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Supplementary CPC/SWA Resistance Ratio

Evening all, as the title above i am just trying to get my head around how a supplementary CPC will help lower the Zs to achieve a sufficent 5s disconnection time. My thinking is that the SWA and CPC are supplementary mechanically connected at both earth bars (supply and load) so the fault current will divide based on the resistance ratio (even if it takes the longer route through the supplementary CPC if there was a fault on the amour to line conductor).

 

I have attached a sketch of this for clarity. 


I also know strapping a cpc on the side of the armour is bad practise, however im just trying to get my head around the principles.
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  • Zs = Ze + (R1 + R2)


    Therefore, if R2 is reduced, all other things being equal, Zs will be reduced.


    Note that if Ze is high, R1 and R2 don't matter a jot; and if R1 is high, it may not be possible (or desirable) to get R2 low enough to make Zs compliant.


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  • Zs = Ze + (R1 + R2)


    Therefore, if R2 is reduced, all other things being equal, Zs will be reduced.


    Note that if Ze is high, R1 and R2 don't matter a jot; and if R1 is high, it may not be possible (or desirable) to get R2 low enough to make Zs compliant.


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