volt drop calc

[lockhes]

is this right

mv on chosen cable from regs x length of cable run x design current of circuit

divided by 1000

is it calulated amps or the breaer your going to use

is that all as im trying to work out for this lighting circuit but to get the volt drop under 3% the cable needs to be 6mm


ta for help

 

[i=p/u]

you can get ib down with diversity and length of your run could be shortened

 

[Octopus]

Cripes, 6mm on a lighting circuit. How big is the building?

 

[i=p/u]

split the lighting circuit up or

 

[lockhes]

its big ish

got a 60m run

its got 4 400w lights off the circuit which are low bay ones which do you need to x by 1.8 as arent they a bit like fluoescents

 

[Johnny-G]

Then 6mm sounds about right.
Using a certain suppliers online calculator -

3kw lighting circuit over 60m, clipped direct =

[TABLE="class: results"]
[TR]
[TD]Required Cable Size[/TD]
[TD]6 mm[/TD]
[/TR]
[TR]
[TD]Voltage Drop[/TD]
[TD]5.71 Volts.[/TD]
[/TR]
[TR]
[TD]Percentage Drop[/TD]
[TD]2.5%[/TD]
[/TR]
[TR]
[TD]Current Load[/TD]
[TD]13.0 Amps[/TD]
[/TR]
[TR]
[TD]Max Cable Load[SUP]*[/SUP][/TD]
[TD]47.0 Amps[/TD]
[/TR]
[/TABLE]

 

[lockhes]

yup thats how i worked it out from my regs and a calc

but when i went onto an online calc it said 2.5 was fine which confused me

not really possible to shorten run as its the switch wire which is causing the problem only option is to make into 2 circuits or just use 2 lights

have my calcs been right though

 

[johnboy6083]

you could use a contactor to switch the lights?

 

[Engineer54]

If this circuit is longish then to get an accurate VD figure you need to calculate at each lamp, something similar to calculating VD and cable sizing on street lighting!!

Remember the first lamp position on the circuit will have the shortest distance, but the largest load, while the last lamp will have the longest distance but the least load...

 

[i=p/u]

or just fuse each light separately, so fused spur before it, reducing cable size to each light but on whole circuit, am i correct i have a L sticker slapped on me

 

[Knobhead]

Personally I’d run a 2.5 ring with as i=p/u says a fused spur to each light. No struggling to get cables in to the fittings.

 

[JUD]

If this circuit is longish then to get an accurate VD figure you need to calculate at each lamp, something similar to calculating VD and cable sizing on street lighting!!

Remember the first lamp position on the circuit will have the shortest distance, but the largest load, while the last lamp will have the longest distance but the least load...

What Engineer said is correct.

Let's say your 4 x 400W lights are spaced equally along the 60m run, 1 every 15m

Assuming 230V supply and 2.5mm² clipped direct...


a) The cable from DB to light 1 will carry the full load current
I = 1600 ÷ 230 = 6.96A
VD = (18 x 6.96 x 15) ÷ 1000 = 1.88V

b) The cable from light 1 to light 2 will carry ¾ of the load
I = 1200 ÷ (230 - 1.88) = 5.26A
VD = (18 x 5.26 x 15) ÷ 1000 = 1.42V

c) The cable from light 2 to light 3 will carry ½ the load
I = 800 ÷ (230 - 1.88 - 1.42) = 3.53A
VD = (18 x 3.53 x 15) ÷ 1000 = 0.95V

d) The cable from light 3 to light 4 will carry ¼ of the load
I = 400 ÷ (230 - 1.88 - 1.42 - 0.95) = 1.77A
VD = (18 x 1.77 x 15) ÷ 1000 = 0.48V


Total VD = 4.73V (2.06%)


As you can see you are well under the 3%.

 

[Spoon242]

Would you calculate like this for a ring final circuit, work out at 5% between each socket etc?

 

[JUD]

Would you calculate like this for a ring final circuit, work out at 5% between each socket etc?

No. This is for "distributed loads" where you know what the load will be at each point in the circuit.

Ring circuits are calculated normally and the result divided by 4 i.e ((mV/A/m) x Ib x L) ÷ 4, L being the total length of the ring from end to end.

The design current of a 32A ring is taken as 26A, 20A at the mid-point of the ring and the remaining 12A shared between each leg (6A each leg), therefore no point in the ring should carry more than 26A if you get what I mean.

So the volt drop in a 50m ring circuit wired in 2.5mm² would be calculated like this:

((18 ÷ 1000) x 26 x 50) ÷ 4 = 5.85V (2.54%)

 

[Spoon242]

Nice one,

32 metres is my last point on the ring and thats my 26A and then divide by 4 because its a ring , ie 4 off 2.5 mm cables.

So ((18 ÷ 1000) x 26A x 32M) ÷ 4 = 3.74v