minimum cpc sizes

[SafetyEICR]

Just running thro some design calcs for minimum cpc size and im getting some unrealistic readings. Example 1.5mm c.s.a live and 4.mm cpc on a lighting circuit?? circuit is Zdb = 0.083


Circuit 1L1


Assuming we can use 1.5mm cpc



R1 + R2 = (12.10 + 12.10) x 1.2 x 3.57 / 1000 = 0.10ohms


ZS = 0.083 + 0.10 = 0.183


Ia = 230 / 0.183 = 1,256a


S = 1,256 squared x 0.1 = 157753 square root this figure / 115 = 3.45mm cannot be right surely??

 

[Geoffsd]

So I need to replace 0.1 with 0.4. as the circuit is 240v <32a. Just a little confused as we were shown a different way by our course tutor. thanks for feedback.

No. 0.4s is the minimum disconnection time for the circuit.

The device will open in 0.1s at 60A

 

[SafetyEICR]

Thanks. Does this look correct for ring circuit cpc Assuming we can use 2.5mm cpc


R1+ R2 = ( (7.41 + 7.41) / 4) x 1.2 x 7.15 / 1000


ZS = 0.083 + 0.031 = 0.11ohms


Ia = 230 / 0.11 = 2,090a


As a 60898-B 32a will trip in 0.1 at 160a this figure will be used


S = 160 x 0.1 / 115 = 0.43mm

 

[Risteard]

Your fault current is not 160A though - according to your calculation it is 2.09kA!

So you need to use this - and as I said confirm with the manufacturer of the circuit breaker that it will infact open in 0.01s!

 

[Geoffsd]

So you need to use this - and as I said confirm with the manufacturer of the circuit breaker that it will infact open in 0.01s!

Are you suggesting that the OP is expected to contact the manufacturer for a college exercise?

 

[SafetyEICR]

Struggling to find data for memshield mcb to see if it will trip in 0.01. Going with 0.1 as stated in regs using above calculation im getting back to using 4.0mm cpc on a 2.5mm ring main.

 

[jamesBoater]

surely you just use the time current graphs for BS EN 60898 in BGB rather than confirming with the manufacturer.

 

[SafetyEICR]

Although using 0.01 gives a cpc of 1.81 therefore 2.5mm

 

[SafetyEICR]

I think ill have to go with 0.1 as stated in regs. Circuits within mains room will have unusually high cpc due to the low impedance on the circuit. All other circuits are coming out about right.

 

[SafetyEICR]

Still getting 4mm cpc on some lighting circuits. Just doesn't happen in the real world. Although cant be marked down wrong by going with figures stated in regs!!

 

[Guest55]

Still getting 4mm cpc on some lighting circuits. Just doesn't happen in the real world. Although cant be marked down wrong by going with figures stated in regs!!

then you must be doing the sums wrong......
getting a 4mm cpc for a 6A lighting circuit by using the adiabatic eq. means your circuit must be 100's of metres long ?

 

[SafetyEICR]

Heres one of the calcs. assuming we can use 1.5mmR1 + R2 = (12.10 + 12.10) x 1.2 x 3.57 / 1000 = 0.10ohms


ZS = 0.083 + 0.10 = 0.186ohms


Ia = 230 / 0.186 = 1,236a


S = 1236 x 0.1 / 115 = 3.40mm

 

[Guest55]

blow me over , your sums appear correct.

what is causing this calculation quirk is your Zdb being incredibly low at 0.08 , which in turn massively increases the circuits prospective fault current.

looks like you need a 4mm cpc then

 

[SafetyEICR]

Yes its the lighting circuit in the main room mate. its the same for the socket circuit within the mains room. 6mm earth needed!! Assuming we can use 2.5mm cpc


R1+ R2 = ( (7.41 + 7.41) / 4) x 1.2 x 7.15 / 1000 = 0.031


ZS = 0.083 + 0.031 = 0.11ohms


Ia = 230 / 0.11 = 2,090a


S = 2090 x 0.1 / 115 = 5.74mm

 

[Baker1988]

Yes its the lighting circuit in the main room mate. its the same for the socket circuit within the mains room. 6mm earth needed!! Assuming we can use 2.5mm cpc


R1+ R2 = ( (7.41 + 7.41) / 4) x 1.2 x 7.15 / 1000 = 0.031


ZS = 0.083 + 0.031 = 0.11ohms


Ia = 230 / 0.11 = 2,090a


S = 2090 x 0.1 / 115 = 5.74mm

I thought it was fault current squared x 0.1 and also your forgetting the square root ain't you

 

[Baker1988]

So it would be so, S = sq root 2090 x 2090 x 0.1 / 115