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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
 
I know that it will be on the appliance BUT they don't have the appliance yet all they're able to do it point at it on line and the only power information on line is in kWh. So how do I work out the total load with that information.

e.g.
Buy BEKO LX5053S Undercounter Fridge - Silver | Free Delivery | Currys
+
Buy BEKO WM74155LS Washing Machine - Silver | Free Delivery | Currys
+
Buy BOSCH EcoLogixx 7 WTW86561GB Condenser Tumble Dryer - White | Free Delivery | Currys

Presumably you would transpose the calculation to work out kWh to get the power and then do the calculation for current as usual.
 
scarey questions by somebody claiming to have passed the 2391

I know how to work out a total load when given the appliance amps and that this info is on each appliance. I thought I had already explained this. I'm confussed about the "annual consumstion" and "kWh" part.
 
i dont need to convert from kwh to kw to tell you that all those appliances are going to well exceed 13A lol.
tumble drier alone is nearly 10amps.
washing machine the same , and around 5-6A for fridge.
 
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Thanks Biff. Then to someone else, how do you convert kWh into kW so I can have a real understanding of whats required not an educated guess.



I'm not sure why I'm getting such a bashing here. This isn't anything that was covered in 2391 and I'm keen to learn how to do this.
 
Yep, I see a 13a fcu is no good but can someone explain how I find out what is required from the information I've been given. Please remember I've not got the appliance to hand and only have info in kWh. ;)
 
Thanks Biff. Then to someone else, how do you convert kWh into kW so I can have a real understanding of whats required not an educated guess.



I'm not sure why I'm getting such a bashing here. This isn't anything that was covered in 2391 and I'm keen to learn how to do this.

No! it wouldn't but you should know this from earlier qualifications you should have or not if you are asking them, are you a Electrical Trainee by any chance?
 
kWh = Power (in kilowatts) x time (in hours)

You do the rest...

Now that rings a bell, thanks! So (and this is from memory so I may very well not be getting this right) if E=Pxt then because I've already got the watts per hour then the math for the tumble dryer is...
E=Pxt
1.4kWh=1400x1

A=w/v
5.8A=1400/240

So then a 6A fuse should take that load. Obviously I have other loads to add to that in this situation BUT from what you guys are saying that seems very low? And now I've just seen the manafactorys details which prove my math to be wrong. Bugger this, I give in and will now just google the details.
 
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?

what are you drinking tonight?
 
Now that rings a bell, thanks! So (and this is from memory so I may very well not be getting this right) if E=Pxt then because I've already got the watts per hour then the math for the tumble dryer is...
E=Pxt
1.4kWh=1400x1

A=w/v
5.8A=1400/240

So then a 6A fuse should take that load. Obviously I have other loads to add to that in this situation BUT from what you guys are saying that seems very low?

230 is the voltage used in calculations.
 
I think he now knows that his cheap ring circuit FCU approach to supplying these types of white goods, isn't an option open to him. It's going to have to be a separate circuit out to this shed, or forget the idea altogether. lol!!
 
I think he now knows that his cheap ring circuit FCU approach to supplying these types of white goods, isn't an option open to him. It's going to have to be a separate circuit out to this shed, or forget the idea altogether. lol!!

I don't think the OP is in the mood for advice.

May be he should employ a competent sparky!!!!
 

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