It's not your day, is it...At the risk of anotherbut I've been waiting for someone else to say it.
No.
So 1A along 1m will drop 11mV.
R = V/I
.011/1 = _______ ?
Err.....
12v lights
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R = V/I
.011/1 = _______ ?
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That sounds about right. However, as I'm sure you really understand, Stoday's point is that BS7671 regards that (75%, or whatever, less output than at the 'normal/average' supply voltage {240V}) as acceptable for an incandescent lamp run off the mains - so it is not reasonable to consider the same percentage (relative to what it would be at a 'normal' 12V) of light output acceptable for a 12V lamp?
However, whilst I agree with that concept, I think Stoday is using the wrong resistivity/VD figure for his calcs - such that the 'VD' on his 4mm² ring would be a lot more than 12.8%.
Kind Regards, John.
4.6mΩ is the resistance of a 1m long single piece of 4mm² copper wire at 20°C.
Which is a good point - tabulated VD figures are for conductors at 70°C, but since this 4mm² cable will be carrying at most 10A if it's a radial, less if it's a ring, it won't be running at 70°C.
My honest opinion is that I really can't be @rsed right now to work out how hot it would be.
So VD will be less than 11mV/A/m.
Sadly though, what Stoday's calculations gain on the swings they lose on the roundabouts, because a single piece of 4mm² copper wire is not much use - you need 2 to tango, so at 20°C you're looking at a resistance of 9.2mΩ/m.
But even then we aren't out of the woods...
5A along 24m of cable with a VD of 4.6mV/A/m drops _______V ?
Should have been mΩ, but what's wrong with the rest of my post?
It's not 4.6 it's 11 That was my point.
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Ah, I see.
Maybe not - but I've never seen a VD discussion here where people didn't use the BS7671-tabulated 70 degree VD figures, even though very few of the cables in question will ever be anywhere near 70 degrees.
Anyway, as you go on to say, even if one uses the 20 degree figure of 9.2mΩ/m for two conductors, Stoday still got his sums wrong.
Kind Regards, John.
When you wrote:
... it appeared that you were saying that 4.6mΩ per metre was the same as 11mV/m/A.
Kind Regards, John.
No, I was correcting Stoday (albeit with the typo) and stating what it actually is.
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Never mind all of that. Will the SELV luminaires take off though?
Well he has a strange mix of values there.
Normal can't be both 230V and 240V....
And it doesn't work like that anyway.
It is true that you're allowed to drop 3% of 230V, i.e. 6.9V, and it is true that the supply itself might drop to 216.2V, and that therefore the lamp has to make do with 209.3V.
But that does not mean that when the supply voltage is 240V your installation is allowed to drop 30.7V, i.e. 12.8% of 240V, just because that still gets you 209.3V at the lamp.
Permissible voltage drops are % values, not absolute ones.
And I should point out that all I've really said about an acceptable VD is that I didn't think 4mm² was large enough and to ask the OP why he thought that a 16% drop was OK.
If we take his scenario of a 48m 4mm² ring at 20°C with all the lights in the middle the voltage drop would indeed be 1.1V
(0.0092 x 10 x 48)/4
But...
You know how in exams they always tell you to show your calculations so that if you make a mistake they can still tell that you understand it?
I think this might have cost Stoday marks:
4.6mV/A/m is wrong, and if you use that figure you don't get 1.11V....
I now realise that (and I now understand where the 4.6 came from), but that wasn't how I read your words. Anyway, we now all understand one another!
Kind Regards, John.
I agree with all that, but I think we both understand what Stoday was probably trying to say, even if his words were less that ideal...
... to the best of my knowledge, every UK incandescent lamp I've seen has been 'rated' at 240V. As you say, the regs consider it acceptable for such a lamp to be supplied with 209.3V - which is 30.7V (12.8%) less than its rated voltage - which, as you said before, probably means about 75% of output it would have at its rated voltage.
12V lamps are rated at 12V, so I think one can reasonably apply the same criteria, and accept a situation in which such a lamp receives 12.8% (1.54V) less than it's rated voltage (and, again, probably gives out about 75% of the light that it would at 12V).
It's not very surprising that Stoday's 4mm² ring with load all at the centre works out 'OK' (VD<12.8%) by those criteria (even if the 'working' he presented was wrong), since that's equivalent to an 8mm² radial.
Kind Regards, John.
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There seems to be a general thought that he is using a battery! This makes it more of a problem. If for example the small down lighters are used the 12 volt type used in houses are different to the 12 volt type used in caravans because the battery in a caravan may be on charge or be discharging so the voltage in real terms may vary from 12 volt to 14.8 volt.
So not only does one need to know how the lights are positioned but also what is the lamp type and supply voltage.
I used a wind generator to power lights in the Falklands and it was made from an old wagon alternator. My problem was excitation as when using a battery the rotor had to be in parallel to the output and could draw more current for excitation than the unit produced. So the rotor was rewound and so was the stator and the rotor was in series with the output and was feed directly to the bulbs. A large ex motorcycle zener diode stopped the voltage raising above the value which would blow the bulbs. There was no battery and of course no wind meant no lights but that was rare on the Falklands.
