Radial or Ring

[JohnD]

never thought about it.

I just like rounded things.

(o)(o)

 

[davy_owen_88]

Like grommets?

 

[jj4091]

How can you balance the current in a ring main when you don't know what appliances are going to be connected to it, thats the whole idea of it surely.

True, but you need to make the balancing of load on each leg possible by ensuring that as close to the same number of sockets are on each leg rather than having all the sockets on one leg and then a long return leg from the furthest socket.
It is hard to gain an exact idea of loading but you have to make some assumptions.

Long live 4mm² radials!

Sorry ricicle, but there is no "return leg" on a ring main, you have to calculate the cable size required for the length of the run, having more sockets in-line does not increase the load carrying capabilty of a cable in any particular direction.If your ring main were 50metres long with sockets every 2 metres it would be no different than having 1 socket at 5 metres, 1 at 7m, & the last 1 at 10m. It is hardly ever practical to equally split up sockets in lengths apart.

 

[Chri5]

Could someone confirm who is right?

I've always installed rings with a long leg between the final socket and the CU, always use the OSG for cable size when distance becomes an issue.

It's obvious that if:-

CU ----5m----socket 1----5m---- socket 2----5m----socket 3----10m----socket 4-----20m----CU

Ring is 55m long, 4mm TE

Socket 1 demand will take the route of easiest resistance, but is balanced by virtue of the ring.

Surely engineering in ring 1st / last sockets to have similar lengths back to the CU isn't required by virtue of the ring.

Am I having a blonde moment

 

[Adam_151]

Surely engineering in ring 1st / last sockets to have similar lengths back to the CU isn't required by virtue of the ring.

Its not quite as simple as that, its not just about similar end leg lengths, its about complete balence of the rfc (which a radial with an extra return cable added is unlikely to have)


If you have a socket 1/10th of the way along the rfc, then the current will be divided up in a similar way (but 'flipped' around way), with 9/10ths taking the short leg and only 1/10th taking the long leg, you are always going to get this, but if the sockets are spaced as evenly as possible then it'll balance out in the end rougly enough, but if you have them all clustered towards the end of the rfc, then it can almost be as bad as a broken ring (using the idea of 1/10th again, three 10A loads, and we have maxed out one leg to 27A and only have 3A on the other), such a thing can happen if you put the kitchen on a ring shared with other rooms, and the kitchen is close to one of the 'ends' of the ring.

Have a read of this: http://www.iee.org/Publish/WireRegs...s_7671_amendment_1_2002_and_ring_circuits.pdf

The electrical theory of why the currents divide into a radio based on the reciprocal of leg lengths is simply that the resistance of each leg will be in proportion to its length, think about the fact that the voltage at the outlet will be a tad smaller than that at the DB by virtue of volt drop, stick ohms law on each leg and you get I=V/R

I hope I've answered the right question here...

 

[Chri5]

Thanks for the detail

 

[Steve]

I just like rounded things.

(o)(o)

Mucky get.

 

[JohnD]

sorry, typographical error. Should have said:

I just like rounded thighs.

That better?

 

[plugwash]

Socket 1 demand will take the route of easiest resistance, but is balanced by virtue of the ring.

No it is not, the proportions of current flowing down each leg are proportional to the position on the ring.

rings must be designed so that in the expected use of the installation they are likely to be roughly balanced (the sizing of the cable in a ring allows for a moderate imbalance, thats why we use cables rated at arround 20A rather than ones rated arround 15A)

also moderate overloads only make cable age faster they don't generally cause catastrphic failure so short term imbalances (say because of a big party causing lots of high power gear to be plugged in close together in the sockets close to the garden) aren't a huge deal.

getting back to the problem at hand if the two legs (legs here being the bits from the active part of the ring back to the supply) are of significantly different lengths we can fix the imbalance by making the longer leg proportionally thicker. So for example if the return leg is roughly half as long again as the outward leg we can wire the return leg in 4mm and hence end up with the two legs balanced.

 

[jj4091]

Surely engineering in ring 1st / last sockets to have similar lengths back to the CU isn't required by virtue of the ring.

Its not quite as simple as that, its not just about similar end leg lengths, its about complete balence of the rfc (which a radial with an extra return cable added is unlikely to have)


If you have a socket 1/10th of the way along the rfc, then the current will be divided up in a similar way (but 'flipped' around way), with 9/10ths taking the short leg and only 1/10th taking the long leg, you are always going to get this, but if the sockets are spaced as evenly as possible then it'll balance out in the end rougly enough, but if you have them all clustered towards the end of the rfc, then it can almost be as bad as a broken ring (using the idea of 1/10th again, three 10A loads, and we have maxed out one leg to 27A and only have 3A on the other), such a thing can happen if you put the kitchen on a ring shared with other rooms, and the kitchen is close to one of the 'ends' of the ring.

