New Zealand calculating main earth size

[jamesBoater]

i am just a little unsure if i need to use 5 seconds as my D/T or 0.4? PFC obtained through Ze test at origin.

i have been told 2 different answers and any guidance would be helpful.

thanks james

 

[D Skelton]

On TN systems the maximum permissable disconnection times are 0.4s for final circuits up to and including 32A. 5s for distribution circuits and final circuits over 32A.

For TT systems it is 0.2s and 1s respectively.

If you are trying to work out the 't' in an adiabatic equation then you should be using manufacturers data or at worst the regs book and the time current graphs to tell you what the D/T will be for the OCPD when exposed to the PEFC you have measured. Using 0.4 or 5s for your calculations will more often than not give you an earthing conductor size which is far higher than required.

 

[telectrix]

as above ^^^^^^ .

 

[jamesBoater]

Cool thanks for the reply. This info has cleared that up for me.

 

[SirKit Breaker]

You need to work out the 'S' in the adiabatic for the size of your conductor, not t, you cant work out the 't' as this is determined by the maximum disconnection time allowed for the particular circuit, 0.1, 0.4 etc etc and is a constant.

 

[jamesBoater]

The bit i was not sure about is what the max DT is. My PFC is taken at the origin through a Ze test. would this be a distribution circuit and be 5 seconds?

thanks james

 

[telectrix]

no. t is the time the ocpd takes to trip with the pfc that you have measured. usually be 0.1sec. taken from appendix3 BGB. time/current curves.

 

[GLENNSPARK]

its the fault current squared X the disconnection time (0.4 or 5 sec)...apply the square route to that sum then divide the result by the factor of K....which for copper is 115

 

[telectrix]

its the fault current squared X the disconnection time (0.4 or 5 sec)...apply the square route to that sum then divide the result by the factor of K....which for copper is 115

. nope, glenn. get the value of t from time/current curves, and k is 143 if it's a single cable like as wot a main earth is.

 

[GLENNSPARK]

. nope, glenn. get the value of t from time/current curves, and k is 143 if it's a single cable like as wot a main earth is.

i`l have a look at that ...
i`m going by what i was taught at college....

- - - Updated - - -

anyway Tel...i`ll send you that thing in a mo...

 

[telectrix]

k is 115 in a multicore cable or cables bunched, but it's 143 for a single cable in free air or clipped etc., as it has better cooling.

 

[GLENNSPARK]

k is 115 in a multicore cable or cables bunched, but it's 143 for a single cable in free air or clipped etc., as it has better cooling.

yeah...thanks Tel....

 

[ajelec10]

Telex and D are right here. You have to work out using the graphs/manufacture data, the actual time in which it takes the over current protective device to disconnect the current that would flow in the circuit during fault conditions. (PFC) so if the pfc was lower it would take longer to disconnect! And vice versa! Hope this helps!!

 

[GLENNSPARK]

Telex and D are right here. You have to work out using the graphs/manufacture data, the actual time in which it takes the over current protective device to disconnect the current that would flow in the circuit during fault conditions. (PFC) so if the pfc was lower it would take longer to disconnect! And vice versa! Hope this helps!!

i`m aware of disconnection times....

 

[stigwart]

You could just refer to table 54.7 of the regs book, where the sizing of the earth has been calculated for you I believe?