Normally wind generators use permanent magnets for the field and old hub dynamos (OK really alternators) work well. However today it would seem odd to use a wind generator with tungsten lamps they use far too much power for the light given especially in a garden where one does not need the heat. I would guess a 20W LED lamp would have a large voltage range and may well work with voltages as low as 9 volt. I have a radio (FT290R) designed to work from batteries and this will work from 9 volt to 16 volt without a problem. Discharge lamps may cause a problem but again the standard 8W and 16W units designed for a caravan use a switched mode inverter to get the 80 volts required and will again run with a wide voltage range.
So really until "seabreeze1" returns and says exactly what he is using and how it is to be run it is rather pointless making up spread sheets to work out the volt drop when we have so little information.
So not only does one need to know how the lights are positioned but also what is the lamp type and supply voltage.
I used a wind generator to power lights in the Falklands and it was made from an old wagon alternator. My problem was excitation as when using a battery the rotor had to be in parallel to the output and could draw more current for excitation than the unit produced. So the rotor was rewound and so was the stator and the rotor was in series with the output and was feed directly to the bulbs. A large ex motorcycle zener diode stopped the voltage raising above the value which would blow the bulbs. There was no battery and of course no wind meant no lights but that was rare on the Falklands.
Normally wind generators use permanent magnets for the field and old hub dynamos (OK really alternators) work well. However today it would seem odd to use a wind generator with tungsten lamps they use far too much power for the light given especially in a garden where one does not need the heat. I would guess a 20W LED lamp would have a large voltage range and may well work with voltages as low as 9 volt. I have a radio (FT290R) designed to work from batteries and this will work from 9 volt to 16 volt without a problem. Discharge lamps may cause a problem but again the standard 8W and 16W units designed for a caravan use a switched mode inverter to get the 80 volts required and will again run with a wide voltage range.
So really until "seabreeze1" returns and says exactly what he is using and how it is to be run it is rather pointless making up spread sheets to work out the volt drop when we have so little information.
Sorry — I should have been more explicit. I didn’t want to teach grandma how to suck eggs.
The cables are not run at their maximum current carrying capacity, far from it. So the table 4D2B in BS7671, which gives the volt drop of a 2-conductor 4mm² cable as 11mV/A/m is inappropriate. The resistance of a 4mm² copper conductor at 20°C is 4.6mΩ/m, so I’ve used this figure. Of course, there are two conductors per cable so the volt drop per m of cable is 9.2mV.
So volt drop = 5A X 24m X 4.6mΩ/m X 2 conductors / 1000 = 1.1V
Ericmark makes some useful comments about his experience with wind generators in the Falklands. I assumed that the battery in the OP’s installation would be connected to the generator and the lights through some sort of controller to a modern design rather than “an ex motor cycle Zener”.
The OP asks if there is any way of reducing the cable size. Ericmark has put his finger on the answer: use LED lighting.
Indeed. As you will have seen, we did eventually work this out ourselves in your absence. However, the approach you've suggested does raise a very interesting question - since (reasonable though it sounds) it's not an approach I've seen being used ....
Lighting circuits in domestic installations invariably run their cables at far less than their maximum current-carrying capacity. With all CFL's, the load is often going to be no more than 10% of so of CCC, and even traditional incandescent installations probably rarely more than 25% or so. Even if one assumes that a 5A/6A lighting circuit is loaded right up to the In of its OPD, we're still in the 35%-50% (of CCC) range with 1mm² cable and maybe 25%-30% with 1.5mm².
However, when it comes to doing VD calculations, people (other than you, it seems) almost invaraibly seem to look to Table 4D2B for their resistivity/VD figures. As you have illustrated, the difference between 20 degrees and 70 degrees is not trivial (9.2 mΩ/m vs. 11m Ω/m for two 4mm² conductor, a difference of 16-20% {depending on which way you look at it}), so can lead to quite significantly different VD calculations.
I would certainly question to appropriateness of designing on the basis of 20 degrees - even in the UK, and even with no current flowing in the cable, ambient temperatures of 30 degrees or more are far from impossible. However, it would seem that, now you have put the thought into my mind, it would nearly always be reasonable to calculate on the basis of a temperature considerably less than 70 degrees. Do people actually do this, since I can't say I've ever seen it done? It might be of particularly value to people who have long outdoor lighting circuits.
Kind Regards, John.
The idea of using series connections and a higher voltage is a good way to reduce current and hence cable diameter.
Put a suitable zener diode across each lamp so that in the event of a lamp buring out the series chain to other lamps is maintained.
Instead of using an invertor to convert 12 volts to a higher voltage use two or more 12 volt batteries in series to get the voltage. This is not going to be much more expensive as the batteries can be smaller capacity. Four batteries would give 48 volts which is regarded as ELV and therefore safe for outdoor use
The 12 volt generator is connected across one battery at a time. Charge battery A for 15 minutes, then move it to battery B and charge for 15 minutesm then C then D.
Four relays and a simple timer circuit will do the connections automatically.
You could then also have a fifth battery charged for other equipment.
Put a suitable zener diode across each lamp so that in the event of a lamp buring out the series chain to other lamps is maintained.
Instead of using an invertor to convert 12 volts to a higher voltage use two or more 12 volt batteries in series to get the voltage. This is not going to be much more expensive as the batteries can be smaller capacity. Four batteries would give 48 volts which is regarded as ELV and therefore safe for outdoor use
The 12 volt generator is connected across one battery at a time. Charge battery A for 15 minutes, then move it to battery B and charge for 15 minutesm then C then D.
Four relays and a simple timer circuit will do the connections automatically.
You could then also have a fifth battery charged for other equipment.
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