Have a read of this: http://www.iee.org/Publish/WireRegs...s_7671_amendment_1_2002_and_ring_circuits.pdf

The electrical theory of why the currents divide into a radio based on the reciprocal of leg lengths is simply that the resistance of each leg will be in proportion to its length, think about the fact that the voltage at the outlet will be a tad smaller than that at the DB by virtue of volt drop, stick ohms law on each leg and you get I=V/R

I hope I've answered the right question here...

When I asked davy owen in my original post if this was a new theory, it was a serious question, not being flippant. This is a different electrical theory than I was taught almost 50 years ago, which was that current drawn on each leg of a ring main was equal. In other words if you connected an ammeter in each leg of a ring main at the consumer unit, you would measure the the same current being drawn on both meters. I can accept what you are saying may be true for a fraction of a milli-second at switch on, but after that then sorry I find it difficult to understand your theory. Which is likely to be the the leg with the higher resistance a 20m length of continuous cable, or a 10m length with several connections at sockets(in practice)?

 

[C&GStudent]

This is interesting.

Isn't there someone here with an electrical engineering degree to clarify what goes on in a ring final circuit?

 

[JohnD]

I'd be very much happier to see an experimental measurement with a milliamp clamp meter.

 

[C&GStudent]

I'd be very much happier to see an experimental measurement with a milliamp clamp meter.

This one is interesting!

 

[plugwash]

When I asked davy owen in my original post if this was a new theory, it was a serious question, not being flippant. This is a different electrical theory than I was taught almost 50 years ago, which was that current drawn on each leg of a ring main was equal.

then you were taught by someone who didn't understand basic electrical principles. Ohms law and Kirchhoff's circuit laws have been known for over 150 years.

Which is likely to be the the leg with the higher resistance a 20m length of continuous cable, or a 10m length with several connections at sockets(in practice)?

the cable by far! a contact with a resistance as high as a meter of cable would be smoking when run at full load. The fact that there aren't tables for connection stuff in the volt-drop section of the regs clearly shows that properly made connections have negligable resistance.

Isn't there someone here with an electrical engineering degree to clarify what goes on in a ring final circuit?

Well i'm actually studying electronic systems engineering but we have a common first year with the electrical engineering people and this a simple application of first year material.

for our purposes we can approximate a wire by a resistor whose resistance is lp/A where l is the length, A is the CSA and p is the resistivity of the metal used. If we assume p and A are constant then resistance is proportional to length.

lets initially assume that we place a single appliance on the ring. clearly the live terminal of its socket can only be at one potential so we have a voltage drop between the live at the origin and the live at the socket. Since V=IR and the volt drop must be the same on both paths we can say that IR must be the same for both paths. In other words assuming that all the cable is the same the current share taken by a path is inversely proportional to the length of that path.

exactly the same reasoning applies in the neutral.

Now we get on to multiple appliances on a ring. First we make the assumption that the voltage drop is small enough that the current drawn by an appliance is not significantly affected by cuircuit conditions. That allows us to model the appliance as a current source and that allows us to apply the principle of superposition and consider the affects of each appliance seperately.

so for example lets take a 50M ring made of uniform cable with a 10A load at 5M, a 12A load at 10M and a 10A load at 20M where 0M is the end of leg A and 50M is the end of leg B.

the load at 5M has a 5M path to leg A and a 45M path to leg B. So leg A will get 9 times as much current from it as leg B does. 9A to A and 1A to B

the load at 10M has a 10M path to leg A and a 40M path to leg B so leg A will get 4 times as much current from it as leg B does. 9.6 amps to leg A 2.4 amps to leg B.

the load at 20M has a 20M path to leg A and a 30M path to leg B so leg A will get one and a half times as much current from it as leg B. so leg A will get 6A from it and leg B will get 4A from it.

totaling those values up gives us

leg a: 9+9.6+6=24.6A
leg b: 1+2.4+4=7.4A

 

[jj4091]

Pugwash, many thanks for that reply, as you say it is totally at odds with what I was taught, but I am amazed that this is the first time in all these years I have had to question what I learnt then. I have to be honest that I cannot remember Kirchoffs or whether I was ever told about it,but I presume this is the lp/A part of your answer. It would seem from other responses to this thread that I am not alone in being confused about this theory so at least my ignorance may have helped others aswell.