Voltage drop

Hi, I'm a first time user of the forum. I am trying to calculate the voltage drop to determine the cable to use for garden lighting. I have a run of 300m from the fuse board to the last light on the circuit. The total wattage of lights that will run off this circuit is 1000w.

I will be using amoured cable running on the surface with a junction box at each light intersection.

Can anybody help as my brain has now short circuited trying to work this out.thanks

Basic volt drop for the total circuit (assuming no other loads) is
Vd (mV/A/m) *A (amps) *m (metres) / 1000
In this case Vdtot = Vd*4.34*300/1000 = Vd*1.302
Where Vd is the voltage drop per ampere per metre for the size of cable.
Lighting must be less than 6.9V (3%) drop.
rearranging you can then state max volt drop (mV/A/m) for a cable is 6.9 *1000 / (4.3*300) = 5.35 so 10mm2 SWA (4.4mV/A/m)

If this is too large a csa for the cable then you can divide the circuit up into sections for the lengths of cable between each light, and calculate the total volt drop as the sum of the volt drop for each section using the current drawn at each point.

........here's a question related to the OP, I know these calcs are all assuming 230v on a single phase domestic supply.

What if the origin happens to provide an actual V of 254Volts?



Always botherd me that one. The pount being you could use a much smaller CSA , get a huge drop and still have well over 230V.

Im pre empting one response which might be you must assume the actual V may very well mot be 254 tomorrow and 230 is guaranteed but thats cobblers- we've been supplied much less for weekd.

P.s wouls an undervoltage knacker SELV lighting? 4 or ours (the tranny) has popped and house srm while this under voltage happened...?


(
sorry OP slightly off topic)

The nominal voltage supplied is defined as 230V so that is what we have to calculate the values from. Doesn't make a lot of sense, I can't actually remember ever measuring a voltage less than 235V in practice.

Wouldn't have thought undervoltage would cause damage to a ELV transformer, but they are a bit temperamental.

SPARTYKUS said:

........here's a question related to the OP, I know these calcs are all assuming 230v on a single phase domestic supply.

What if the origin happens to provide an actual V of 254Volts?





Always botherd me that one. The pount being you could use a much smaller CSA , get a huge drop and still have well over 230V.

Im pre empting one response which might be you must assume the actual V may very well mot be 254 tomorrow and 230 is guaranteed but thats cobblers- we've been supplied much less for weekd.

P.s wouls an undervoltage knacker SELV lighting? 4 or ours (the tranny) has popped and house srm while this under voltage happened...?


(
sorry OP slightly off topic)

Click to expand...

Apologies for my appalling spelling in that post.

Many thanks for the reply Richard,

I didn't quite understand the last paragraph. I was planing on just running the cable in a daisy chain. Distance between each light is between 5m and 50m and the lights vary between 70w and 150w. Are you saying to run more than 1 cable instead of a daisy chain. Is it possible to use 2.5mm SWA as I have about 200m of it already or will this overload it?

Thanks.

Richard Burns said:

Basic volt drop for the total circuit (assuming no other loads) is
Vd (mV/A/m) *A (amps) *m (metres) / 1000
In this case Vdtot = Vd*4.34*300/1000 = Vd*1.302
Where Vd is the voltage drop per ampere per metre for the size of cable.
Lighting must be less than 6.9V (3%) drop.
rearranging you can then state max volt drop (mV/A/m) for a cable is 6.9 *1000 / (4.3*300) = 5.35 so 10mm2 SWA (4.4mV/A/m)

If this is too large a csa for the cable then you can divide the circuit up into sections for the lengths of cable between each light, and calculate the total volt drop as the sum of the volt drop for each section using the current drawn at each point.

Click to expand...

Andyg7 said:

Many thanks for the reply Richard,

I didn't quite understand the last paragraph. I was planing on just running the cable in a daisy chain. Distance between each light is between 5m and 50m and the lights vary between 70w and 150w. Are you saying to run more than 1 cable instead of a daisy chain. Is it possible to use 2.5mm SWA as I have about 200m of it already or will this overload it?

