Discuss Example of use of Adiabatic Equation. in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

so im trying to work out the minimum size main earth for main supply ze=0.11 its a tncs supply i would say k=143 and for the main consumerunit i would say its 5 second disconection time so
230/0.11=2091
so doing the caculation i get 32.9mm and this cant be right what am i doing wrong it must be T
a bs 1361 will go quicker than 0.1 s at that fault curent.
so taking pg 244 time/current characteristics for 1361 60 a fuse at 5 seconds is 330A so doing the calc again i get 5.16mm this is better but what is right or for a consumerunit disconection time is it 0.4
please help i done it on a pir with a 6mm main earth which i doubled up to 12mm but minimum size is 5.16mm now this is safe just not conforming to bs7671 2008


You are incorrect with your 5 second disconnection time. The disconnection time is worked out by finding the fault current and using the time current curves in the Green book.
 
This equation allows you to calculate the CPC size only this must be used as a part of a cable calculation not just by its self. As you also need to take into account your R1 + R2 & Zs calcs.
You can not use this for main earth or bonds. You need the green book for that.
 
Just a quick question on the rules in maths if this thread is still live. In the equation do you apply the sq root before before deviding by the K factor ? If you do the sq root last, the answer is vastly different.

Mick
 

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