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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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    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
Reply
  • 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
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