-all-sheds]

By working out what current, drawn at the mid-point of the ring, would result in a ring length limit of 106m.

Basically the KISS principle - why look for a complex method of arbitrary assumptions if there's a simple one?

But it turns out that that was a flawed approach:

 

[JohnW2]

... but that uses an ultra-simplistic figure of 26A, rather than your 24 or my 24.5, and, with the 18mV/A/m, that results in a maximum length of about 83m and that then turns into 106m if one takes temperature correction into account.

However, that approach is 'ultra-simplistic', and essentially ignores the fact that we are talking about a ring. In other words, it igores the fact (which my more rigorous calculation takes into account) that a quarter of the 6A of load (i.e. 1.5A) will take the 'long route' around the ring, leaving only 4.5A of the distributed load in each leg to return to the CU (and with the two 'long route 1.5As cancelling, since they travel in opposite directions).

So, as I've said, unless I've made a mistake I think that the 'correct' answer is ~104.3m (not 98.3m) without temperature correction, and ~113.1m (not 106.5m) if one applies a Ct of 0.923 (i.e. effectively changing 18mV/A/m to 16.6V/A/m).

I see no real reason for using a calculation based on an assumption which produces 'incorrect' answers when it is just as easy to undertake the 'correct' calculation - do you? Either way, one ends up with:

L = (11.5 x 4) / (k x 0.018) ... (without temp correction, and with the load distribution we've been discussing))

... so the only difference is the value of k - so why use the 'wrong'' one, since it's no 'easier'?

 

[EFLImpudence]

Wow - alot to take in.

Just to say firstly - the 18mV figure, as do all the VD values, does include the temperature correction.

 

[JohnW2]

Wow - alot to take in.

As I said at the start the (proper') calculation is far from totally straightforward. As I said, everyone seems to be forgetting that it's a ring. I may be wrong, but I think what that book, BAS and probably also you have effectively done is a calculation for just one half of the circuit - i.e. as if it were a 16A radial with a 10A load at the end and an additional 6A distreibuted along its length. As I said, that seems to ignore the fact that when it IS a ring, a quarter of the distributed load in the leg (i.e. 1.5A of 6A) will take 'the long route' back to the CU.

Just to say firstly - the 18mV figure, as do all the VD values, does include the temperature correction.

That's doesn't seem to be what the book (a copy of which I have somewhere!) posted by BAS thinks. It uses 18mV/A/m when it says Ct=1, but then goes on to apply a 'correction' of Ct=0.923 for the situation in which It is 20° and Ip is 70°.

I may, of course, be wrong - on any of these counts!

Kind Regards, John

 

[JohnW2]

The thing that always occurs to me when this issue gets discussed is that, even in my very large house, none of the ring finals are anything approaching 106m in length. How often does this actually happen in practice, in domestic properties (even when all the sockets are on 'drops')?

I thought that a 'rule of thumb' was that a rewire of an 'average' house (probably including 2 or 3 ring finals, and maybe other things using 2.5mm² cable) would take something like 100m of 2.5mm² T&E. Is that not roughly correct?

Kind Regards, John

 

[EFLImpudence]

All I did was use (11.5V, 26A & 18mV)x4 = 98.3m

10mm² copper has resistance of 1.83mΩ/m @ 20°C (my starting point for all such calculations)

So, for 2.5mm² it is 1.83 x 4 x 2 x 1.2 (Ct) = 17.568mV/A/m (it comes to 100.7m if 17.568 is used so perhaps the calculations aren't that critical)

 

[JohnW2]

All I did was use (11.5V, 26A & 18mV)x4 = 98.3m

Indeed, that is eacxtly what the book does for Ct=1.

In passing, the first part of their "example" is just plain daft, since it would be impossible for the current throughout a 32A ring to be 32A!

10mm² copper has resistance of 1.83mΩ/m @ 20°C (my starting point for all such calculations) ... So, for 2.5mm² it is 1.83 x 4 x 2 x 1.2 (Ct) = 17.568mV/A/m (it comes to 100.7m if 17.568 is used so perhaps the calculations aren't that critical)

You seem to be tryiong to re-invent the figures in 4D2B from first principles - but we are expected to assume the figures in that Table. In any event, there seems no consensus about the precise temperature coefficient of resistance for copper. If I understand your "1.2" correctly, I think that it implies a coefficient of 4.0 x 10^-3 (aka 0.004) per degree C. However, the first four hits I got on Google gave figures ranging from 3.93 to 4.29 x 10^-3 per degree C. Therefore one has to "pay one's money and take one's choice".

However, I'm starting to get very confused by that "Ct" used in the book (they cite 0.923 for 20-70 °C). I'm not convinced that it has anything to do with temperature coefficient of resistance and, since they are multiplying Ib by it, I wonder if it might be the de-rating figure for CCC. By dividing by 0.923, they are increasing the maxim circuit length by a factor of ~1.083. As I said, I'm a bit confused about what they are doing - and maybe we need to see more pages of the book to see the definitions (I will try to find my copy)!

Kind Regards, John

 

[JohnW2]

Is there some flaw in the more extensive calculation I presented in post #24 .....?

No-one has jumped on me yet, (despite the passage of best part of 24 hours), but I've just spotted the flaw myself!

There was clearly a fundamental flaw - since, with my calculations, whilst there was clearly a total of 32A worth of loads, there was only 29A coming out of the CU!

Whilst I still think I was correct in saying that the two 1.5A currents ('taking the long route') would 'cancel' in the two 'middle quarters' of the ring (so each just carried a net 10A - half of the central load), what I had overlooked was the fact that the same is not true in the two 'outer' quarters (the ones connected to the CU), since each of them carries the 1.5A 'long route' current from the other side of the ring. The current in those two 'outer quarters' is therefore (as I should have realised it had to be!) 10 + 4.5 + 1.5 = 16A - i.e. half of the 'central 20A load' plus 4.5A of the 'short route' 6A load on 'this' side plus 1.5A of the long route' 6A from the other side. My final expression therefore should have been ...

L = (11.5 x 4) / (26 x 0.018) = 98.3 metres

... which, of course, is exactly what 'the book' (and others) say (but I still don't know where BAS's 24A came from), and it is the temperature correction of that which takes it up to around 106 metres ('around', since it depends upon exactly which correction factor one chooses!).

Apologies for the confusion.

Kind Regards, John

 

[EFLImpudence]

If I understand your "1.2" correctly, I think that it implies a coefficient of 4.0 x 10^-3 (aka 0.004) per degree C. However, the first four hits I got on Google gave figures ranging from 3.93 to 4.29 x 10^-3 per degree C. Therefore one has to "pay one's money and take one's choice".

Yes but they are all around 0.004. Certainly a lot higher than 0.0017 (for 0.923 Ct)

However, I'm starting to get very confused by that "Ct" used in the book (they cite 0.923 for 20-70 °C). I'm not convinced that it has anything to do with temperature coefficient of resistance and, since they are multiplying Ib by it, I wonder if it might be the de-rating figure for CCC.

How can or why would they derate the cable ccc after it has been designed with a temperature factor?


The 18mV figure is without doubt 1.2 times the accepted resistance at 20°C corrected for 70°, i.e. 0.004 x 50.
It might include approximations and rounding-up but that is it.