Circuit design - SWA R1+R2

[pencilpusher]

Hi all

This has got me thinking. I am designing a circuit using 3 core 2.5mm SWA and using table 11 in the OSG on page 182.

With an R1 of 2.5mm and the same for R2 I know what the milli ohms per metre is. That's all well and good but I also know from experience that when uisng the armour as an additional CPC (so to speak) that the R2 resistance will be much lower due to the additional conductor csa in circuit. However, in order to design my circuit more accurately, how do I work out the actual R1+R2 for the SWA used?

Cheers for any help on this.

PP

 

[telectrix]

there's a table on a thread giving equivalent copper size to steel armour. e.g. for 6mm SWA 3 core the armour is 34mm eq. to 15 Cu. think it's a sticky.

 

[pencilpusher]

Thanks Tel. I have found it but it does not help with a milli ohms per metre calc. It gives an armour value of 19mm csa.

http://www.electriciansforums.net/a...99484673-zs-swa-csa-tables-swa-csa-armour.pdf

I can use a 2.5mm csa line conductor value and I could probably use the 16mm csa value as the cpc as that is the closest listed value I can find??

 

[Richard Burns]

Once you have the copper equivalent csa then you can use the formula
R= rho * l / A
where rho is the resistivity of the material. For copper it is 1.7 * 10^-8. (It varies for different types of steel being about 50*10^-8 so use the copper value) then you can work out the resistance of the armour and therefore the total R2.
R is in ohms, l is in metres, A is in m².

remember that if you are running a separate cpc and using the armour the regs recommend using a separate CPC of sufficient csa to carry the entire fault current because the split between the two conductors may not be accurately known.

 

[spark 68]

A simple way of doing it is by finding the miliohm per M for 1mm^ and dividing by 19

 

[pencilpusher]

Thank you Richard. To clarify though, my CPC is within the 3 core and is the same size as the line conductor so I presume that complies anyway re fault currents?

 

[pencilpusher]

A simple way of doing it is by finding the miliohm per M for 1mm^ and dividing by 19

Good point.

 

[spark 68]

This simple method only works up to a point, I think once you get on to the larger cable sizes impedance plays a part and changes the calcs.

I was talking about the copper equivalent csa of the steel.

IIRC Steel has 8x the resistance of equivalent copper, or IOW needs to have 8X the csa of an equivalent Copper conductor , but I will stand to be corrected on this point.

 

[Richard Burns]

Thank you Richard. To clarify though, my CPC is within the 3 core and is the same size as the line conductor so I presume that complies anyway re fault currents?

Certainly should comply yes, i was being a bit pedantic, sorry!

 

[pencilpusher]

Certainly should comply yes, i was being a bit pedantic, sorry!

Not at all. I am grateful for the help.