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R

Robo

Iv been calculation voltage drop. Could some see if what im doing is correct.

So from looking in the regs i have found that your aloud 6.9v voltage drop for lights and 11.5v for sockets etc. S i take it that i need to go buy the 6.9 v as i have lights???

Im intending on feeding a new board which has lights and one socket. so 6a and 16a mcb.(Ib=22a)

Customer has already had a 4mm 3 core installed.

so calc is ((mv/a/m) x Ib x L) / 1000

(9.5 x 22 x 60)/1000 = 12.54v So voltage drop is to high and i can not use this cable.

Would it be ok to do this. Use a 10a mcb for the socket and fit a fcu with a 5 a fuse to protect socket.
Ib= 6a + 10a)

(9.5 x 16 x 60)/1000 = 9.12 This is under the voltage drop for sockets but what about lighting part as vd is only 6.9.

I would label socket and say 5a fuse only saying that if i was using a 5a fuse then i could use a 6 mcb and voltage drop would be ok for lights and power.
She intends on running a battery charger from the socket.


So any help would be appreciated, some cow boy clearly didnt do the relevant calcs and now im left to pick up the pieces and try explain. Ideas?? Robo
 
can't see it pulling 22A steady unless some sort of heating is plugged in.Even so, a couple of 1kw halogen heaters is only 9A. for calculating volt drop, you can apply diversity. that should leave you with an acceptable figure. Personally, I can't see a problem with 4mm, as far as volt drop goes
 
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As telectrix says i can't see there being a problem with 4mm. Your design current is not 22amps. A single socket is 13A and a couple of low energy bulbs is about 0.1A so even if you had the max you are only just over 6.9 volts and anything else would be fine.
 
also, as the supply is 240, you've got a free 10 volts anyway!!!!!!
 
Diversity

Lighting : 66% of total demand

Cookers : 10A + 30% balance + 5A for socket outlet

Electric showers : 100% for 1st shower
100% for 2nd shower
25% of the full load of any remaining showers

Immersion heaters : 100%

Underfloor heating : 100%

Storage heaters : 100%

Sockets & permanently connected equipment : 100% of demand of the largest load 40% of demand at every other point.
 
so for say 3 60w fittings. 120w/230 =0.8a 66% of this is 0.53?

1x 13a socket say its total load is 12a so theres no diversity for this. so could i then say 12.53 is my design current (Ib)

So vd = (9.5 x 12.53 x60) / 1000

So vd is 7.14.

Now do i use 6.9 for lights or 11.5 for power to see if the Vd is acceptable. Sorry for the questions just wana get it straight. Thanks
 
Need to use the 6.9V as there are lights on the circuit.
 
so for say 3 60w fittings. 120w/230 =0.8a 66% of this is 0.53?

1x 13a socket say its total load is 12a so theres no diversity for this. so could i then say 12.53 is my design current (Ib)

So vd = (9.5 x 12.53 x60) / 1000

So vd is 7.14.

Now do i use 6.9 for lights or 11.5 for power to see if the Vd is acceptable. Sorry for the questions just wana get it straight. Thanks
3 X 60 watt fittings = 180watts me thinks, getting the initial values right may help you.Think you just type error as you summed it as 180w.
 
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Cheers sintra. So because my vd is a little to high what would you do. \thers no way the customer will wanna replace this cable.thanks
 
TBH i would think you could save yourself/the customer some money and do without the submain and just fuse the supply down to 10A for the socket (and label it) and down to 3A for the lights and then you will be well within any volt drop limits. What do others think?
 
Sounds like a plan to me pushrod. As long as its well labelled i think its ok. I didnt kno u could get 10a fuses. Excuse my ignorance
 

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