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wade88

Hi chaps,

Can someone please give me a quick lesson on adiabatics? When to use the equation, detailed description of what it is etc.

Been scouting any literature i have here but not been able to pin something down.

Thanks
B
 
Right, so if i was trying to determine what sized CPC to use over a 30 meter run with a PFC of 1256 i would carry out this sum:

1256 x 1256 = 1,577,536 x 30 = 47,326,080 / k factor?
 
It's the square root of the fault current required to operate the CPD within the required time squared then multiplied by the requred time, then that result divided by the k value.
 
well i was taught like this , the square root of the current squared multiplyed by the current squared multiply by the disconnection time dived by k factor

so example say 6A mcb, 0.4 disconnection time would be as follows square root of 6x6 x 6x6 x 0.4 = dived by k factor in big red book bu say we use 115 for k factor so.... now ive no calculator but you try it
 
Ok.
Now as you have used a PFC value, I take it we are checking an existing installation, and determining whether the earth/bonding conductors are correctly sized.
First you determine the type and rating of the CPD, and then determine whether the PFC is high enough to operate the CPD in the required time.
For instance the value you used of 1256A will cause a 100A BS1361 to operate in 5, 1 and 0.4 secs, but not however in 0.1 and 0.2 secs.
You then choose the highest value in which the CPD will operate, in this instance 1200A, square it giving a value of 1440000, multiply it by the time which in this instance is 0.4secs which gives a value of 576000 and then determine the square root, which is 758.94.
This value is then divided by the k value for the type and material of the CPC.
If we use the value for single conductors we would use 143 for k, if however the conductor was in T&E, we would use a value of 115 for k.
So 758.94 divided by 143 gives a minimum CSA for the CPC of 5.3mm², if it was T&E and the value for k 115, we would require a minimum of 6.6mm².
 
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