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150A Ring Final Circuit

Neil Annetts

I am after some help in calculating the curent carrying capacity of a ring final circuit wired in 16mm Tri rated singles!

From the book 16mm tri rated will take 100A. circuit no more than 10 meters long, all in steel trunking / DB enclosures etc.

Supplied from a Schnieder NSX160F with a 100 - 160A TMD unit.

I know there is a way of calculating this but for the life of me i cannot remember how to do it and no amount of Googling is being productive!

Some background on the reason - 8-row schneider Prisma panel with 6x 80A Multiclip bus bars. cannot wire each back seperatley without having 6 cables per phase at the MCCB so the thought was to wire a ring around the lot.

Any help would be greatly appreciated

 

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    Assuming the cable is the same cross-section all the way round, the thing to realize is that the current from the origin to any point on the ring divides round the two paths in the ratio of those path resistances - which are in effect path lengths.

    So from half way round the path lengths are equal and the current splits perfectly 50/50 . In this case with 100A cable you could have a 200A total  load, so long as the load was positioned symmetrically.

    Now if you had your load ¾ of the way round then the path lengths and current division would be in the ratio of 1:3 , with clearly the larger share going down the shorter path. Now if there were no other loads you could have a total load of 133A, with 33A going the long way and 100A the short way.

    The 160% thing works for13A sockets, becasue they are spread around the ring, and are not all fully loaded. This probably deserves a better analysis

    In general you want to avoid the situation where most of the load is near one end, adding dead length to the shorter path is actually worthwhile, not in terms of VD, but maximum current.

    If you can sketch the layout with the real distances, then we can chat you through a proper analysis.

    Mike.

     

     

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  • 0

    Assuming the cable is the same cross-section all the way round, the thing to realize is that the current from the origin to any point on the ring divides round the two paths in the ratio of those path resistances - which are in effect path lengths.

    So from half way round the path lengths are equal and the current splits perfectly 50/50 . In this case with 100A cable you could have a 200A total  load, so long as the load was positioned symmetrically.

    Now if you had your load ¾ of the way round then the path lengths and current division would be in the ratio of 1:3 , with clearly the larger share going down the shorter path. Now if there were no other loads you could have a total load of 133A, with 33A going the long way and 100A the short way.

    The 160% thing works for13A sockets, becasue they are spread around the ring, and are not all fully loaded. This probably deserves a better analysis

    In general you want to avoid the situation where most of the load is near one end, adding dead length to the shorter path is actually worthwhile, not in terms of VD, but maximum current.

    If you can sketch the layout with the real distances, then we can chat you through a proper analysis.

    Mike.

     

     

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