Planning Kitchen Wall Electrics - Is This Good? [PIC]

[ban-all-sheds]

It's a radial, there are no unfused spurs, they're all fused by the origin.

So what are we to make of all the rules/guidelines here?

 

[John D v2.0]

Yes, we agree.

I've done another diagram.
This may not be practical but it is just an illustration.

Left is a bungalow with the ring in loft and drops to sockets.

Right is how it could be wired with 4mm² drops.
This could actually make it better for the ring when there are high load on outer sockets:

View attachment 121070

Obviously - connecting the 4mm² 'L's the other way round would not be a good idea - you have to think about it.

We agree! Stop the thread now! Oh wait, JohnW2 has turned up, I'm about to learn even more

Thanks for the diagram that's really useful. I totally get the thing about the Ls the other way around.

 

[JohnW2]

17th Edition 2001 AMD 1: OSG Appendix H: The total number of fused spurs is unlimited, but the number of non-fused spurs should not exceed the total number of socket outlets and items of stationary equipment connected directly in the circuit.

Unless it were citing an actual regulation in BS7671, the OSG cannot mandate anything. At best, it is a suggestion/guidance.

... and nor do I see that such 'guidance' really makes much electrical sense, since, as far as the ring is concerned, there is no material difference between an unfused spur and a "socket outlet or directly connected load" - the only difference being that the spur connects the socket (or whatever) to the ring via a single cable (at least adequate for the permitted load).

I suppose it's possible that they are assuming (not necessarily correctly) that all spurs originate from sockets or FCUs, and that the guidance therefore relates to their feeling that it undesirable to originate more than one spur from the same point on the ring (although, as often discussed, that makes little sense, either, so long as terminals can accept the necessary number of conductors).

Kind Regards,
John

 

[John D v2.0]

Hi folks. It has been suggested that I might have something to offer this discussion, so I have come out of hibernation (severe busy-ness) to see if that's true. I haven't yet read the whole thread, but, in terms of this recent posting ...
I think Mr Kirchoff would agree with you. If one connects a 'bridge' between two parts of a ring (or, indeed, multiple bridges between parts of a ring), then I think it is inevitable that the current will reduce in the leg of the ring which was previously ('pre-bridge') carrying the higher current, and that the the current will increase in the leg of the ring which was previously ('pre-bridge') carrying the lower current - i.e. 'balance' of the ring would be improved.

We've discussed this matter many times before, and I think the consensus has generally been that, in functional terms, turning a ring into a figure-of-eight is advantageous, the main downside being the nightmare that may result in terms of testing and fault-finding, particularly for someone unaware of the presence of the bridge(s).

Kind Regards, John

Do you define a bridge as being symmetrical? If so you're right
I don't want to drag up old debates though... but consider this textbook ring (worse cases are available!):

ring final, CU at 12 o clock
single load of a) 20a at 6 o clock
2no load of 6a at b) 3 o clock and c) 9 o clock
current in each leg at the CU is 16a

the ring happens to be physically like a lollipop shape, so the cables between logical 3 o clock and 6 o clock positions are physically run together.
I bridge between the 3 o clock and 6 o clock positions

consider the current in each leg now?
a) 20a will go 2/3 anticlockwise (13.3a), 1/3 clockwise (6.6a)
b) 6a will go 2/3 anticlockwise (4a) , 1/3 clockwise (2a) (as it's connected to a by negligible impedance)
c) will go 1/3 anticlockwise (2a), 2/3 clockwise (4a)
anticlockwise total at the CU: 19.3a
clockwise total at the CU: 12.6a

So the textbook ring been reduced in balance by the bridge, contrary to the many previous discussions - so where has the discrepancy come in?

 

[John D v2.0]

So what are we to make of all the rules/guidelines here?

Looks like that applies on radials where some of the cable has a reduced CCC insufficient for overload. Which cables in your drawing were smaller?

 

[JohnW2]

Do you define a bridge as being symmetrical? If so you're right

I don't know whether I'd call it 'defining', but I was thinking of a bridge as having its ends connected on opposite sides of the midpoint of the ring. Your example below is not so much a 'bridge' in that sense but, rather, effectively reducing the CSA of one side of the ring.

ring final, CU at 12 o clock ... single load of a) 20a at 6 o clock ... 2no load of 6a at b) 3 o clock and c) 9 o clock ... current in each leg at the CU is 16a

Agreed.

the ring happens to be physically like a lollipop shape, so the cables between logical 3 o clock and 6 o clock positions are physically run together.

I don't really understand that - what cableS between 3 o'clock and 6 o'clock - there is surely only one cable between those two positions?

I bridge between the 3 o clock and 6 o clock positions

I'll ignore the previous comment and will assume we're just bridging between 3 o'clock and 6 o'clock ....

consider the current in each leg now? ... a) 20a will go 2/3 anticlockwise (13.3a), 1/3 clockwise (6.6a) ... b) 6a will go 2/3 anticlockwise (4a) , 1/3 clockwise (2a) (as it's connected to a by negligible impedance)
c) will go 1/3 anticlockwise (2a), 2/3 clockwise (4a) ... anticlockwise total at the CU: 19.3a ... clockwise total at the CU: 12.6a

I disagree with your figures, but not with the conceptual conclusion you are drawing. However, as above, I think this has happened because you are 'bridging' a bit of cable which is all on one side of the ring.

