If you are trying to determine this size of cpc to use, then how are you calculating Zs which is Ze +R1+R2. R2 is cpc so how are u getting your Zs to find your fault current.
Great video, just watched it for a second time. I take it this can be applied for main earth sizing as well? Just use a Ze instead of a Zs and it’ll all be the same? That would be the only time I would use the equation i think.
Great video! Question:- Does the factor of K @ 30 degrees not get used? If the initial temperature isn't >70 degrees? In which case 143 would be used, not 115? (As 115 is assumed initial temperature of >70, but in a house etc >70 is potentially unrealistic?)
Hi Ricky1396. Thanks for watching. Good question. My understanding is, when dealing with fault current calculations the value of K is to correspond with the final temperature.
Wow! This is video very informative and the best tutorial on the topic for me so far. I really appreciate it. Thanks for sharing it with us.
If you are trying to determine this size of cpc to use, then how are you calculating Zs which is Ze +R1+R2. R2 is cpc so how are u getting your Zs to find your fault current.
As a lecturer of this subject the content in this video was excellent. Great job 👏
Thanks Connor, appreciate your kind words
Very well presented thank you, Im an old timer retired but like to keep m finger on the button and your vids are exemplary...
Thats really kind of you to say, thank you.
Can the Cmin be excluded or is it recommended to keep it in?
Thanks for the explination
Thank you. Great video, really appreciated the explaination of calculating the fault currrent for the adiabatic equation.
Thanks Gavin, please you found it of some help
Very good lecture as usual.
Thank you
Hi great video.
Have you got a example ref adiabatic if i was to calculate earth fault if using a 3core swa of say 16mm? ie k1/k2 x S
With 35m run.
Great video, just watched it for a second time. I take it this can be applied for main earth sizing as well? Just use a Ze instead of a Zs and it’ll all be the same? That would be the only time I would use the equation i think.
Well explained 👍thanks
Thanks Martin
Great video!
Question:-
Does the factor of K @ 30 degrees not get used? If the initial temperature isn't >70 degrees? In which case 143 would be used, not 115? (As 115 is assumed initial temperature of >70, but in a house etc >70 is potentially unrealistic?)
Hi Ricky1396. Thanks for watching. Good question. My understanding is, when dealing with fault current calculations the value of K is to correspond with the final temperature.
@@JPElectric Yeah that sounds the best route to go down. Been trying to get my head into these for a while, thanks for the content & reply!
Thanks for the vid!
Thank you Joeldrummer79 appreciate it.