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Design Current and Maximum length and Radials.

Discussion in 'Electrics UK' started by studentspark, 16 Dec 2020.

  1. studentspark

    studentspark

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    Just trying to calculate maximum lengths of cable for standard circuits (Just CPD)

    OSG tells us...
    A1 Ring - 100m2 A3 Radial 50m2

    Now I don't find m2 a useful number, as it doesn't tell us the conductor length. Its not necessarily the run of the cable.

    So I thought I would use volt drop to find the maximum length so

    VD = Mv/A/m /1000

    So need the design current. Not sure what this is on circuits which have fluctuating loads

    Maximum loads assumed per circuit (OSG)

    Lighting - 6A Load 5A

    Ring final - 32A - Load 26A

    Radial - 20A - Load - 16A


    So Ib could be this assumed maximum load? and not the rating of the protective device?

    I get that, with the RF circuit Assuming 2.5mm flat cable has a CCC of 27A protected by a 32A MCB.

    But Radial? why not 20 A


    If I calculate at 20A for a 2.5mm radial protected by a 20A MCB

    and transpose the formula.

    So maximum volt drop 11.5v


    11.5 / 18x20 x 1000 =31.9 m

    and 16A

    11.5 / 18 x 16 x 1000 = 39.9m

    So which value do we use for radial and ring final Ib ?

    and the maximum length of a radial circuit is at most 39.9 meters or 31.9 m

    Which is not that long is it really if you consider the circuit starting in CU down stairs, and then upstairs and around a couple of rooms. 31 m can soon be used up?

    So as we move away from rings. If you consider a 'standard' three bed house, and wanted to wire in radials. I understand this is difficult to say, just very basic minimum design basics. (And before correction factor are used)

    I would think you would need a radial for downstairs sockets (Living room - dining room) (Mainly low demand TV, lamps, media...) two radials for upstairs longer length (More use of power (Hair dryers, tongs etc...) (Kitchen sockets on 4mm) Appliances on own radials.

    Would you consider one 20A radial adequate for the two downstairs rooms. I can't really see what large loads would be plugged in, unless the central heating failed and you needed to use electrics heating for a short while.

    Appreciate your thoughts

    Thanks
     
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  3. ericmark

    ericmark

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    The maximum length of 2.5 mm² in a ring is considered to be 106 meters. However when I tried to calculate that I at first could not get the figures, it was 80 something meters when volt drop was 4% and at the time of the change I was an IET member so I asked at the meetings.

    So it is down to how they arrive at "Design current for circuit Ib" they consider it is 20 amp at centre, and 12 amp even spread, so Ib = 26 amp not 32 amp. You also need the Correction factor Ct, which will be around 0.9177626606198034 think you can guess I wrote a program to work it out, done in Java Script.

    So with a radial also 2.5 mm² and 20 amp or less at furthest point, if we take 16 amp for example we end up at 42.3 meters because 16 is higher than ½ 26 =(13) so using radials you need 3 radials to replace one ring.

    You talk about a 20 amp radial that has maximum length of 32.4 meters, this starts to get rather restrictive. Move to 4 mm² and 20 amp is 55.6 meters which is more reasonable, but still not great, 6 mm² with 32 amp using 26 amp as Ib and 64 meters.

    You can't get over 6 mm² into a 13 amp socket, and getting 3 wires into the terminal hole with 2 x 6 mm² and 1 x 2.5 mm² is hard, so back to basic the 13 amp socket was designed to use 7/0.029 wire with a 30 amp fuse, in a ring formation and it worked well, that was 2.9 mm² the reduction to 2.5 mm² and the reduced heat carrying capacity of the 13 amp plug pins due to necking to fit insulator on live pins, has degraded the British ring final design, however it is still better than the other options.
     
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  4. AndyPRK

    AndyPRK

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    When the answer is to replace 1 ring with 2 radials, there is no advantage, except less testing!
    And if the answer is to use 3 radials, that is now using 3 mcb's or rcbo's which is getting ridiculous
     
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  5. JohnW2

    JohnW2

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    As you imply, "... 20 amp at centre, and 12 amp evenly spread ..." is essentially arbitrary, and probably a bit over-cautious, give that it is pretty unlikley that there would be 20A of load at the centre.

    However, as I've often said, given that someone came up with that formula (which, as you know, does not appear in the regulations) for ring finals, I don't see why they didn't do/think the same for radials - in which case a 20A radial would (pro-rata) become 12.5A at the end of the radial and the remaining 7.5A evenly spread along it's length (i.e. 'effectively 16.25A' for VD calcs) - and for a 16A radial, 10A at the end and 6A evenly spread (i.e. 'effectively 13A'). For a 32A radial, it would, of course, be the same as you quote for a 32A ring final (20A at the end plus 12A evenly distributed).

    Having said all that, as often discussed, what BS7671 says about voltage drop is only guidance - so, unless you can think of any loads whose 'safe functioning' will be impaired by a low voltage supply (the only think that interests the regs), one does not have to stick religiously to those guidance's - so, at the very last, I think it would be perfectly reasonable to make the same assumptions (as above) for distribution of loads on radials as on ring finals..

    Kind Regards, John
     
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  6. EFLImpudence

    EFLImpudence

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    Yes but you can have several 31m. routes because of the branches - in effect a tree.

    How long would a ring be that supplied a similar arrangement?
     
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  7. JohnW2

    JohnW2

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    That's true, but one can also have multiple ring finals - even (although I've never see it done) two or more fed from the same OPD.

    Kind Regards, John
     
  8. EFLImpudence

    EFLImpudence

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    I don't see how that alters anything.
     
