High Voltage Forum

General electronics => Electronic Circuits => Topic started by: Solhi on November 09, 2022, 12:11:04 PM

Title: Induction cooker without electronics
Post by: Solhi on November 09, 2022, 12:11:04 PM
Hi all,
What made me join this forum was a post* by Downunder35M I stumbled over in my search engine, when I was looking for confirmation that using caps to resonate a coil was unnecessary.
* https://highvoltageforum.net/index.php?topic=437.msg2709#msg2709
As you all know has any conductor a natural self-resonance.
According me, the only thing you have to do is to feed it a frequency preferably close to it's resonance or a part of it to avoid harmonics or opposing the natural frequency.
I have to disclose that I'm extremely poor in using all the available formulas (just seem not to be able to use them), and not having any equipment (living in the bush on a tropical island) So all my assumptions are pure theoretical and based on bits and pieces what  others have discovered and tested.
Meaning that I have have used a lifetime to read me up and try to understand giants like Tesla and the maybe as giant underwood of all the people using their time to experiment, explain and even improve these theories.

Now I wish to test my theory about an alternative induction cooker, and hope you guys can learn me a thing or two and correct me if I seem to be way off. In advance, excuse me for when it seems I'm abusing your time.

Now my induction cooker:
I came in to this HV, HF world through reading Patrick Kelly and from there Don Smith this of course after all Nikola Tesla left behind. So my theory is based on this.

An induction cooker needs a working coil that  has a strong enough alternating magnetic field at high enough frequency to induce powerful Eddy current in the ferromagnetic cookware.

Looking around I'm seeing they mainly use frequencies in the 20-80 KHz range.
My first question is now: does it need to? Would frequencies in the 100 MHz not be as good? This is of course the frequency of an working coil in self resonance with standing waves, giving maximum current and magnetic flux.

This cooker is ment to be used local here with a square 30*30 cm 20 liter stainless steel vessel to cook rice.
This requires an output of 6-8 Kw according my estimate.

My thought solution is a square pancake coil made of 2 mm copper wire. Since,  as far as I know, current at these high frequencies travel outside the wire, I do not need to worry about heat.
Such a coil would approximately need using Coil32 - Multilayer coil on a rectangular former

Input data:
Number of turns N = 54.0
Width of the former a = 100.0 mm
Height of the former b = 100.0 mm
Winding length l = 2.0 mm
Wire diameter d = 2.0 mm
Wire diameter with insulation k = 5.0 mm

Result:
Inductance L = 1381.958 µH
Winding thickness c = 270.0 mm
DC resistance of the coil Rdc = 0.282 Ohm
Wire length without leads lw = 51.3 m
Weight of wire m = 1444.027 g
Number of layers Nl = 54

So wire length 51.3 + leads = 51.5 m
Would give a self resonance of
1/4 wavelength / wirelength = Mhz
75,29/ 51.5 = 1.46 Mhz

Maybe the gap between the turns could be a bit bigger?

Next feeding this coil in the same freq. range
and enough current would suffice with 80V and 80 A

Which I think to get from an air coil transformer 24=>80V with a split secondary delivering the same amount of volts and amperes.

I hope my explanation was clear and not the least obtainable.
Title: Re: Induction cooker without electronics
Post by: klugesmith on November 09, 2022, 05:11:34 PM
Welcome, Solhi.

Frequency must be chosen to avoid interfering with wireless communication.
So induction heaters/cookers (and microwave ovens and diathermy apparatus)
generally use ISM bands, which are scattered over many decades of the radio spectrum.

Your estimate of 6 to 8 kilowatts seems high for cooking rice.  Can you get away with 1 or 1.5 kW, available from ordinary household wall plugs, and wait a little longer to boil water in your 20 liter pot?

It's not clear how you will get power into a high frequency resonant circuit, without using electronics.
It was done for many decades without semiconductors, and I have tested a semiconductor-free microwave oven.
Title: Re: Induction cooker without electronics
Post by: Twospoons on November 09, 2022, 10:36:53 PM

Looking around I'm seeing they mainly use frequencies in the 20-80 KHz range.
My first question is now: does it need to? Would frequencies in the 100 MHz not be as good? This is of course the frequency of an working coil in self resonance with standing waves, giving maximum current and magnetic flux.


You can answer this with a simple thought experiment:
If the frequency goes to zero, then no energy is transferred - so we can infer there is a minimum frequency which will effectively transfer energy.
If the frequency becomes extremely high, then the skin depth in the receiving metal becomes extremely small - as a consequence the resistance to the circulating current becomes very high and virtually no current flows. So very little energy gets transferred.  From this we can infer there is a maximum  frequency to effectively transfer energy.

