Calculating the correct cpc size

[Chris C]

Thanks again for the replies.

From fig 3A4 the current that would cause a 20 A rcbo to trip in 0.1 to 5 seconds is 100 A
T = 5 Seconds (distrubution circuit)
K =115

Putting these figures into the adiabatic equation, I get a minimum cpc of 1.9 mm2

Could someone please advise if i have used the correct figures in the equation to get a cpc of 1.9mm2

Thanks for your help

 

[johnduffell]

you need to use the actual PEFC.

 

[Baddegg]

And I still can’t see why it’s a distribution circuit? Apologies if I’ve missed something

 

[Wilko]

And I still can’t see why it’s a distribution circuit? Apologies if I’ve missed something

Hi - I'm too sure what's proposed here exactly (lm a bit worn out tonight) but if there is current using equipment attached to this 2.5mm cable directly (lights, outlets etc) then it's a final circuit. If it only feeds another board, then it's a distribution circuit. But at 2.5mm it's a bit anorexic .

 

[Baddegg]

Op only states it’s 2 sockets and a light, I assumed he was fusing down the lighting point,which makes it a final circuit so disconnection time 0.4?
OP I am reasonably new to this myself so if I’ve quoted anything wrong I do apologise

 

[johnduffell]

duplicate, sorry

 

[johnduffell]

I'm not sure the OP understands why the equation exists and what exactly it's trying to prevent. Basically in simple terms it's preventing a high current of short duration from damaging the cable insulation.

The energy per metre of the cable is what heats it and the volume is what buffers the heating due to thermal mass. I.e. twice as much copper takes twice as much heat to get to, say 100c.
This is measured as a constant number of joules "j" per cubic metre of copper required to heat the conductor by the required number of degrees. j joules per metre^3
This is the same as j watt seconds per metre^3
or j volt amp seconds per metre^3
or j ohm amp amp seconds per metre^3.

The resistance of the cable (ohms) is proportional to the length (metres) so we end up with amp amp seconds per metre^2 times another constant "r". Let's make a new constant "k" which is just r.j From that blurb we just derived the adiabatic equation:
CSA = k (I^2)t.

So what is I? the actual current that will flow in the worse case, an earth fault at the origin? ie. the PEFC.
So what is t? the time the current will flow before being cut off. You can read that from your protective device chart. Make sure you read the worse case, not the average.
Can't read t because the disconnection time is less than 0.1 seconds? There is another protective device chart, you have to check the energy class of the MCB, and look up based on the upstream protection. Then you can just read the I2t straight off the chart.

 

[Michael Davis-McCoy]

Hi Chaps,

Just been reading the thread. There seems to be some confusion.

The maximum disconnection time is 0.4 seconds. As it classed as a final circuit and is under 32A. See reg 411.3.2.3

You need the Zs of the circuit. That is the Ze +R1 +R2 for the length.

Then calculate the fault current. The nominal voltage 230/Zs = If

Then you need to check the disconnection time from the protective device curve, or the associated table on the curve figure. t

Now you can do the adiabatic equation.

S = minimum cpc size
If = the result of the above calculation
t = disconnection time from the figure
K = the constant for copper 115

Don't forget the squaring and square rooting. I think you maybe surprised how small the cpc can be calculated to.