  9. ericmark

    ericmark

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    @EFLImpudence is correct, to take a ring final around the house and back it goes to every socket, but with a radial out like a tree and also you can take to first socket as 6 mm² so reducing volt drop, but it needs to be designed as a radial not a ring split into two radials.

    I remember working in some offices, and there was a 32 amp supply to a 4 mm² radial, which some one had split half way and added some extra sockets using 2.5 mm² so due to that whole radial had to be reduced to 20 amp, could not find 25 amp MCB's for that fuse box, and clearly the circuit was supplying maximum load as after it started to trip. (Fan heaters)

    Since offices and a resident electrician can't blame it on DIY, it would have been done by an electrician, so it seems even electricians jump to conclusions and assume ring finals without testing. Be it broken ring, spur on spur, or other installation faults, it is down to not testing, not something wrong with the ring final system. I think there is a place for both.
     
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  11. JohnW2

    JohnW2

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    I wasn't suggesting that it did. I was merely pointing out that if one is concerned about VD (which I generally am not), there are many ways of reducing/minimising it, whether with radials or ring finals.

    Kind Regards, John
     
  12. flameport

    flameport

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    Only if all of the load is at the end of the circuit, for something like a cooker or electric shower.

    For other circuits such as socket outlets and lighting, the load is usually distributed along the cable, and the total length for volt drop can be significantly longer.

    Maximum load assumptions in the OSG and elsewhere may be far away from reality - a typical domestic lighting circuit with LED lighting will never get anywhere near 5A.
    Loading on socket outlets depends on where those sockets are - those for bedrooms will be entirely different for those in a kitchen.
     
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  13. JohnW2

    JohnW2

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    As I've said, it really depends upon the situation.

    For example, at least in my opinion, there are some situations in which a 'lollipop' circuit probably makes sense - which is essentially the ring-final-equivalent of your idea of taking a 6 mm² feed to the first socket of a radial. I have that in my cellar/workshop (which has many woodworking machines etc.) - since I made use of an unwanted pre-existing 10 mm² cooker circuit to feed a 2.5 mm² ring final (which is pretty distant from the CU).

    Kind Regards, John
     
  14. studentspark

    studentspark

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    Thanks for all the replies.

    Don't really understand this.

    So for a 20 A radial and the load on each socket is in the brackets. and the -----20---- is the load on that section of cable

    20A MCB ------20---------[ 1A ]-----19-----[ 3A ]------16------[ 5A ]------11-----[ 0A ]-----11-----[ 1A ]-----10-----[ 10A ] Total load 20A


    Is it evenly spread, I would have thought the load on the cable would gradually decrease along its length.

    How do we know how the load is going to be distributed. One day you might have 20A at the furthest point, the next 20A at the first socket


    ----------------------31m---------------------------[ 13 a ]
    [
    20A
    [
    -----------------------31m-------------------------[ 13 A ]

    You can't have this situation ?
     
  15. JohnW2

    JohnW2

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    Indeed it will - so the voltage drop (per metre of cable) will gradually decrease as you move along the length of the circuit. At any socket, the total VD will be the sum of the VDs in all the sections of cable (from socket to socket) upstream of the socket in question. What don't you understand?
    With a sockets circuit, one can never know for sure what is going to happen - so one just has to make assumptions (basically 'intelligent guesses') about what is likely to happen. Were that not the case, one could never have a 32A sockets circuit with more than one double socket, or a 20A sockets circuit with more than just one single socket.

    Kind Regards, John
     
  16. ericmark

    ericmark

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    I am not saying I agree with the idea of working out Ib, however we have considered a twin socket is unlikely to get a 26 amp load, and we normally consider 20 amp more probable, so it is reasonable to consider the end socket of a radial or mid socket of a ring final will have a 20 amp load unless some thing stop this like a 16 amp over load device.

    However of course there many be variations, in a collage it would seem reasonable to expect the bench sockets all to have about the same load, so one could say Ib was 16 amp for a 32 amp ring final.

    I know my son and I have argued over splitting sockets up/down or side/side. At any one time it is unlikely to have a heavy load both up and down stairs at the same time so splitting side to side will likely mean more even load to the two ring finals, more even cable use, so neither circuit will have a high loop impedance, and should one circuit fail, the home can continue to function without any extension leads running up/down stairs, but my son says should split up/down as that is the way we expect it to be split, and he does have a point.

    However the result can mean the load is concentrated to one end, specially if no dedicated kitchen ring final. We should consider each home, and not do it one way because we have always done it that way. We sign the installation certificate to say we have designed the home, so we should spend a little time considering best design.

    Since the 32 amp ring final was designed with the idea of electric heating, I suppose we should consider likelihood of an electric heater in every room, it is the only way that if the advice on items over 2 kW which are fixed having a dedicated supply that a ring final will ever be loaded to maximum, so we should consider heating in the day on lower floor and night upper floor, so again side to side split makes sense.
     
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  17. JohnW2

    JohnW2

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    Yes, but you're talking literally about the socket at the end of a radial or the socket exactly at the mid-point of a ring.

    There's nothing to say that there can't be one, or even two (or more!), further double sockets within a few inches of that end- or mid-socket - so, in terms of 'what could be plugged in', even if we assume only 20A per double socket, there could be 40A, or even 60A, of load plugged into the circuit as close as makes no difference to the end of a radial or the middle of a ring circuit.

    As I said, with a sockets circuit, a designer has no option but to apply common sense and 'intelligent guesswork' to decide what is likely to be plugged in (and where) - but there will always be a risk that some user could subsequently prove them wrong!

    Kind Regards, John
     
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