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??

 

[tony mc]

Have a read of this Example of use of Adiabatic Equation. - Electricians Forums

 

[i=p/u]

Square root being used in correctly

 

[SafetyEICR]

Pretty sure im using the calculator correctly. As im getting same reading as on the example on the link above.

 

[Geoffsd]

S = 1,256 squared x 0.1 = 157 square root this figure / 115 = 3.45mm cannot be right surely??

You have done the maths right but it should show 157,000.

However, a 6A mcb (if that's what you have) will trip in 0.1s at 30A so that is the current that should be used.

30² x 0.1 = 90
√90 / 115 = 0.08mm²

 

[SafetyEICR]

Sorry type error. Yes 6a type c. Thanks that's a great help

 

[Geoffsd]

As it is a type C you will have to use 60A (6 x 10)

 

[SafetyEICR]

This design project doing my nut in. Its got to be in Tuesday!!

 

[SafetyEICR]

So irrespective of all circuit characteristic as long as Ia is 60a or above, we can use 60 as the Ia value to determine cpc size?

 

[Risteard]

Bear in mind that with sufficient current the device may open in the definite minimum time of 0.01 seconds rather than the 0.1s "instantaneous" figure in the Appendix.

 

[Geoffsd]

So irrespective of all circuit characteristic as long as Ia is 60a or above, we can use 60 as the Ia value to determine cpc size?

Yes, that's what Ia is - the current which causes the device to open in the required time.

The required time is actually 0.4s.

 

[Risteard]

You should be using the actual earth fault current and the actual disconnection time of the device.

Yes, that's what Ia is - the current which causes the device to open in the required time.

The required time is actually 0.4s.

 

[SafetyEICR]

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.

 

[SafetyEICR]

So that calculation now becomes Assuming we can use 1.0mm cpc



R1 + R2 = (12.10 + 18.10) x 1.2 x 3.57 / 1000 = 0.129ohms


ZS = 0.083 + 0.129 = 0.21ohms


Ia = 230 / 0.21 = 1,095a


As a 60898- C will trip in 0.4s at 60a this figure will be used


S = 60 x 0.4 / 115 = 0.32mm

 

[Geoffsd]

You should be using the actual earth fault current and the actual disconnection time of the device.

Getting confused - it's not Ia in this calculation, is it?

Then you have to find the actual disconnection time for the actual current.