Between these extremes there will be an optimum working frequency, one which matches the impedance of the source (the work coil) to the impedance of the load (the cook pot) in order to get the best power transfer.  Industry has clearly decided this is in the 20kHz to 80kHz range. 
There are other considerations, of course - kW level drivers in the sub 100kHz region are much simpler than kW level drivers in the 100MHz range.  The coils are simpler to build too.  Radio frequency compliance has already been mentioned.

Since,  as far as I know, current at these high frequencies travel outside the wire, I do not need to worry about heat.

This is fundamentally incorrect - the current still travels in the wire, but only in the surface at high frequencies ( look up 'skin depth'). This makes self heating of the wire worse at high frequencies, not better.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 10, 2022, 09:01:31 AM

Welcome, Solhi.

Frequency must be chosen to avoid interfering with wireless communication.
So induction heaters/cookers (and microwave ovens and diathermy apparatus)
generally use ISM bands, which are scattered over many decades of the radio spectrum.

Thanks klugesmith
I was aware of that problem with RF interference and therefore included a aluminum pan/case under and round the coil. The cookware shields the top

Quote
Your estimate of 6 to 8 kilowatts seems high for cooking rice.  Can you get away with 1 or 1.5 kW, available from ordinary household wall plugs, and wait a little longer to boil water in your 20 liter pot?

The locals boil on gas and for a similar size pan (10 Kg rice) they use 40-50 minutes.

My calculation of 6-8 KW is based on formulas from:

https://www.batteryequivalents.com/how-long-does-it-take-for-water-to-boil.html

Which indicates a need of 6 KW to boil 15 liters of water in the same time or preferrable shorter.
I'm still of the opinion that the K watts are easy to fetch following Tesla.

Quote
It's not clear how you will get power into a high frequency resonant circuit, without using electronics.
It was done for many decades without semiconductors, and I have tested a semiconductor-free microwave oven.

As you already write it is done before.
I did write:"Which I think to get from an air coil transformer 24 => 80V with a split secondary delivering the same amount of volts and amperes."
But maybe I need to elaborate that?



If the frequency becomes extremely high, then the skin depth in the receiving metal becomes extremely small - as a consequence the resistance to the circulating current becomes very high and virtually no current flows. So very little energy gets transferred.  From this we can infer there is a maximum  frequency to effectively transfer energy.

Hmm, this past my radar, need to dive into that.

Quote
There are other considerations, of course - kW level drivers in the sub 100kHz region are much simpler than kW level drivers in the 100MHz range.  The coils are simpler to build too.

I agree if you are talking about electronic drivers. But using air core transformers it is just a matter of adjusting conductor length to hit the right frequency.

Quote
"Since,  as far as I know, current at these high frequencies travel outside the wire, I do not need to worry about heat."

This is fundamentally incorrect - the current still travels in the wire, but only in the surface at high frequencies ( look up 'skin depth'). This makes self heating of the wire worse at high frequencies, not better.

I will check this up once more, but according to what I remember it would be just the opposite.
So called "cold electricity"

Thanks so far guys, you gave me some challenge, I take it :)
Title: Re: Induction cooker without electronics
Post by: klugesmith on November 10, 2022, 05:49:36 PM
>> I did write:"Which I think to get from an air coil transformer 24 => 80V with a split secondary delivering the same amount of volts and amperes."  But maybe I need to elaborate that?

Yes please.  How will the air core transformer do anything useful when fed by DC or mains frequency AC, without some switching?  There is a limit to energy transferred with each opening or closing of the switch. In SGTC the spark gap is the switch.

>> I was aware of that problem with RF interference and therefore included a aluminum pan/case under and round the coil.

How many dB of attenuation are necessary?   The shielding in a household MWO can be enough to meet safety limits, and still leak enough to jam nearby WiFi without violating any emissions regulation.
What fraction of your induction heating power will go into your aluminum pan/case instead of the cookware?   

>> My calculation of 6-8 KW is based on formulas from:
https://www.batteryequivalents.com/how-long-does-it-take-for-water-to-boil.html

I agree with your heating time and power predictions.  That's a nice site you pointed to, but it does include some mistakes.  For example, regarding electric kettles: "On average, in order to boil 1 liter of water, 1500W kettle requires ~4 min and 2500W kettle requires 2-3 min. An electric kettle is not as an efficient heater as a microwave oven, since there is a large "parasite" mass to be heated, but there is no danger of superheating the water."  The writer overlooks the elephant in room: power loss in conversion from mains frequency power to microwaves.  When MWO is running, the hot exhaust air is carrying waste heat from the magnetron.
Title: Re: Induction cooker without electronics
Post by: petespaco on November 10, 2022, 06:13:40 PM
Quote
I will check this up once more, but according to what I remember it would be just the opposite.
So called "cold electricity"

My experience is limited to 1 to 2.5 KW ZVS induction heaters, but I can tell you from lengthy experience that the work coils DO heat up substantially,  These systems  (and the much larger ones) DO require that the work coils be made of copper tubing and that copper tubing must be water cooled.
I don't know why those HOBs (household induction cookers) manage to keep their work coils cool enough.
Title: Re: Induction cooker without electronics
Post by: Twospoons on November 10, 2022, 09:05:07 PM
Induction hobs typically use Litz wire coils so the useable copper cross-section is much larger than with copper pipe, and hence the resistance, and self-heating, is much lower.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 11, 2022, 05:26:32 AM

Yes please.  How will the air core transformer do anything useful when fed by DC or mains frequency AC, without some switching?  There is a limit to energy transferred with each opening or closing of the switch. In SGTC the spark gap is the switch.

