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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.
Parents
  • The idea is indeed to provide a parallel path and lower the resistance that way, of course if the resistance in the live path is also high, then there is the additional  advantage  of lowering the voltage at the point of fault to less than half of the line voltage. (if the R1 and R2 are equal then the voltage at fault will be exactly half, but generally it isn't, and if you can get it down to less than 50v then there is no need to disconnect at all.)
Reply
  • The idea is indeed to provide a parallel path and lower the resistance that way, of course if the resistance in the live path is also high, then there is the additional  advantage  of lowering the voltage at the point of fault to less than half of the line voltage. (if the R1 and R2 are equal then the voltage at fault will be exactly half, but generally it isn't, and if you can get it down to less than 50v then there is no need to disconnect at all.)
Children
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