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Cable calculations

MrJack96
Hi guys i wanted to calculate the maximum run when spuring of a ring final circuit is there any equations or is it just as simple as calculating VD to the furthest point in the ring and then VD to the end of the spur. The reason for the question is I want to spur of an existing socket using a 10meter run of 2.5 for an outside socket. Just want to Double check it’s sufficient 


thanks guys
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    Of course having read it I apply it meticulously.
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    The Onsite Guide puts it much more succinctly:


    7.2.2 Socket-outlet circuits

    As a rule of thumb for rings, unfused spur lengths should not exceed 1/8 the cable length from the spur to the furthest point of the ring.
  • 0
    Sparkingchip:

    The Onsite Guide puts it much more succinctly:


    7.2.2 Socket-outlet circuits

    As a rule of thumb for rings, unfused spur lengths should not exceed 1/8 the cable length from the spur to the furthest point of the ring.


    So if the spur is adjacent to the DB, that gives 80 m to the furthest point and 160 m all the way round. That's a long ring!


    In fact all may be well if the load is not great. Perhaps an FCU in the spur would help?


    ETA: IMHO spurs should be avoided. We are not aiming to make a coronavirus! ?


  • 0
    The spur could be out of the fuse or MCB.
  • 0
    The ring serves a room of approximately 20meters x 10 meters. It has only 3 x double RCD sockets. Protected by a BS88 type 2 fuse. ZS at the furthest point is 0.40ohms there is no Spurs on the circuit I was just going to come of the furthest socket at run a 2.5 to an outside RCD socket think it’s more like 8 meters from the furthest socket.
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    That fails the 1/8 rule of thumb design rule, there’s a few other potential issues as well.
  • 0
    well I'd suggest you just do it..

    But  some sums to make you feel easier. (and  no this is not a method in the OSG, but will not result in a dangerous installation)

    If we have a no load voltage of say 250 (for easy sums) we coud for example deduce a PSSC from your Zs of of appro x650A, and pretend it is the same for LN and LE. ( but you could put the Zs meter between L and N and get the exact figure)


    Now that all means is there will be 0.4 V drop per each amp of load between the load at that point and the substation, and the one we worry about, volt drop from the origin in the building will be a few volts less as some is in the street.

    So if your new spur  draws 13A at that point, you will get 0.4*13A=  5.2V volt drop for example ~ 2.2 %

    Add in 10m of 2.5mm each way, is ~  16*2*10 /2.5 = 130 extra milliohms in L and N combined  (*)

    at 13A , an extra 1.7 V drop or so.


    Total VD to new spur at 13 A load ~ 6,7V  about 3%. Even if you took the full 30A at that point its not going to lead to massive volt drop issues, but the  2.5mm cable may prefer you did not.

    Like all my worked examples random rounding errors may apply. ?

    Convince yourself it is not miles off.


    BUT it sounds like no RCD at the origin, so be aware it is not ideal, you need some sort of RCD covering the socket and in the current regs to rely on an RCD socket to do this is strictly a non-compliance.


    Mike



    EDIT

    PS after a query.

    (*) 16 milliohms - another rule of thumb I'm afraid - I do not carry a full set of voltage drop tables in my head for all cable sizes. But I can recall one number, 16 milliohms for cold copper, 1mm2 cross -section 1m long.
    (also 1/16 inch is about 1.6mm and 16 Swg sheet metal thickness too, which is sometimes useful as well but not here today)


    More like 19 milliohms when hot.


     


    you will have 10m of live core out, and 10m of neutral core back would be 2 lots of 160 millioms  if it was 1mm2, but its 2.5 times fatter than that so divide through.


     


    you could try with 19milliohms  if you think the copper will run hot.


    it does not alter the main conclusion.


    This rule is not quite right, but is certainly close enough to allows me to walk around looking up at things and decide if it passes on resistance for volt drop and tripping times by miles or if the proper sums need doing, while younger folk are booting their fondle slabs and older ones are running downstairs and out to the van for a dog eared set of tables.


    Don't under estimate the time saving powers of being able to walk round and go 'marginal - needs checking', 'fine', 'fine',  'oh dear that's miles out..' at a brisk trot  without waiting for any obvious assistance.



    regards,


    M
  • 0
    Sparkingchip:

    That fails the 1/8 rule of thumb design rule, there’s a few other potential issues as well.


    If you look in the Commentary, you'll see where the 1/8 rule came from ... it's to do with loop impedance not "mis-loading" or anything else.


    You can see that from my previous post.


    Whether the socket-outlet is added on the ring itself or a spur won't affect the loading of the ring at all - but too many outlets or connection points or spurs at one point of course can do so, unless you know the loads will be small (e.g. concentration of outlets near a TV).


  • 0
    Twenty years ago I sat sat in a classroom doing a loop impedance calculation for a lighting column in a car park supplied from source by three circuits each decreasing in size.


    I was getting quite involved working it out for each section of cable and the lecturer told me to just work out if the smallest cable would be okay for the total length, which it was.
  • 0
    A few days from now MrJack96‍ will get an email from the IET asking if his question has been answered, I wonder what he’s going to say?

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