The "switching" is created by the 24V AC feed to the primary coil at any frequency. The length of the primary is tuned to the secondary at self resonance which will overrule any other frequency.

My "textbook" you can download here:
http://www.free-energy-info.tuks.nl/PJKBook.html

Quote
The shielding in a household MWO can be enough to meet safety limits, and still leak enough to jam nearby WiFi without violating any emissions regulation.

To jam a frequency you need to broadcast the same frequency with a stronger signal. WiFi is in the 1.6 - 5 GHz. What can be interfered is AM shortwave radio. But I doubt the signal will be strong enough.

Quote
What fraction of your induction heating power will go into your aluminum pan/case instead of the cookware?

The aluminum shielding interferes with the radiation downwards, which is lost anyhow.

Quote
That's a nice site you pointed to, but it does include some mistakes. 

I was only using the calculation how much K-joule => Kw I needed to heat the water itself.
The examples how much time an utensil uses is maybe only based on that not including possible losses and can therefore be invalid.

The other posts referring to how traditional induction heaters are build have no relevance, since this proposal is trying to eliminate this.
The post from Twospoons regarding skin effect is very useful, since it forces me to dive into the cold electricity again.

BTW, interestingly the skin depth at 60 KHz is close to that at 1 MHz for steel. As you can see ca. 1 KHz to 1 MHz is pretty flat, so I doubt that your claim that "Industry has clearly decided this is in the 20kHz to 80kHz range" has any value.

https://incompliancemag.com/article/skin-effect-and-surface-currents/
Title: Re: Induction cooker without electronics
Post by: Twospoons on November 12, 2022, 12:42:54 AM
What you are seeing with the steel is the effect of complex permeability.  Remember the induction hobs are probably also try to cope with copper, stainless steel and aluminium cookware as best they can.

I looked at your "textbook". It has a chapter on chemtrails  ::) . Massive red flag.  There maybe be other parts of this ebook which have correct information, but you really need to get some other books too, and not just from other  'free energy' sites of dubious provenance.

Quote
The "switching" is created by the 24V AC feed to the primary coil at any frequency. The length of the primary is tuned to the secondary at self resonance which will overrule any other frequency.

So you want to feed 24VAC 50Hz or 60Hz into a coil with a self-resonant frequency ~100MHz, and you expect the coil to just start oscillating at 100MHz?  Because thats not going to happen.  Because physics.

Quote
To jam a frequency you need to broadcast the same frequency with a stronger signal.
No, all you need to do is raise the in-band noise floor enough to disrupt the data flow.  Doesn't have to be a stronger signal.  (Further reading: look up Shannon's Law, which relates bandwidth and noise to theoretical maximum information throughput).  Also most receivers include some form of down-conversion to an intermediate frequency, and you can sometimes create a jamming signal at that frequency too (depending on how well constructed the receiver is).
Title: Re: Induction cooker without electronics
Post by: Solhi on November 16, 2022, 11:43:49 AM

@Twospoons I might comment later.
In the meantime your previous comment made me dive into the alternative world. More precise the TBPC or Tesla Bifilar Pancacke Coil.
One of it's merits is that it according to many years testers it cancels out impedance so only the wire resistance is left.
Anyhow I did an attempt to visualize it in the simulator app on my phone, and yes the result is scary, anyhow for the capacitor.
Guess it will explode.
BTW the little cap should give 40 KHz
A Screen video you find here
https://filetransfer.io/data-package/WwnPHxnv#link

HMM, something went wrong. Can not repeat the results in the app. Not good, have to look in it.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 18, 2022, 05:21:52 AM
I am playing around with a simulator but do not really understand the results I get. Maybe the more capable members can guide me.
The simulator depicts a TBPC in series feed with a DC source and a capacitor that should control the frequency to 40 kHz.
The attached screenshots show the results. Does this indicate very strong pulsing magnetic fields?
Title: Re: Induction cooker without electronics
Post by: Twospoons on November 18, 2022, 06:29:23 AM
No, it indicates you need to learn about SPICE simulators.  You've drawn a circuit with 'perfect' components - your L and C have no resistance, and there's no resistance anywhere else either. Its also supplied with a  voltage source, which in simulator world can supply infinite current.   So you have a circuit with infinite Q, which will happily ring forever.  The result you are seeing is nonsense, and bears no resemblance to the real world.
The simulator doesn't care though, its just doing what you asked it to.
Title: Re: Induction cooker without electronics
Post by: klugesmith on November 18, 2022, 06:54:55 AM
How did you model the bifilar pancake coil?
Suppose we think of it as two windings occupying nearly the same space, so tightly coupled, each with inductance L.
A model with two identical L's in series, for total of 2L, is wrong.
If the inside of one coil connects to outside of the other coil, the combination gives you 4L. (Realizable in simulation by drawing two identical L's and setting a coupling factor of 1 or 0.999.)

