Relationship between electric arc and heat?

[bidouille23]

Bonjour,

I have a question for you, and if you can give me an answer I would love to .

I would like to know how to translate the thermal power supplied by an arc as a function of the current and intensity supplied to the primary of the THT?

Thank you in advance


fred

[dedeleco]

Quite poorly asked question so difficult to answer !!!

Indeed, the arc does not have a constant current and voltage but with very high frequencies up to those of a cell phone (see Hertz in 1886 with its spark gap! http://fr.wikipedia.org/wiki/Heinrich_Rudolf_Hertz ) and therefore it is necessary to integrate the current product by voltage with steps of 0,1 nanoseconds to measure the total power !!
This is possible with current but expensive ultra-fast microprocessors !!
At the primary level one can hope that the useful frequencies are lower but it is necessary to check with a fast oscilloscope 300MHz at least, especially powerful arc !!

If we want to measure the total power we put everything in a calorimeter (well insulated box) and we look at the temperature T and the power necessary with a resistance alone in it to obtain the same T!

Otherwise, given the complexity of the discharges we risk making a mistake and finding crazy results of all kinds !!!

With a fairly fast oscilloscope we can try to evaluate, but the discharges from one to the other are not reproducible (generalized chaos) and therefore we have an unreliable estimate.

A simple method is to feed by a filter which lets pass only the very low frequencies (the arc will be probably modified) with self of several Henrys (large winding on large transformer iron core) followed by a capacitor (to be calculated ) and to measure these slowed-down currents and voltage with an ordinary oscilloscope and to integrate the product current by voltage with steps of 10 to 100microseconds following the observed curves. There are integrated circuits to do these calculations.
http://en.wikipedia.org/wiki/Bandpass_filter

Such a filter (cutoff around 100Hz) allowing only a few harmonics of 50Hz to pass through can make it possible to measure the power with an oscilloscope on few harmonics by integrating the observed I and V curves;

Check the consistency with the calorimetric measurement.

[elephant]

Del, doesn't it seem like you're cutting your hair in four? Probably well beyond the intention of the question asked?

A P = UXI approximation less losses in the cables and the transformer would it not already shed some light on our friend's lantern?

[dedeleco]

That he knows, the simplistic P = VxI each V and I supposed to be perfectly sinusoidal (only case of a perfect resistance), but given the harmonics and complexity of the superimposed discharges on the 50Hz, he goes right into the measurement error if discharges are powerful, the case of arcs !! And it is not only the cos (phi) !! which always assumes the perfect sine wave !!

This is also an explanation of a lot of false measurements with over-energy on complex, non-linear and chaotic phenomena !!
EDF measures integrals, because otherwise we could cheat the meter !!
A solution measure on the EDF meter !!
At least that's what gets paid !!

Must have watched on oscilloscope the reality of I and V to become more aware of errors !!

The filtering to have on the power supply only the first harmonics at 50Hz is the simplest way to integrate the arc discharges through the PI filter.
Without that errors guaranteed, even more on a digital voltmeter disturbed by high frequencies !!

[Arnaud M]

for an automobile spark plug there have already been many measurements made, and it is the heat supplied which is used to ignite the mixture.
You owe of course on Google the power absorbed at the primary and that restored by the arc.

[dedeleco]

And there is as much dissipated in the induction coil as there is in the spark from the spark plug. For a candle we do not worry about the energy balance too much.
The spark rises to 2000 ° C (very white color) transiently by breakdown of the air with a voltage greater than 30000V / cm over very short times a few microseconds so that the heat of the current in the very resistive air filament (plasma) does not have time to diffuse and remains in place for the time of the spark (0,1ms and less) over a diameter of 0,04mm, the length of heat diffusion in the air over 100micoseconds.
This is made visible by the white colored area of ​​the spark.
http://de.wikipedia.org/wiki/Temperatur ... %A4higkeit
You get millions of degrees with fine wires or lasers or lightning pouring a lot of energy into a tiny volume.
The explosive expansion of the superheated air in this tiny volume makes the boom afterwards.

Beware of overly simple energy measurements on the internet which can be wrong.

[bidouille23]

Good evening,

The Dedelco is a detailed explanation on a specific subject, hats off to the artist, only problem you lost me on the way (well almost;)), indeed you have gone a little too far in the explanation, I don't 'did not expect so much;) but I admit that the question is not necessarily very clear (I tried to make it understandable I lost;)).

I explain a bit more where I wanted to come from but I think the answer is included in the Last post from DEdelco:

Explanation of the question:

At the moment I hack THT to make arcs in order to make an ignition.

So by doing some testing (in safety I reassure you I am not suicidal even if I have already taken good potatoes of 40 Kv with a horse fence, the cow fence no longer stopped the horses the great means were put on and it worked well lol),

In short, during these tests, I put a metal part between the two phases and an arc is therefore created on the part (one on each normal side).
I let thirty seconds and, I touch the metal part which was very hot, in fact hotter than what I had imagined, for barely 1.5A under 12 v DC at the primary circuit, and indeed there is the transistor which also heats up during the transformation therefore loss effectively and good in addition, the resistance too (ceramic 5 w for that to hold;) lol).

Naturally I said to myself (not knowing much about the properties of THT) that there might be a transformation into heat which could be economically advantageous?

