Example of use of Adiabatic Equation. | Page 3 | on ElectriciansForums

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

Resistivity of the material the conductor is made of, taking into account temperature and heat capacity of the conductor material.
There is a more complete (and technically accurate) definition just above table 43.1.
 
Very good example thanks for sharig it i am due to take my exam next month, and i am told it is 50/50 if this question comes up

\thanks once again Ian.
 
Its been a while but I have my NAPIT inspection tuesday week and jumped on here for a refresher....lovely example of the adiabatic and easy to understand when put the way it was...thanks Dave....i'll keep reading....best sparky forum IMO.
 
Hi If I may chip in what you say is correct but the driving force behind pme is to save the REC money on not having to provide an earth conductor .
my view on pme is its dangerous as it relys on the consumer having to install larger than normal earthing conductors in case the pme neutral fails otherwise you have a very dangerous situation give me TN-S any day
 
If the PME neutral was to fail then no matter what size earthing conductor you had fitted the earth would still fail as the return path for the earth is the neutral. (Main bonding conductors are sized according to the neutral)

I was under the impression that if the DNO did not supply a earth connection then it was down to the customer who would then have to go down the TT route.

There have been a few posts recently re high ZE readings on TNS systems which turned out to be the suppliers sheath broken!
 
Hi Tony mc

Im sorry I have to disagree with you pme installations have multiple earth spikes as back up for just such a senario they usually fit a earth rod at the transformer & another where the service enters the building so loss of a neutral will divert to the earthing system back to the transformer hence the need for large earth wires & bonding as every customer fed of that transformer has a neutral return via the earth. ( very dodgy but the rec don,t follow Bs7671 )
 
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.
 

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