In previous post you talked about canceling inductance. That would happen if inside of one winding connected to inside of other winding, so current in adjacent turns flows in opposite directions.  Net inductance is zero.   Realizable in simulation by keeping the coupling factor of 1 or 0.999 but reversing connections to one winding.  Or erase both L's and just place a single R to represent wire resistance, as you said in previous post.  Of course it won't resonate with any capacitor !

What online (?) simulator are you using?  Looks like current is shown by animated moving dots.   I want to see if they represent electron drift direction (bad) or direction of charge transport (good) or user gets to choose (even better).
Title: Re: Induction cooker without electronics
Post by: Solhi on November 18, 2022, 11:38:44 AM
How did you model the bifilar pancake coil?
Suppose we think of it as two windings occupying nearly the same space, so tightly coupled, each with inductance L.
A model with two identical L's in series, for total of 2L, is wrong.
If the inside of one coil connects to outside of the other coil, the combination gives you 4L. (Realizable in simulation by drawing two identical L's and setting a coupling factor of 1 or 0.999.)

Ås there is nowhere a TBPC included in any of these Apps. I just thought wrong and inserter 2 L with the same inductance. Indeed you are right it should be 4 L. So I adjust the model by choosing a transformer model with the coupling factor just given by you and connect the coils in series?

@Twospoons thanks for your info, learns me not to be so gullible :/
Edit: Found hidden in the app a wire resistance switch which was off. Diametral different result.

Quote
In previous post you talked about canceling inductance. That would happen if inside of one winding connected to inside of other winding, so current in adjacent turns flows in opposite directions.  Net inductance is zero.   Realizable in simulation by keeping the coupling factor of 1 or 0.999 but reversing connections to one winding.  Or erase both L's and just place a single R to represent wire resistance, as you said in previous post.  Of course it won't resonate with any capacitor !

I did write it canceles impendance not inductance. This should be according Tesla himself. Aka resistance will never be more than the wire resistance.

Quote
What online (?) simulator are you using?  Looks like current is shown by animated moving dots.   I want to see if they represent electron drift direction (bad) or direction of charge transport (good) or user gets to choose (even better).

I used the android app "PROTO" I will check what the animation represents.
Title: Re: Induction cooker without electronics
Post by: klugesmith on November 18, 2022, 05:48:00 PM
>> I did write it cancels impedance not inductance. This should be according Tesla himself. Aka resistance will never be more than the wire resistance.

You're making progress, Solhi.   Your transformer model will work, with the well-coupled inductors in series.  You can get 4L with no canceling of impedance.  Or you can get 0L by reversing the series connection to cancel impedance.  Please try it both ways & report result.

Do you want TBPC to behave as an inductor in resonant circuit, or as a noninductive wire resistor?   Tesla (pbuh) never expected both in the same coil at same time.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 19, 2022, 06:38:00 AM
>> I did write it cancels impedance not inductance. This should be according Tesla himself. Aka resistance will never be more than the wire resistance

Do you want TBPC to behave as an inductor in resonant circuit, or as a noninductive wire resistor?   Tesla (pbuh) never expected both in the same coil at same time.

My idea was generating very strong magnetic pulses without generating the heath. I still think I need to generate 6 KW in the cookware
Title: Re: Induction cooker without electronics
Post by: Solhi on November 19, 2022, 02:30:04 PM
My results for the moment are not loyal to the title of the thread :-[
But to get some result I borrow some electronics which I later must try to substitute with a sparkgap of some sort.
Maybe a gas discharger.
Anyhow my question now is: do the spikes as a result of back EMF shown in the sim. create enough EM pulses to create useful induction and how would one convert that to power?
I have heard that electric waves that go in the negative phase converts to magnetic energy.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 20, 2022, 10:00:13 AM
Found out that I need close to the Max allowed magnetic flux of 26 uT. Have not been able to find any example of how much aTBPC can generate. For sim users I found a post describing how to model a Tesla coil in Sim. I need a desktop computer to do that.
https://overunity.com/17186/the-bifilar-pancake-coil-at-its-resonant-frequency/msg505402/topicseen/#msg505402
Title: Re: Induction cooker without electronics
Post by: klugesmith on November 20, 2022, 07:17:36 PM
Please explain where you got 26 microtesla limit.   Do you have very sensitive instruments, to detect any heating from induced currents in 26 uT field at any frequency?