IN fact I imagined in this case to make a kind of boiler with as heating core two arcs or one also which strikes a metal part which is responsible for dissipating the heat thus created. boiler but can be that for small kind instantaneous water heater ??

Does this clear up the question?


If I do not connect an EDF meter before the power supply. I would have its consumption (and therefore that of the total circuit).

This power supply. therefore sends 12V dc and the circuit requires 1.5 A = 18 W

And depending on the time and the mass and size of the heated room and the place where it is (cooled, confined etc.), if I measure the temperature, I get the transformation balance to a few things, but this 'is precisely not to do that that I asked the question;). In case there would be a realistic approximation of the thing what;). That in fact it may be the only solution to be sure of something actually seen the complexity of calculation when talking about THT as Dedelco said well (well it's that he lost me the day was hard and my brain has trouble keeping up this sorry evening;))

I hope I have been clear enough in my explanations, if anyone has an opinion please;).

And thank you again to everyone for the time you have already taken to respond

[dedeleco]

Bidouille23 may have found a source of over-energy !!!!

Electric fences are designed not to be dangerous, even with 40KV !!
Indeed they are limited in current
What counts is the average current which must not exceed the mA, even with 300000V as one obtains with the sparks of a Thermolactyl knitting (more than 10cm long) !! !!
But your HV generator is much more dangerous by outputting more than the mA, and therefore you risk dying at the slightest mistake of the same type !!
Unless your HV reaches 18000V with a current of maximum mA to make 18W = 18000Vx0,001A !!
but if the short-circuited HT causes an increase in the current in the power transistor on 12V, you can go up to well over 100W, to be measured !!

Otherwise the slightest ordinary phenomenon is very complex and once the art of the scientist was to simplify it to make it clear and precise !!
Currently with the chaos and the progress he dares to attack the appalling complex like an electric shock and the plasmas, as for ITER, a chaos which we hope to master in 50 years !!.

So first in my basic explanations where hack 23 was lost ????

On the other hand the piece of metal was in what metal, copper, steel, of what weight, with what increase in temperature observed (lukewarm, hot, burning hands?) Because then we can calculate the energy necessary to heat it as well with heat capacity and see if it is less or more than 18W on 30s or 540Joules.

It is possible that the current during this short intense juice is exceeded the 1,8A ??

Copper has a capacity of 0,382KJ / Kg ° C and steel or iron has a capacity of 0,465.
http://de.wikipedia.org/wiki/Temperatur ... %A4higkeit
So the half KJ of the discharge for 30s can heat 20g of copper metal 540J / 0,02x0,382 = 70 ° C to fix ideas ?????

If it was 1 kg of copper or steel there was over-energy and repeat the experiment and measure all the conditions with great care!


Does the discharge heat the air around it more than the metal?
Steel more resistive than copper should heat up more.
If the HV voltage is supplied by a circuit on 12V, by putting a good big capacitor on the 12V (10000microFarad), damping the fluctuations, we can measure the energy by the value of I and V DC during the discharge.


Formerly we lit powerfully with electric arcs between two coals and very strong intensities which heated more than lit !!
For powerful lasers we do high voltage and high intensity discharges !!

So these dumps are well known and very studied, even still for ITER !!

[bidouille23]

Hello, Dedelco after re-reading effectively the history of the low frequency filter is not that losing; lol I was on my knees yesterday it's clear .

So thank you for the answer full of irony on energy;).

I never spoke about over_energy;), it was out of curiosity and parse that in thermal I don't know much about it, really not much (I'm going to put myself in the burner but slowly calm the cat or the ;)).


Then thank you for calling her back. Dedeclo THERE IS DANGER OF DEATH (it's very serious, so I'm careful;), (I also modified the circuit yesterday, the arc starts with 8mm of acrt between the electrodes; ) and it's already hissing a lot more;), so even more aïlle aïlle you may be dead if you touch, but I do not touch and I use the switches set by me to be safe;)).

So thank you, I have plenty to worry about with the end of your explanations super thank you, in fact that's what I was looking for I realize, the CAPACITY of metals.

And beisn for info it was iron a small iron pliers;).

And there was no over energy and that's not what i was thinking , this is just to know if it would be profitable in comparison with a heating resistance for example as in an instantaneous water heater. it also heats up hard lol but it eats juice), so I imagine that if have kept the basic resistance it is not for nothing;).

I'll take it all over again;) the taffff.

see you more and again Thank you Dedelco.

have a nice week end

[bidouille23]

Good evening,

Lol and re lol, and be it who is the craziest;) in his ideas well it's not bibi;)

http://www.worldlingo.com/ma/enwiki/fr/ ... nstruction

there are micro-ovens for dentists and laboratories, and the interior can go up to 1800 ° c, good on the other hand it does not speak of the powers of the small ones, just of that of the big ones for the steel industry and the ouhahahha it is crazy stuff 44.000 A

dixit "A modern mid-rated steel furnace would have a transformer rated at around 60.000.000 volt-amperes (60 MVA), with a secondary voltage between 400 and 900 volts and a secondary current above 44.000 amps."

Come on, it stings hard in your body, only once after you are picked up with the vacuum cleaner.

In short, the idea was not stupid cqfd

see you