The Adiabatic Equation

[@danielteyehuago1633]

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.

[@connormcgrath1152]

As a lecturer of this subject the content in this video was excellent. Great job 👏

[@JPElectric]

Thanks Connor, appreciate your kind words

[@angelofranklin1]

Very well presented thank you, Im an old timer retired but like to keep m finger on the button and your vids are exemplary...

[@JPElectric]

Thats really kind of you to say, thank you.

[@gavinhibberd1708]

Thank you. Great video, really appreciated the explaination of calculating the fault currrent for the adiabatic equation.

[@JPElectric]

Thanks Gavin, please you found it of some help

[@NaciMahmut-rf5xf]

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.

[@iqbalnuthoo1566]

Very good lecture as usual.

[@JPElectric]

Thank you

[@martinburgess2322]

Well explained 👍thanks

[@JPElectric]

Thanks Martin

[@BTC_hoven]

Can the Cmin be excluded or is it recommended to keep it in? Thanks for the explination

[@garythespark6295]

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.

[@joellaight]

Thanks for the vid!

[@JPElectric]

Thank you Joeldrummer79 appreciate it.

[@harry24780]

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.

[@LukeRichards15]

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?)

[@JPElectric]

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.

[@LukeRichards15]

@@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!