[edit] My previous comments about bifilar pancake coils, and coupled inductor/transformer model, neglected the capacitance between turns.  That does contribute to energy storage, and behavior associated with transmission-line wave propagation in spiral paths.

But the circuit model shown in that overunity post represents a single ordinary symmetric transmission line, appropriately modeled as a ladder of discrete L's and C's. I see no "cross coupling", that is to say any magnetic or capacitive coupling between turns of the bifilar wire pair.   In the course of a century, there must have been serious attention to theory and practice of TBPC behavior, but none appears in that post. Don't rush to get a desktop computer just to repeat that lame exercise.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 21, 2022, 03:09:26 PM
Please explain where you got 26 microtesla limit.   Do you have very sensitive instruments, to detect any heating from induced currents in 26 uT field at any frequency?

From bfs.de
"At common distances to the hobs the appliances usually comply with the reference value of 27 µT for the magnetic flux density developed for the relevant frequency range by the International Commission on Non-Ionizing Radiation Protection (ICNIRP Guidelines 1 Hz – 100 kHz, 2010)"

https://www.bag.admin.ch/dam/bag/de/dokumente/str/nis/faktenblaetter-emf/b-field-exposure.pdf

Quote
[edit] My previous comments about bifilar pancake coils, and coupled inductor/transformer model, neglected the capacitance between turns.  That does contribute to energy storage, and behavior associated with transmission-line wave propagation in spiral paths.

There I am completely blank (empty gaze)


Quote
In the course of a century, there must have been serious attention to theory and practice of TBPC behavior, but none appears in that post. Don't rush to get a desktop computer just to repeat that lame exercise.

Well, the strange thing is, if it was, it is damned good hidden. I can't find anything. The best source is the ppl. in these forums and they do a serious amount of work without me being able to find anything that definitive closes the case.

I should have disclosed that I'm situated on a topic island without easy access to anything and in addition minimal funds to use. Further do I not have any useful place to work and even less the instruments. The most advanced is the sim on my phone and a old Fluke multimeter.
I admit, it can, until I have found out the what and how, only be a theoretical exercise.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 24, 2022, 01:45:33 AM
I found out it is easier to work back from how much μT I need than just trying to get some output from coil, since it is the strength of electromagnetic pulse that do the work, not how much energy the coil consumes.
I found the standard (27 μT) which is used, but it seems that that is when a vessel covers the induction area. This does not tell anything about how much is needed to heat the vessel satisfactory to get reasonable cooking time. So to solve that first, should I just calculate back in a form of KW needed => A/turn times amount of turns used?

Edit: calculation A/t solved = 50 A.
My coil at the moment emits theoretical 650 μT. TBPC should then double that.

Trying to find a conversion from μT to energy, unless somebody can confirm that total A/t = Ampere.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 24, 2022, 08:08:48 AM
I found following in
https://www.bfs.de/EN/topics/emf/lff/application/induction-hob/induction-hob_node.html

Basic threshold values
− Low-frequency fields at 50 Hz: current density of 2 mA/m2
− Medium-frequency fields: the permissible current density depends on the frequency and ranges from 50 mA/m2 at 25 kHz to 140 mA/m2 at 70 kHz.

Reference values
− Low-frequency magnetic field: 100 µT
− Medium-frequency magnetic field: 6.25 µT

Again not knowing if that is the maximum allowed magnetic power.

Further I am confused about the input on this site: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html

Do I really input the amps from the sim times the windings as the A for the calculator?
Title: Re: Induction cooker without electronics
Post by: Solhi on November 25, 2022, 07:53:52 AM
The A/t issue is solved, answer is indeed.
This gives me a theoretical magnet field of 650 μT at 40 kHz.
Since this is a TBPC that number should double.

All the references I found before, about legal magnetic fields, were all related to exposure during use.

I need only to find a conversion from μT to energy, unless anybody can confirm that A/t = amperes.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 25, 2022, 12:05:50 PM
How did you model the bifilar pancake coil?
Suppose we think of it as two windings occupying nearly the same space, so tightly coupled, each with inductance L.
A model with two identical L's in series, for total of 2L, is wrong.
If the inside of one coil connects to outside of the other coil, the combination gives you 4L. (Realizable in simulation by drawing two identical L's and setting a coupling factor of 1 or 0.999.)

Can not define coupling between 2 independent coils. So I keep it as is only to get an indication of power in use. Then double it to simulate a TBPC.
I have no idea if, how far off I am.

Re sim:
Quote
"I want to see if they represent electron drift direction (bad) or direction of charge transport (good) or user gets to choose (even better)."

Not been able to find out, though I  guess it is the electron drift.

@klugesmith, I hope you can guide me more since I feel I'm stuck now.