As for your figures (just for completeness) .... If we assume that the ring is symmetrical in terms of cable lengths (ie. the impedances of each of the four logical quarters of your ring are the same) and that the bridging cable is of the same CSA as the ring cable (and of the same length as the bit of ring cable which it 'bridges', then, by my reckoning ....
(a) the ratio of impedances of the two paths is 2 : 1.5, which leads to figures of approx 8.58A clockwise and 11.42A anticlockwise
(b) the ratio of impedances of the two paths is 2.5 : 1, which leads to figures of approx 1.71A clockwise and 4.29A anticlockwise
(c) mirror image of (b), hence 4.29A clockwise and 1.71A anticlockwise.

That all works out as a total of approx 14.58A clockwise and 17.2A anticlockwise.

Kind Regards, John

 

[John D v2.0]

Thanks John, actually we're on the same page but the extra cable in my model has negligible impedence.
This is because they run physically close to each other due to the physical lollipop shape, even though electrically it is a one dimensional closed shape (so could be physically a square, circle, triangle, keyhole, lollipop)
The hours of the clock I mentioned are purely electrical, so each "hour" has 1/12 of the total impedence regardless of where it happens to be physically.
The only reason I went for that lollipop example is because it matches the original situation "part of the ring has two cables to the same place" so they can become one cable of greater cross section (ie add a connection/bridge of negligible impedence at that point)

 

[ban-all-sheds]

Looks like that applies on radials where some of the cable has a reduced CCC insufficient for overload. Which cables in your drawing were smaller?

As many as the ones identified in the rules/guidelines as being smaller.

 

[ban-all-sheds]

Do you define a bridge as being symmetrical? If so you're right
I don't want to drag up old debates though... but consider this textbook ring (worse cases are available!):

ring final, CU at 12 o clock
single load of a) 20a at 6 o clock
2no load of 6a at b) 3 o clock and c) 9 o clock
current in each leg at the CU is 16a

the ring happens to be physically like a lollipop shape, so the cables between logical 3 o clock and 6 o clock positions are physically run together.
I bridge between the 3 o clock and 6 o clock positions

consider the current in each leg now?
a) 20a will go 2/3 anticlockwise (13.3a), 1/3 clockwise (6.6a)
b) 6a will go 2/3 anticlockwise (4a) , 1/3 clockwise (2a) (as it's connected to a by negligible impedance)
c) will go 1/3 anticlockwise (2a), 2/3 clockwise (4a)
anticlockwise total at the CU: 19.3a
clockwise total at the CU: 12.6a

So the textbook ring been reduced in balance by the bridge, contrary to the many previous discussions - so where has the discrepancy come in?

 

[John D v2.0]

As many as the ones identified in the rules/guidelines as being smaller.

If you have a point it would be quicker if you just explained rather then making pointed comments.

In your opinion maybe, but feel free to draw a picture if you think it would help.

 

[JohnW2]

That is often the case, if one can't understand what is being said without pictures. However, in this case, I would have thought that any even remotely intelligent person could understand what is meant by a ring with loads at "3, 6 and 9 o'clock" relative to a CU.

Kind Regards, John

 

[aptsys]

I would have thought that any even remotely intelligent person could understand what is meant by a ring with loads at "3, 6 and 9 o'clock" relative to a CU.

Kind Regards, John

Yes, those with plug in timers?

 

[deadshort]

A cooker point, two single sockets and a double. We are on page 7. FFS get a grip.

DS

 

[JohnW2]

Thanks John, actually we're on the same page but the extra cable in my model has negligible impedence. This is because they run physically close to each other due to the physical lollipop shape ...

I still don't really understand, so maybe (despite what I just wrote!) BAS was right in saying that we need pictures. What are these things which "run physically close to each other". If you're talking about cables and one has negligible impedance, it must also have negligible length - so how could that "run close to" a cable of appreciable non-negligible length.

Are you perhaps suggesting that your loads (b) and (c) are physically very close, even though logically "90 degrees" / "3 hours" apart on the ring? If you are saying that (b) and (c) are joined by a cable of negligible length and impedance, you are effectively creating a situation in which those two loads are connected to the same point in the ring, with the original cable (of non-negligible length/impedance) between those two loads having become irrelevant (i.e. carrying essentially no current).

Kind Regards, John

 

[John D v2.0]

I still don't really understand, so maybe (despite what I just wrote!) BAS was right in saying that we need pictures. What are these things which "run physically close to each other". If you're talking about cables and one has negligible impedance, it must also have negligible length - so how could that "run close to" a cable of appreciable non-negligible length.

Are you perhaps suggesting that your loads (b) and (c) are physically very close, even though logically "90 degrees" / "3 hours" apart on the ring? If you are saying that (b) and (c) are joined by a cable of negligible length and impedance, you are effectively creating a situation in which those two loads are connected to the same point in the ring, with the original cable (of non-negligible length/impedance) between those two loads having become irrelevant (i.e. carrying essentially no current).

Kind Regards, John

Yes you're spot on, and the original cable is indeed no longer carrying any ring current, but it would still be carrying current relating to any accessories on that part of the ring, if I had included any in my example.
This is exactly the opposite process to "opening out the ring" to flow through a new set of accessories, as contrasting with just creating a 4mm spur at that point.
Although again, in my example, I didn't draw any current from those new accessories.