Discuss Lighting spur off a fuse spur in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

M

Mr. Anderson

Hi guys. The situation is I'm pricing a kitchen re-wire and the owner also has a new shed (6ft away from kitchen) that they would like to supply 4 single sockets to feed the white goods and a light. Price is a factor (when is it not) and the best way I can see to meet the requirements without having to go over board is to stick a 13a switched FCU in the ring to supply the 4x sockets in the shed (2.5mm radial) and then take a 5a switched FCU to hit the light requirements (1.5mm). I can't find any reason why I can't do this, reg 433.1.103 doesn't cover a second spur of a fused spur. The CU is RCD protected, the ring will be on a 32a MCB and it's a small kitchen (9 sockets total). I've considered loads on the 13a spur and there are no concerns there. ~~~Second question, why can't I hit "enter" and space out my post? Is it my lap top or a setting on the forum? Many thanks, Chris.
 
assuming the feed to the shed will be in SWA.
 
I was going to use PVC conduit as the vertical runs on the shed and wall are solid and the horizontal run will be under the decking run through the frame work. Does this sound a little dodgy?
 
If your pricing for a kitchen rewire, it's not going to cost much more to run a separate supply out to the shed. At least then if their was a problem in the shed, it's not going to affect the kitchens installation!!

As Tel stated, either way the shed supply is going to need a SWA cable, if this cable is going to be buried. Don't use T&E for external underground installations!!
 
but he's got a bit of timber to surface clip the t/e to. :ban:
 
Good point about the decking, cheers. As for the separate supply, I would go down this route but there isn't the space on the CU for a new circuit.
 
Sorry telectrix, what do you mean? I wasn't going to clip direct to decking but run conduit through the frame work under the decking reducing the risk of a trip hazard.
 
I'm doubting my math now. To work out the amps when all I've got is the annual consumption I take the annual and divide by 365 (days) then by 24 (hours). This then gives me my watts per hour, I can then add up each appliance and divide by voltage to give me my total amps to find my breaker size?
 
Last edited by a moderator:
So is this really your house and it's a DIY job?
If not then your last post shows your level of skill in your trade.
If you're honest people they will guide you down the right path if you're willing to listen and take it in.
 
I'm doubting my math now. To work out the amps when all I've got is the annual consumption I take the annual and divide by 365 (days) then by 24 (hours). This then gives me my watts per hour, I can then add up each appliance and divide by voltage to give me my total amps to find my breaker size?

Me thinks you need to review this a bit more. Not the way to calculate the load IMHO
 
I thought I knew how to work out the load. (Amp x volts x power factor) then work out you cable and breaker sized based on this info.

But in this situation I'm only getting the catalogue showing the data for each appliance which shows an "annual consumption in kWh" for each appliance rather than its amps (which will be shown on the appliance).

So with the information I've got (the annual consumption of each appliance in kWh) how do I work out the total load for that circuit?

Sorry I'm confusing everyone here, if I'm still not making my self clear forget it and I'll ask someone who knows what I'm trying to say.
 
if the total current consumption of the appliances in the shed exceeds 13A , then the shed will need its own circuit of 20A or greater.
in other words , youll get away with a fridge and freezer , but wash mach & tumble drier use alot more power.
 
Thanks Biff, I do understand that but I guess what I'm now asking is how do you work out the total load when the information you've got for each appliance is measured in kWh?
 
the data stamped on the back of each appliance will be kW , not kWh's.
for example , a 2kw heater.......

2000w divide by 230v = 8.7 A
 

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