Until now I have concluded this:
My TBPC is by feeding it with 24 V AC in it's resonant frequency 42,7 kHz, producing a magnetic field of 1200 μT.

Even in my search to convert that to KW and finding a site that probably solves it. I have for now problems in calculating a result.
The site:
https://pressbooks.bccampus.ca/introductorygeneralphysics2phys1207/chapter/24-4-energy-in-electromagnetic-waves/

If I can round off this satisfactory, then it looks like I am close to the intend of this thead.
Title: Re: Induction cooker without electronics
Post by: klugesmith on November 27, 2022, 01:13:28 AM
Quote
Until now I have concluded this:
My TBPC is by feeding it with 24 V AC in it's resonant frequency 42,7 kHz, producing a magnetic field of 1200 μT.
Even in my search to convert that to KW and finding a site that probably solves it. I have for now problems in calculating a result.
The site:
https://pressbooks.bccampus.ca/introductorygeneralphysics2phys1207/chapter/24-4-energy-in-electromagnetic-waves/
If I can round off this satisfactory, then it looks like I am close to the intend of this thead.
In OP you said you want 6 kW of heating power.  Does your design include 24 V source, and associated wires, able to deliver 250 amperes continuously?  If it works with smaller current, then you will become famous as discoverer of free energy.

Let's set the magnetic field questions aside for now, and talk about electrical circuits.   You recently spoke of 43 kHz instead of 100 MHz.    Do you still intend to get there with no discrete capacitor?  What coil inductance L and capacitance C will give you that frequency?    Sqrt(L/C) will give you the resonant circuit impedance in ohms (tank V / tank I), chosen by equipment designer to match the power source.   The product of RMS tank voltage and tank current is necessarily higher, usually many times higher, than the heating power delivered to load.   Give us some L, C, V, and I numbers; then we could talk quantitatively about V and I induced in the cookware.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 27, 2022, 01:24:05 PM

In OP you said you want 6 kW of heating power.  Does your design include 24 V source, and associated wires, able to deliver 250 amperes continuously?  If it works with smaller current, then you will become famous as discoverer of free energy.

It looks like you are thinking of direct resistance heating.
There it would be correct to use 250 A to obtain 6 KW.
We are though taking about inductance. For my way of thinking I would try to find out how strong a pulsing magnetic field we need (independent of how we produce that),  to get the same energy in the vessel.
Since in most vessels only the bottom sheet of 1-2 mm is ferrous and in my case with the dimension of 2900 x 2900 x 1.5 mm, we need to find out how strong a pulsing magnetic field is required to create an eddy current big enough to create heat in said plate to boil a certain amount of water in a certain time. So we talk about 6.000 KJ and say 30 min.

My problem now is that I seem to be unable to calculate that (sure it can be calculated).

Quote
Let's set the magnetic field questions aside for now, and talk about electrical circuits.   You recently spoke of 43 kHz instead of 100 MHz.

The 43 kHz is a result of the self resonance in the quarter wave determent by the length of the wire of the coil (square coil).

300k / (3,5 + 3,5) = 43 kHz

Input data: *
Number of turns N = 8.0
Width of the former a = 80.0 mm
Height of the former b = 80.0 mm
Winding length l = 3.3 mm
Wire diameter d = 1.5 mm
Wire diameter with insulation k = 6.6 mm

Result:
Inductance L = 15.419 µH
Winding thickness c = 52.8 mm
DC resistance of the coil Rdc = 0.034 Ohm
Wire length without leads lw = 3.51 m
Weight of wire m = 55.582 g
Number of layers Nl = 8

*there is the problem that the sim does not have a TBPC model, so I improvised by defining a coil with room between the windings to give place to an identical coil in series. Then doubling the results of the sim, hoping to get close to the truth.


Quote
Do  still intend to get there with no discrete capacitor?  What coil inductance L and capacitance C will give you that frequency?

The TBPC IS the capacitor, double power vs a single PC.
Fed with 24 V A/C 43 kHz, the sim shows a 3 A draw.
With 16 turns would give 48 A/m
Would give 600 μT x 2 = 1200 μT, because it is a TBPC
The frequency I explained over.

Quote
Sqrt(L/C) will give you the resonant circuit impedance in ohms (tank V / tank I), chosen by equipment designer to match the power source.   The product of RMS tank voltage and tank current is necessarily higher, usually many times higher, than the heating power delivered to load.   Give us some L, C, V, and I numbers; then we could talk quantitatively about V and I induced in the cookware.

Since the TBPC cancels impedance the Rdc = 0.034 Ohm which is neglectible?

The rest is given over.

Rest my question how many KJ would this 1200 μT alternating field produce in given plate?


P.s. I was partly wrong regarding the calculation of the power needed to heat the water. Forgot the time constant.
E = 4.2 * 18 * (100 – 20) = 6.048 kJ
6.048 kj = 1.667 kwh
30 min = 1.667 * 2 = 3,4 KW. = 12.097 kj
In addition comes the energy to heat the slab.

