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hoppy

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Hi guys got a job doing a kitchen / diner / utility room extension.

Customer wants a 2nd cu located in new utility room to feed extension.
Circuits = 1 x 32A cooker, 1x 16A boiler, 1 x 32a utility room sockets, 1 x 32A kitchen sockets, 1 x 6A lights.

Plan is split supply tails, feed 60A switch fuse, then 16mm swa clipped around outside of house from meter cupboard to new cu location, run approx. 20 meters.

My question is do you think 16mm is ok or should I run 25mm with regards to VD.

My calcs for 70 degree SWA =
2.8 x 60 x 20 / 1000 = 3.36V

then lighting circuit, roughly 20min length
29x6x20 / 1000 = 3.48V

3.48 + 3.6 = 6.84V

I am just under the 6.9V @ 3% VD allowed for lighting.

What do you guys reckon,is this a bit too close, do my calcs look ok.

Are my intended methods ok?

Guess I just lack a bit of confidence when it comes to design?

Thanks
 
What is it with people wanting to install additional separate CU's for extensions, why not just replace the existing CU for a larger one that can incorporate the extra circuits. If the truth be known, the existing CU is probably a 3036 that could do with upgrading/changing anyway!!
 
Haha tel, it's not that bad round here!

Engineer - the rest of the house is decorated with laminate floors through out so no route back to existing cu! So easiest to run sub main from external meter to extension and take new circuits from there rather then struggling to get loads of cables back to new cu!

do my calcs look ok?
 
You have used for the calcs this:
'29x6x20 / 1000 = 3.48V' (same with the 60 in the previous calc?)
Where the 6 is the rating of the CB.

You will not be using 6 Amps on a typical lighting circuit. You add up your expected load,
add on power factor, correct it upwards if necessary if you think there may be additions in the future, and use that figure for your calculation.
Here is a different example that shows how your calcs will not comply, but in reality, it would comply:.

Lighting circuit,on a 6A CB, 1mm T+E, 30 metres long.
44mV/A/m VD, x 6A CB x 30m = 7920 = 7.9Volts drop
That does not comply.

But, you have used the full load of the CB, yet only have 4 x 100w lamps on the circuit.
So do it again with the actual load. Round up the load for PF etc, so a 1000w load = 4.34A.

44 x 4.34 x 30 / 1000 = 5.7Volt drop, now the exact same circuit does comply.

You dont want to be using the CB rating as the current usage figure (unless it is expected to run at full rating), firstly as it is very rare to use a CB up to its capacity, secondly, doing that will make you use a larger cable, where clearly it will not be needed.

Doing the Volt drop calcs will also show you where the Zs is likely to comply too. Have a play with the VD/Zs figures, and you'll see they are quite closely related, if the VD does not comply, it is likely that the Zs will not comply too.

 
as above. also the 60A figure you have used is pessimistic. rare that you'll see more than 20-25A through that 16mm. id use 25A max. as Ib for calculating VD.
 
I don't think i'd be a happy bunny if this was my house and i had a cable/conduit running externally round two sides of my house. In fact i damn well know i wouldn't!! lol!

It's ok it's only Liverpool it won't look out of place. :rofl:
 

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