Rules and Guidelines
Want to participate in the discussions? Please read our Community Rules and Guidelines. Need some help? Visit Support and Answers.

  • State Not Answered
  • Locked Locked
  • Replies 7 replies
  • Subscribers 32 subscribers
  • Views 790 views
  • Users 0 members are here
Options
You may also be interested in
This discussion has been locked.
You can no longer post new replies to this discussion. If you have a question you can start a new discussion

Supplementary CPC/SWA Resistance Ratio

fiftyhertz
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
  • 0

    mapj1:

    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.)




    Mike great explanation, so if you get voltage at the fault to 50v via a low resistance cpc, there’s no ‘need’ to disconnect, however by nature of the short circuit between live and earth won’t it will still automatically disconnect as the fault current will trip the breaker (even if the majority of the voltage is dropped in the line conductor)

Reply
  • 0

    mapj1:

    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.)




    Mike great explanation, so if you get voltage at the fault to 50v via a low resistance cpc, there’s no ‘need’ to disconnect, however by nature of the short circuit between live and earth won’t it will still automatically disconnect as the fault current will trip the breaker (even if the majority of the voltage is dropped in the line conductor)

Children
No Data

Community Rules & Guidelines