Edit:
Steel 470J/(K kg)
E = 0.47 * 0.91 * 85 = 36 kj = 10 watt

Further this paper seems to give the answer, without me being able to follow the mathematics, it is far over my head.
https://www.osti.gov/pages/servlets/purl/1433514

Page 2,3,4 see attachment

Title: Re: Induction cooker without electronics
Post by: klugesmith on November 27, 2022, 06:05:52 PM
Let's tackle a few easy steps.   You sound like a sincere inventor, not a forum troll.
It looks like you are thinking of direct resistance heating.
There it would be correct to use 250 A to obtain 6 KW.
We are though taking about inductance. For my way of thinking I would try to find out how strong a pulsing magnetic field we need (independent of how we produce that),  to get the same energy in the vessel.
Induction heating is always less efficient than resistance heating. You can't generate a sufficiently strong pulsing magnetic field, in presence of the cookware, without at least 6 kW of electric power from source.  Unless, as said before, you are first to discover the key to free energy.

>>Since in most vessels only the bottom sheet of 1-2 mm is ferrous and in my case with the dimension of 2900 x 2900 x 1.5 mm, we need to find out how strong a pulsing magnetic field is required to create an eddy current big enough to create heat in said plate to boil a certain amount of water in a certain time. So we talk about 6.000 KJ and say 30 min.

That's more like it.  You want 3333 watts of power from eddy currents in the resistive metal plate.   We could go there, figuring eddy ohms and amps and volts, but let's not do that now.  Did you mean 290 x 290 mm?  Is round spiral coil close enough for first estimate?
   
Quote
The 43 kHz is a result of the self resonance in the quarter wave determent by the length of the wire of the coil (square coil).
300k / (3,5 + 3,5) = 43 kHz

Uh, please repeat that calculation, paying attention to the units of measurement for each factor on the left side.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 28, 2022, 05:55:16 AM

That's more like it.  You want 3333 watts of power from eddy currents in the resistive metal plate.   We could go there, figuring eddy ohms and amps and volts, but let's not do that now.  Did you mean 290 x 290 mm?  Is round spiral coil close enough for first estimate?

You are correct with regard both to the power and the dimensions.
Excuse my sloppiness.
   
Quote
~The 43 kHz is a result of the self resonance in the quarter wave determent by the length of the wire of the coil (square coil).
300k / (3,5 + 3,5) = 43 kHz~

Uh, please repeat that calculation, paying attention to the units of measurement for each factor on the left side.

Edit:
A quarter wavelength calculation is
78,28/length wire = freq in MHz or
78,28*freq = length of wire (meter)
78,28 * 0,043 = 3,5


Here I'm way off, meaning that I'm not even close to any radio wave length. The next formula does do the job. After all, I'm looking for the conductors self resonant frequency.

The other I used is fr=(c/2L) where c is the speed of light.

300.000.000/(3,5+3,5) = 43.000 Hrz

Quote
Is round spiral coil close enough for first estimate?

I think it does not make such a big difference,  square or round. Square makes it a little more effective with a square bottom.

So OK with round

Title: Re: Induction cooker without electronics
Post by: klugesmith on November 28, 2022, 11:22:51 AM
>> The other I used is fr=(c/2L) where c is the speed of light.
>> 300.000.000/(3,5+3,5) = 43.000 Hrz

You lost a factor of 1000 right there!   
Driving too fast, and depending on guardrails to keep you on the road.
The guardrails might not hold up much longer.

Does your design concept make sense after correcting the 1000?
We need to see whether you dramatically change frequency or wire length,
before taking another step with pancake coil analysis.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 28, 2022, 11:38:26 AM
>> The other I used is fr=(c/2L) where c is the speed of light.
>> 300.000.000/(3,5+3,5) = 43.000 Hrz

You lost a factor of 1000 right there!   
Driving too fast, and depending on guardrails to keep you on the road.
The guardrails might not hold up much longer.

Does your design concept make sense after correcting the 1000?
We need to see whether you dramatically change frequency or wire length,
before taking another step with pancake coil analysis.

Hmm, having a bad day not seeing the right.amount of zeros.
Back to the drawing table.
Thanks for pointing out.

I was rather driving to slow moving meters instead of km :/
Title: Re: Induction cooker without electronics
Post by: Twospoons on November 28, 2022, 10:09:56 PM
The tool you really want for this is FEMM4.3 . This is a free finite element magnetic/electrostatic/thermal simulator which works for planar and solenoidal problems. FEA simulators will do many millions of calculations, taking into account your chosen geometry and the properties of the materials in the model, and provide a reasonably accurate result. You can readily extract things like field strength and power dissipation from the results.  But you will need a PC to run it, which I believe you said you don't have.

