CSA of protective conductors (16th)

[dunny123]

I know that the CSA of a protective conductor can be selected by calculation [543-01-03] or selection [543-01-04]

As selection uses table 54G, the CPC for typical domestic circuits would be the same size as the phase conductor. As this is not usually the case for T&E wiring, e.g 2.5/1.5, what method is being used ?

Assuming the "method" is the OSG, then what figures would have been used for the adiabatic equation for say a 32A ring final in 2.5mm T&E protected by a type B 60898 ?

 

[GaryMo]

Depends on PFC.

 

[dunny123]

Depends on PFC.

From a design point of view, how can it , as you don't know what the PFC will be. This is really my point, as if the max permissible Zs (1.5ohm) of the circuit is used to evaluate the value for I then a very small CPC will be calculated, and if a very low Zs is used, 1.5mm could be calculated to be undersized.

 

[GaryMo]

You need to factor in Ze though along with R1+R2 of the circuit.

I = 230/Zs

 

[dunny123]

The regs state The CSA of a protective conductor can be selected by calculation [543-01-03] or selection [543-01-04] (table 54G)

My point is if the cable is selected from the OSG then neither of the above have been directly conformed to.

However, the OSG cable charts will have been created surely by adiabatic calculations using some values for K,t and I.

So my question remains does anyone know what values they are based on, e.g. max permitted impedance??, max breaking rating of the OPD??..etc

 

[Spark123]

The cpc in a flat twin and earth is calculated [543-01-03] by the manufacturer using an adiabatic equation.
You can work it out for yourself if you like, for example if you take a 15A rewirable fuse connected to a radial of 2.5mm² and look at the current required to operate it at the minimum time on the chart i.e. 0.1s (180A) and the maximum 5s (43A) and do the calculations for s=root(I²t)/K for both these the result is both less than 1.5mm², using a K of 115 for 0.1s it is 0.5mm² and for 5s it is 0.83mm².
You can try working it out for a 2.5mm² ring final circuit using both the 0.1s disconnection times and the max 0.4s disconnection times, both result in a CSA of 1><1.5mm² which is why the old 2.5mm²/1mm² twin and earth is a bad idea with rewirable fuses, especially on any spurs!

 

[dunny123]

Thanks, thats what I was after. The bit that doesnt make sense to me is that the value of I you are using is for (0.1s) an instantaneous operation of the device, not the PFC. Its not uncommon to see a Zs on a radial in 2.5/1/5mm of say 0.3ohms, which would require a CSA of over 2mm using the same 0.1 seconds for "t" in the equation.

 

[Spark123]

Generally as the current increases the disconnection time decreases - the fuse would hopefully be well on the way to disconnecting before the current reaches 800A. It is also a bit off the scale of the graph for a 15A BS3036 so if you wanted to you could phone the manufacturer of the fuse wire and ask them.

 

[Adam_151]

You should really do the calculation for both ends... far end, with the lowest fault current, and highest time to operate, and near end, with higher fault current, and consequently lower operating time

Its important to understand that fuses and breakers work differently, the amount of energy needed to vapourise a piece of fusewire is roughly costant once it happens quick enough that heat cannot escape from it, and thus, the lowest values of I²t occur at the near end and the larger ones at teh far end, for a breaker, its the reverse as its the operating time thats constant, and this I²t rises in proportion to the square of the current and the highest values occur at the start of the circuit where the fault current is greatest

For breakers, best generic thing you can do is t of 0.1secs and I² of whatever calaculation from Zs gives, if you want a more workable value, then the data for the device will give I²t plotted against fault level