Thanks.

Click to expand...

From the voltage drop calculation I have done which is the worst case scenario you would need to use 10mm2 SWA to comply with the voltage drop requirements, as this is a very long run. Obviously the current carrying capacity of the cable would be way over the top and so the cable would not be fully loaded which would give you a slight advantage and may allow a 6mm cable (I have not done this calculation so this is an idea not a recommendation), however this is a very complex calculation to allow for the reduced load.

I attach a picture below to try to explain what I mean about the break down of the circuit, this does not mean running extra cables just calculating each section of the circuit individually and then summing.
The last section of cable with just one light on it will only carry the current for that one light so the volt drop will be much smaller on that section, as you move closer to the CU the current will increase, as you have more lights on the cable and so voltage drop will increase.
I do not know the spacing and wattage at each light but assuming a ten 100W light string only the run to the first light will carry the full current, dropping down thereafter at each light


B to K are the distances between each light, the values of the current carried are shown in the calculation.
The Vd for 2.5mm2 SWA is 18mV/A/m
It is unlikely that you would be able to comply with the volt drop for the whole run with 2.5mm SWA as the total basic volt drop would be 23.4V, however you may be able to start the first 100m with a thicker cable and reduce to the 2.5mm at the end.

Work out the distances between each light and the total current carried in each section (I=P/V), if you are using different csa cables then you would need to do the complete calculation for each section and not use the shortened form I show in the diagram.
i.e. Vdsection = Vd * A * m /1000 for each section and sum the results, this should give you a much lower volt drop.

Richard your a star,Thank you for the detailed explanation. I now understand how the calculations work so thanks again. I will do exactly that and use a 10mm cable for at least the first hundred meters or so.Just out of interst, with a voltage drop of 23.4v as you clearly explained, what are the concequenses of using 2.5mm cable? i.e. would some lights appear dim, would cable overheat or would more power be consumed to compensate? Sorry about the quiz but I am interested.Regards, Andy

Mainly this is the requirements of the regulations limiting the maximum volt drop to 3% for lighting and 5% for power. (Though you could increase these by 0.5% for a 300m run)
The problems can be that equipment may fail to work as intended at the reduced voltage; the cable will not overheat as it will be running well below its maximum capacity.
More power will be consumed with the additional resistance of the cable but this will not be significant.

nice work richard ... so say you getting 3v as vd at say fitting g and its 200m on 6mm cable, how do you incorparate the sum when you change cable size.. do you add that on to that start of the next sum???? sorry if you have typed this clear as day .. you know me...

will pull more current , therefore pushing your design calcs out therefore disrution and faulty supply to be... i think

i=p/u said:

nice work richard ... so say you getting 3v as vd at say fitting g and its 200m on 6mm cable, how do you incorparate the sum when you change cable size.. do you add that on to that start of the next sum???? sorry if you have typed this clear as day .. you know me...

Click to expand...

OK just before I go to sleep!
The attached diagram shows the calculation in full using some assumed values for the csa used and the distance between each light. These values are shown below each letter.
The calculation is then worked out in full and then I have calculated (but not written out) the effect of reducing the csa for two sections from 4mm to 2.5mm.

Hope it makes sense!!

If you conduct the calculation correctly you will know from which point(s) you can reduce cable size, in fact you can size each section as required if you so wish. This string calculation is basically the only calculation that will give you true VD for a string of lights on a single radial circuit, or split phase radials, such as ''street/road'' lighting layouts...

It all depends what type of lighting you have on the end.

If using standard GLS lamps, then you probably will not notice anything different, they will be dimmer.
If you use fluorescent bulbs, probably start getting issues.
LEDs will probably fine as most are running on a buck converter, so whatever voltage you stick in the front you will get the voltage you need out of the back.

The cable will not be stressed by volt drop at all, but do not forget to rate cable for current (A) use. Can't be arsed to calculate that precisely, but think you have miles spare.

Volt drop is for regs compliance, if you do not care about regs compliance the 2.5 should work fine.