Pros...cons...preferences..

[Baddegg]

after my rather vague thread Friday about finding the 30 mtrs of 10mm 2 core swa In the back of me van.....after some views on earthing to the shed, 3 options as far as I’m aware are
1. Run a single 10mm from main met upto the shed and use that
2. Use the armour of the swa, adiabatic permitting..
3. Connect the swa to the main met at source end and TT the shed
Each option is appruthe same cost wise and it’s my own house so Labour is not a big issue, as per title just after others views and preferences....

 

[Squid]

You think the cable is amusing....I’m off to do the adiabatic! Wait till I post them maths up ha ha ha

I hear Croatia is nice............

 

[Wilko]

Hi - if no bonding required then option 2 too

 

[Charlie_]

Talking of red and black..
I extended a 6mm rb t&e last week with some 6mm rb that I’ve had in the workshop for years...
Just had to get it off my chest, please don’t judge me or tell my mum

 

[Baddegg]

Talking of red and black..
I extended a 6mm rb t&e last week with some 6mm rb that I’ve had in the workshop for years...
Just had to get it off my chest, please don’t judge me or tell my mum

 

[Charlie_]

It did save on having to put ‘careful now’ stickers all over the place though

 

[Baddegg]

Oh god! Here goes......

If = 230x0.95 (c min) / 0.36 (Zs) = 606 a

t = 0.1s (40a 60898 mcb)

K = 115 (copper line conductor) / 51 (steel armour) x 10 size of line conductor = 22.54

Sooooooo......

S = (sq root symbol ?) 606a 2 x 0.1/22.54 = 8.5mm

CSA of 2 core 10mm2 swa is 41mm2 so all is good?

Kill me now!

 

[Deleted member 26818]

Why does t = 0.1s?
It’s a distribution circuit, so t = 5s.

 

[Baddegg]

Ahhhhh yes.....

 

[Deleted member 26818]

Ahhhhh yes.....

Also K for steel armour is 51 For 70 degrees C and 46 for 90 degrees C.

 

[Deleted member 26818]

square root of 606 x 606 x 5 divided by 51.
26.56mm2.
CSA of SWA for 2 core 10mm2 is 40mm2

 

[Baddegg]

Did I not need to make the calculation for using different materials, copper and steel?
Thanks for taking the time spin I do appreciate it

 

[Deleted member 26818]

No, you don’t need to make a calculation for the different materials, unless you wanted to know what the copper equivalence of 40mm2 SWA would be. 17.7mm2.

 

[Baddegg]

Bugger! So close.....ish

 

[Risteard]

Why does t = 0.1s?
It’s a distribution circuit, so t = 5s.

Actually t is the actual disconnection time. So in this instance it won't be anything like 5 seconds, but probably between 0.01 and 0.1 seconds.

 

[TJ Anderson]

Yeah, 0.1s is as low as BS7671 goes for your calcs, any Zs =/< max Zs for that 60898 would achieve that (as needs at least that fault current). As Zs drops, fault current increases and disconnection will get quicker. Would require manufacturer's data if you needed to go lower than 0.1s to satisfy thermal constraints but you definitely don't need to in the case of this job.