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Thread: Profi 3 kw heater
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19-01-2021, 05:53 PM #21
I'm not sure where you got your Ohm's law calculation from but it is wrong as the higher the voltage the lower the ampage for any given wattage.
The wattage is a constant as it's basically what the element in the heater actually draws to make it work.
Wattage = voltage x ampage
Or if you work it back the other way
Ampage = wattage divided by voltage
So if you have a supply voltage of 250 volts then 3 kw is only drawing 12 amps but if your supply voltage is as low as 215 volts then 3 kw is drawing just under 14 amps.
From my experience I've seen supply voltages as high as 260 volts and as low as 200 volts as it related to how far away from the substation you are and other influences like industrial machinery in the area starting up and shutting down will cause the mains voltage to rise and fall a fair bit even though there is supposed to be maximum and minimum supply voltage limits. That's why surge protection devices are recommended now.
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19-01-2021, 06:31 PM #22
Hi Frimley Koi Keeper,
It looks as though your calculations are based on a fixed power draw of 3kW - the given wattage you mention. If the heaters are fixed wattage then you are correct, higher V means less A and my assumption is wrong.
I worked it out differently using the principle that the heating element has a relatively fixed resistance when operating within its normal parameters.
So for a set resistance, higher voltage will result in higher current as I calculated using (R=V/I Ohm's Law equation).
I used the resistance principle of the old type filament light bulb. If on the bench you hook up a simple bulb circuit (same principle as a heater element) with an adjustable power supply and an Ammeter in series, as you raise the voltage the ammeter shows more current. Higher voltage and higher current means more Watts as in the power equation you have used.
I do know that resistance can vary with temperature (thermistor) and the reason bulbs would tend to blow when first turned on (lower resistance, so more current until warmed up), but I would have thought the difference in the element resistance of a submerged heater would be minimal with the voltage ranges mentioned. However, I could be wrong about that and am happy to be corrected if that's the case and if so aplogise for any misleading info.
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Frimley Koi keeper Thanked / Liked this Post
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19-01-2021, 06:43 PM #23
Sorry Ukzero I did think along those lines working on resistance changes with temperatures so we were both technically correct.
Great explanation BTW
I was going on the basic Ohm's law and you were going on the advanced one.
I knew I should have done the technicians course at college instead of stopped at the approved course
Bloody electronics lol I like a switch to be a switch and a cable to be a cable all this printed circuit board etc and ECU etc are far to technical for their own good
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Ukzero Thanked / Liked this Post
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19-01-2021, 07:13 PM #24
Totally off topic
Do these calculations explain something that is puzzling me?? I have an old enduro motorcycle, runs direct AC lighting. Original lighting system was 6v ( but unregulated supply of varying voltage above 12v dependent on rpm) regulated solely by using 6v bulbs of a certain wattage.
Although I have it registered for road use it was never really intended for such and didn't have a brake light, I fitted one but there was not enough power to light the lights and brake lamp at the same time.
So I had something niggling my brain from years back (just a vague idea sort of thing) and decided to fit a 12v a.c regulator and bulbs of the same wattage but 12v. That done there was sufficient power to operate all three lights at once..
Anyway, I know it works, I kind of knew it would work before I did it based on some distant past knowledge / experience .. but I don't really understand why??
Does that ampage = wattage / voltage equation explain this? Sorry for being thick lol and for going completely away from ponding
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19-01-2021, 07:23 PM #25
Motor vehicle electrics are a bit of a different thing but the basic rules or equations apply I believe? What you may be getting is the effects of the alternator which is generation the electricity to run the lights etc and you often find as you rev the engine the lights gets brighter to a degree. This tends to happen more if the battery is getting past its best as the alternator has to run the electrics and charge the battery assuming it has a battery that is?
So no I don't think you are thick at all.
Maybe Ukzero could give you a better in depth answer here? He seems very clued up going by his reply earlier.
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arceye Thanked / Liked this Post
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19-01-2021, 07:29 PM #26
Cheers Frim, yeah more power the more you rev, its a battery less system originally regulated to 6v solely by the bulbs wattage, i.e if one bulb blows the other overpowers and blows. My thinking was that though the magneto provides in excess of 12v unregulated it is low on amperage???? so do 12v bulbs demand less amps than 6v bulbs at the same wattage?
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Frimley Koi keeper Thanked / Liked this Post
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19-01-2021, 07:33 PM #27
Arceye.
Assuming old gear I would expect the alternator output voltage to be proportional to rpm.
The power output (Joules/sec) would also go up proportionally.
This is off the top of the head theorising so again, open to scrutiny......
A 12V bulb of 12W would draw 1 Amp.
A 6V bulb of 12W would draw 2 Amp.
So on the face of it the power equation is working for you, but the energy being used per second is identical in both cases.
I am guessing that the bulbs may have been behaving like this:
6V bulb requires more current to get hot/bright if it will be consuming the same power (Wattage as you say) and the filament will be generating as much heat in total, but more at lower IR wavelength and not all visible.
12V bulb of same power requires less current to get hot/bright and although it will generate the same heat output, more will be in the visible spectrum. The difference will be filament resistance and cross-sectional area and /or length.
The back of our log burner doesn't glow at all, but radiates a lot of infra red and far more power (heat energy per second) than a bulb
Hope this helped and/or makes sense.Last edited by Ukzero; 19-01-2021 at 07:36 PM.
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19-01-2021, 07:40 PM #28
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19-01-2021, 07:46 PM #29
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19-01-2021, 07:47 PM #30
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19-01-2021, 07:48 PM #31Freddyboy the legend
"we are water keepers first"
Johnathan
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19-01-2021, 07:51 PM #32
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19-01-2021, 08:17 PM #33
I would say higher resistance.
Using the example:
6V at 12W 12W=6Vx2A R=V/I, so 6/2=3 Ohms
12V at 12W 12W=12Vx1A R=V/I. so 12/1=12 Ohms
As I said it will be down to the filament thickness, etc. A higher resistance filament carrying 1A can still glow very bright if it is thin enough.
See it as too many electrons "squeezing down a very thin pipe".
The 6V bulbs will have thicker filaments, so won't get to as high a temperature and glow less brightly with the same power, even though carrying twice the current. They will give out the same total heat as still 12W, just not at such a high temperature and so largely invisible.
See this as twice as many electrons as for 12V, but "squeezing down a pipe 4x wider (3 Ohms rather than 12 Ohms), so less proportional squeeze", so a lower temperature and less visible light as a result.
As an aside, in the USA their mains voltage is about half of ours, so the supply to appliances carries twice the current and needs correspondingly thicker wire.
So if we are getting near to 13A with a heater, they will be nearer 26A for the same output!Last edited by Ukzero; 19-01-2021 at 08:21 PM.
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19-01-2021, 08:38 PM #34
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19-01-2021, 08:40 PM #35
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Frimley Koi keeper, Ukzero Thanked / Liked this Post
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19-01-2021, 08:42 PM #36
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19-01-2021, 09:05 PM #37
If Fred was still here, bless him, he would be scratching his napper with all this technical stuff and would probably put up a photo of his waterproof socket with their lids open.
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19-01-2021, 09:08 PM #38Freddyboy the legend
"we are water keepers first"
Johnathan
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19-01-2021, 09:12 PM #39
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19-01-2021, 09:23 PM #40Freddyboy the legend
"we are water keepers first"
Johnathan
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