Calculating the correct cpc size

[Chris C]

Good evening all,

Some help with following would be much appreciated.

A 20 Amp rcbo in a CU supplies an adaptable box on the outside wall of a house via 2.5mm2 T & E. From the adapatable box a short lenght of SWA supplies a couple of sockets and light in a shed.

My question is does the 1.5mm2 cpc in the T & E compley with regualtions..?

Given that the cpc is calculated either by using table 54.7 or the adiabatic equation.

Using table 54.7 a cpc of 2.5mm2 would be required.

Using the adiabatic equation S=100, t= 5 seconds, K- 115, giving a cpc of 1.9mm2.

Given that the 1,5mm2 cpc does not comply with table 54.7 or the adiabatic equation, would it be best to use 4mm2 cable or run a seperate earth.

Keep hearing that as a rule of thumb that cpc should be at least half the size of the live conductors but cannot find any reference this in the regs.

Thanks for your help

Chris

 

[Sintra]

Where are you getting the 100A current from if the rcbo is only 20A and why are you using a 5 sec disconnection time?

 

[Baddegg]

Try 20 and 0.4 in the equation...

 

[Wilko]

Hi - you need to use the prospective earth fault current in the 543.1.3 equation. But, I'd be surprised if 1.5 wasn't ok as CPC for a circuit protected by a BS60898 20A MCB in a domestic situation. However, there may be other issues with loop impedance or voltage drop limits to consider.

 

[Chris C]

Hi - you need to use the prospective earth fault current in the 543.1.3 equation. But, I'd be surprised if 1.5 wasn't ok as CPC for a circuit protected by a BS60898 20A MCB in a domestic situation. However, there may be other issues with loop impedance or voltage

Hi - you need to use the prospective earth fault current in the 543.1.3 equation. But, I'd be surprised if 1.5 wasn't ok as CPC for a circuit protected by a BS60898 20A MCB in a domestic situation. However, there may be other issues with loop impedance or voltage drop limits to consider.

Hi Thanks for the Replies,

20A rcbo max zs is 2.19. porospective faut current is 105 A. 5 seconds for T as it is a distrubution circuit. puting these figures into the equation gives a cpc of 2.04mm2 far too high i think.

Will have another look at the equation tomorrow evening.

thanks again


drop limits to consider.

Hi - you need to use the prospective earth fault current in the 543.1.3 equation. But, I'd be surprised if 1.5 wasn't ok as CPC for a circuit protected by a BS60898 20A MCB in a domestic situation. However, there may be other issues with loop impedance or voltage drop limits to consider.

 

[Chris C]

Hi everyone, thanks for the replies.

Max zs for a 20A rcbo is 2.19, giving a fault current of 105 A, 230/2.19.Using 5 seconds for t as it a distribution circuit, think this is where i'm going wrong. K =115.

Putting these figures into the equation gives a cpc of 2.04, far too high i think.

Will look at the equation again tomorrow evening.

Thanks again

 

[Sintra]

Hi everyone, thanks for the replies.

Max zs for a 20A rcbo is 2.19, giving a fault current of 105 A, 230/2.19.Using 5 seconds for t as it a distribution circuit, think this is where i'm going wrong. K =115.

Putting these figures into the equation gives a cpc of 2.04, far too high i think.

Will look at the equation again tomorrow evening.

Thanks again

I would class that as a final circuit unless of course there is another over current protective device at the juncture between the t&e and the swa which I would doubt if they are joined in an adaptable box.

 

[johnduffell]

Whether it's a distribution circuit determines the required disconnection time which would let you decide the protective device.
you would then use the actual disconnection time when using adiabatic to calculate the i2t.
This could be based on the PEFC if the ze is high enough to prevent magnetic tripping, or it could be looked up from the let through for the upstream protective devices for low ze.
Either way, for a 20A type B with a 125A or less bs88 fuse, 1.5mm will always be enough.

 

[Wilko]

Max zs for a 20A rcbo is 2.19, giving a fault current of 105 A, 230/2.19.Using 5 seconds for t as it a distribution circuit, think this is where i'm going wrong. K =115.

Hi - to work out CPC sizing we need the PEFC at the source of that actual circuit, not the minimum needed to get ADS. Thats because the fault could occur adjacent to the breaker. Edits with thanks to JD

 

[johnduffell]

Hi - to work out CPC sizing we need the PEFC of that actual circuit, not the minimum needed to get ADS.

Agree although to be clear it's the pefc at the circuit origin. and in fact as above the minimum to comply with ads on a distribution circuit is 5 seconds, regardless of whether the protective device has a magnetic trip.
So in the ops case it would be well under 100A, which would be interesting to put through adiabatic but should still come out below 1.5mm

 

[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