With 6mm² you drop 4.9% by the first light, 10.4% by the penultimate...
12v lights
- Thread starter seabreeze1
- Start date
I suspect that approach (with lights on alternating rings) may not be the most efficient system, since it would presumably result in both rings ('unnecessarily'?) being roughly the same size.
I thought of all sorts of options but, as you say, it depends totally on the positions. If they are roughly in a straight line, with the battery necessarily at one end, then a ring obviously offers no significant advantage over simply 'doubling up' the cable (or using one of twice the CSA) on a radial circuit.
Kind Regards, John.
But each carrying only half the current, and being rings having a lower intrinsic VD anyway.
And you want everything to be as similar as possible so that all the lamps are the same colour...
I was using the calculator in your link but - for first light with 6mm²
7.3mV x 10A x 2m / 1000 = 0.146V drop. 1.2%
Is that not right?
I shall not do the penultimate light as we're now not sure where it is
No - the first light is 8m away, not 2m...
As already mentioned, it depends so much on the layout of the lights. Were they all 'in a straight line', then a radial with 'doubled up cable (or cable of double the CSA) would halve the VD for all the lamps, whereas the VD would reduce by less than half for all but the furthest lamp if one used a ring.
True, and that's quite a challenge. I suppose one could get them all nearly identical by having 3 ('unnecessarily long') rings.
Kind Regards, John.
Time to add more columns to your spreadsheet and calculate VD given lower resistances and lower currents due to parallel paths...
Or 6 radials, all necessarily the same length.
I remember years ago somebody posting on this forum about his wonderful idea of house installations being 12V for safety, with step-up transformers securely tucked away inside appliances. He just could not grasp the idea that you'd need cables as thick as your arm for cookers and showers....
That didn't come from a spreadsheet, but my head! Maybe I need to play with a spreadsheet - since what I wrote seems to make intuitive senseTime to add more columns to your spreadsheet and calculate VD given lower resistances and lower currents due to parallel paths...
Not 'necessarily the same length', if one carefully selects cables of appropriate CSA for each of them!
Kind Regards, John.
True.
I have calculated the appropriate volt drop and yes, you can use 4mm cable provided you feed the lights in a ring, the two ends being connected to the battery.
BAS is wrong. He uses the wrong volt drop %. His calculations are only appropriate when the electricity supply is from the public distribution network.
The correct calculation would give the same worst case (lowest) lamp voltage (as a % of normal) as you would get from a normal (230V) lighting circuit to BS 7671. That is, a volt drop from the meter of 3% of the nominal 230V. So the lamp gets 230V less 6% supply tolerance, less 3% = 209.3V. The normal voltage is (say) 240V, hence as a % of normal the lamp gets 87.2%.
Now in the OP's case, the supply is not from the mains but from a floating battery. That's pretty near constant voltage. So he could drop 12.8% of 12V and the lamp would still get the same % of normal as the mains version.
Worst case is with all the OP's lamps in the middle of the ring, with a worst case "circumference" of 48m, which then has to supply 5A per 24m leg. Volt drop with 4mm cable is 4.6mΩ/m/A =1.11V, well below the 1.5V threshold.
That's the worst case. The geometry of the actual case could mean that 2.5mm cable might be used.
When you write 4.6mΩ/m/A, I presume you mean 4.6mΩ/m aka 4.6mV/m/A - but where does your figure of 4.6 come from? I've been working with 11mΩ/m for 4mm² 2-core cable.
As for the rest of what you say, I agree with the concept, if not your numbers. As I said, many people might regard the 1.9V VD I calculated for a 4mm² 'straight' radial as acceptable.
Kind Regards, John.
No I'm not.
No I don't.
No they aren't - they are appropriate for any lamp which is subject to the laws of physics.
An incandescent lamp at 87.2% of its design voltage will be producing about 75% of its design output.
And if the voltages at each lamp differ by too much you will perceive differences in brightness and colour.
At the risk of another
but I've been waiting for someone else to say it.However, Isn't 4.6Ω the resistance per metre for 4mm².
The Voltage drop is 11mV /m/A as John has been using.
Therefore the result of the calculation should be 2.39 (11/4.6) x 1.11V i.e. 2.65V - well above 1.5V.
Well, here goes -
Oh Dear! EFLI, I'm starting to feel sorry for you this weekAt the risk of another
However, Isn't 4.6Ω the resistance per metre for 4mm².
The Voltage drop is 11mV /m/A as John has been using.
. For a start, when you write 4.6Ω, I presume you mean 4.6mΩ. However, more important ..... as I wrote earlier, mΩ per metre is the same thing as mV/m/A (clue: mΩ=mV/A),so 4.6 mΩ/m is the same thing as 4.6mV/m/A - whereas you and I both thing it should be 11mV/m/A - hence my asking Stoday where his 4.6 came from.
Kind Regards, John.
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