Voltage drop for outside lights

[adamh]

I need to fit an armoured cable to feed a few outside lights the run is 130m and the load will be 6 amp.

from the OSG i can see that 2.5mm multicore pvc single phase will drop 18mV, 4mm 11mV and 6 7.3mv

ive done the calculations
mV x 6 x 130 / 1000 = 14.04V

mV x 6 x 130 / 1000 = 8.58V

mV x 6 x 130 / 1000 = 5.69V so clearly its 6mm cable required on a 6 amp mcb

when i do the calcualtions using the ressiatance of copper at

1.78X10-8 x 260 / 2.5 X10-6 = 1.85ohms x 6 amps = 11.1V

4mm = 6.942

6mm = 3.06V

As its lighting im looking for no more than 6.9V with a Uo of 230V

now would you go with 4mm using the ressistivity values calculated and rounding to 1 decimal place

or 6mm using the OSG table 4D1B i did check the figures in the BGB for pvc swa and they are the same.

ive not used the adiabiatic to check the earth and another possibilty would be a 3A FCU to feed the c/ct and recalculate?

 

[telectrix]

6A is a lot of lights. for VD you use your design current which may well be 3A, irrespective of the MCB rating, which you would calculate according to the load and the casble used. so you would get your VD from 3A, even if you used a 6A MCB.

 

[JUD]

As telectrix said.

Work out your volt drop on the actual load of the circuit, not the size of the OCPD.

....another possibilty would be a 3A FCU to feed the c/ct and recalculate?

Going by this statement your design current is less than 3 amps anyway.

 

[JUD]

Another thing to keep in mind with volt drop on lighting circuits is that volt drop decreases as you get closer to the end of the circuit.

It's not as cut and dry as mV/A/m​ because it's a distributed load.

If you work it out properly, you could probably get away with 2.5mm²

 

[adamh]

ahhh right that sounds better lol my boss would of had a heart attack with 6mm! i didnt know you could use the design current i assumed it was on the OPD, theres gonna be 3 post lights and two fluorescent wash lights for a sign so even at 500W its only 2.17A ill try again i know your gonna say use the tabulated values but is it ok to use the ones i got from the ressistivity of copper?

also these lights will be terminated into galv t-boxes for the lights if i use flexible conduit for a flex out to the post and fittings etc would it negate the use of an rcd?

 

[adamh]

Another thing to keep in mind with volt drop on lighting circuits is that volt drop decreases as you get closer to the end of the circuit.

It's not as cut and dry as mV/A/m​ because it's a distributed load.

If you work it out properly, you could probably get away with 2.5mm²

ok i guess that will be on 2330 level 3 then :s

 

[telectrix]

and if you're using 3 core SWA with the 3rd core as cpc, you can forget the adiabatic, as the cpc is the same csa as the live conductors.

 

[telectrix]

what jud is saying is that the VD at the 1st point will be calculated using the total load. at the second point, calculate with loads 2,3,4,etc. at the 3rd point it's loads 3,4,5,etc. and so on

 

[JUD]

Here's an explanation from an old thread...

Lighting circuits (3% voltage drop load distributed)

"Load Distributed" is the key point.

IF the whole of your 6 amps was right at the end of the 90m cable then you are correct volt drop would be 15+volts.

BUT... lets assume the load is distubuted... try a few easy numbers

Consider if the lighting circuit has 12 x 100watt light bulbs (stuff the green energy saving lark for the moment!)

AND lets say they are equally spaced along the cable run in pairs,
(could be 6 rooms each with 2 bulbs in)

so 90m length divided by 6 is 15m sections.

15m then 200w load [call this bit A]
30m then 200w load [call this bit B]
45m then 200w load [call this bit C]
60m then 200w load [call this bit D]
75m then 200w load [call this bit E]
90m then 200w load [call this bit F]


so bit F is going to carry 200w 0.87A & thus drop 0.38v (29x0.87x15)/1000.
so bit E is going to carry 400w 1.74 & thus drop 0.76v (29x1.74x15)/1000.
so bit D is going to carry 600w 2.61A & thus drop 1.13v (29x2.61x15)/1000.
so bit C is going to carry 800w 3.48A & thus drop 1.51v (29x3.48x15)/1000.
so bit B is going to carry 1000w 4.35A & thus drop 1.89v (29x4.35x15)/1000.
and
bit A is going to carry 1200w 5.22A & thus drop 2.27v (29x5.22x15)/1000.

Add all these volt drops together and you get somewhere around 7.94v

which is still bigger than the 6.9v max that you correctly mention.

But we still haven't taken account of diversity.
e.g. we have calculated EVERY light is on all the time.

if we assume the 66% diversity rule of thumb with lighting circuits
then it now drops down to approx 5.25v
Jobs a goodun as they say!!

 

[telectrix]

very concise explanation.

 

[adamh]

right i see! ive not counted the ressistance of the lamps, and the volts dropped accross those either.

Unfortunately no diversity to apply as it will be fed via a photocell im looking at 3 post lights with cfls and 2 swan neck fittings also fluorescent. so i need to multiply the load by 1.8 as well i guess?

point 1 post light 30m 100w, 0.43A 18x0.43x30/1000 = 0.23V

point 2 post light 30m 200w, 0.86A 18x0.86x30/1000 = 0.47V

point 3 post light 30m 300w, 1.3A 18x1.3x30/1000 = 0.7V

point 4 swan neck 40m 500w, 2.17A 18x2.17x40/1000 = 1.56

Total volts dropped from photocell onwards 2.96V sounds good to me so 2.5 is acceptable!

Thanks for the help everyone thats cleared something jup for me if im right should make my boss very happy with 2.5 lol

 

[JUD]

It looks like you've worked it out in reverse although I don't think there will be much difference.

Point 1 will be carrying the most current.

Point 1 post light, 30m, 500W, 2.17A (18x2.17x30) ÷ 1000 = 1.17V
Point 2 post light, 30m, 400W, 1.74A (18x1.74x30) ÷ 1000 = 0.94V
Point 3 post light, 30m, 300W, 1.30A (18x1.30x30) ÷ 1000 = 0.70V
Point 4 swan neck, 40m, 200W, 0.87A (18x0.87x40) ÷ 1000 = 0.63V

Total VD = 3.44V (1.5%)


Also as you're using CFLs they're unlikely to be anywhere near 100W each so you're golden :clap:

 

[adamh]

yeah your right i had done it backwards lol thanks for that definately learn something everyday!