Without something like this you really are flying blind, so your first priority should be getting yourself a computer. It needn't be anything fancy - a 5 year old basic machine would do just fine.
Title: Re: Induction cooker without electronics
Post by: Solhi on November 29, 2022, 01:25:43 AM
Funny, I was just thinking the same last night. I am already in search mode to try to borrow one, even they are seldom here where I live.
Allas boy, out to work.
Title: Re: Induction cooker without electronics
Post by: Solhi on December 03, 2022, 06:32:14 AM

Not so easy if not impossible to get hold of a PC to use FEM software, which I am afraid also requires a steep learning curve to master (software).

In  the meantime from:
https://pressbooks.bccampus.ca/introductorygeneralphysics2phys1207/chapter/24-4-energy-in-electromagnetic-waves/#navigation

Do I see here a possibly to calculate back, how strong a magnetic field I need for my purpose?
@klugesmith, @twospoons, please be so kind to check my calculations, if my results miss zeros or worse.

The bottom of the cooking vessel is 0.084 m², which needs 3,500 watt to do the job.

I(intensity)= 3500 watt/ 0.084 m² = 43,617 W/m²
Peak = 2I = 83,234

Peak electric field strength =
E₀= √־ (83,234/ (c "m/s" *1)) = 0.01666 * 1.000 = 16.67 V/m

But not sure this is right. I assumed that permittivity of free space, being air = 1

Magnetic field strength =
Β₀= 16.66/ c = 5.55 μT  Hmmm??

Curious to find out if or where I went wrong.




Title: Re: Induction cooker without electronics
Post by: klugesmith on December 03, 2022, 07:43:44 PM
The bottom of the cooking vessel is 0.084 m², which needs 3,500 watt to do the job.
I(intensity)= 3500 watt/ 0.084 m² = 43,617 W/m² [you meant to type 41,617]
Peak = 2I = 83,234

Peak electric field strength =
E₀= √־ (83,234/ (c "m/s" *1)) = 0.01666 * 1.000 = 16.67 V/m

But not sure this is right. I assumed that permittivity of free space, being air = 1

Magnetic field strength =
Β₀= 16.66/ c = 5.55 μT  Hmmm??
Curious to find out if or where I went wrong.
You are in the right ballpark, of wrong game. Because EM radiation in free space is not a useful model for energy transfer in an IH. 

Let's review the EM result anyway.  And practice starting with RMS instead of peak values, since peak power (2 x average power) is not an otherwise useful measure.

For plane EM waves in free space, your formulas should give same answer as these: E (in V/m) = 377 * H (in A/m).  P(in w/m^2) = EH. B (in T) = u0 * H.

Unless I'm mistaken, 41617 W/m^2 corresponds to
E = 3961 V/m RMS (5602 peak).
H = 10.5 A/m RMS (14.9 peak).
B = 13 μT RMS (19 μT peak).

You are not far off on B.   That's an extremely intense EM wave, compared to (say) the field strengths inside a microwave oven.
But for your induction heating goal, you need B stronger by orders of magnitude. E will not go up in proportion, and will probably be smaller, since this is not a radiative process.  Near-field "radiation" might be a fair description for the leakage magnetic field, where you know 27 μT is considered to be tolerable where the operator is standing.
Title: Re: Induction cooker without electronics
Post by: Twospoons on December 03, 2022, 09:51:09 PM
A better approach to get you started would be to forget about field strength calculation altogether, and reduce the problem to an air cored transformer.

This is a huge simplification of the problem, and ignores a lot of things, but will get you some semi-reasonable numbers to start with.  But the math is simple, and doesn't require a PC.
Title: Re: Induction cooker without electronics
Post by: klugesmith on December 04, 2022, 05:17:56 PM
What Twospoons said is exactly how I would suggest that Solhi dig in.
How about starting with a specific homework assignment, full of simplifying assumptions?

Say the work coil, and conducting area on bottom of pot, are annuli with diameters 29 cm outside and 5 cm inside.

Pot is made of 410 stainless steel.   Estimate the resistance of a single turn in the thin annular sheet carrying current.  If that's too hard, take ring resistance to be 0.05 Ω.  What amount of induced voltage and current will develop 3500 watts of heat?

[edit] Here is a puzzle for some reader more practiced than me in EM wave physics.  In a previous post we figured the E and H field strength in a wave far from some transmitter (e.g. WWVB, 60 kHz digital time broadcast in USA). Suppose that wave is normally incident on a sheet of stainless steel that's substantially thicker than the skin depth (as figured for induction heating exercise). How much of the power is reflected, and how much is absorbed?  What's the form of induced current in the sheet, if the wave polarization is typical for terrestrial radio transmission?
Title: Re: Induction cooker without electronics
Post by: Solhi on December 05, 2022, 03:21:00 AM
Thank you guys. It might be true that the only right way to get a result is what you proposed. But I'm still in my rabbit hole to just find an indication what amount of μT magnetic field I need, so bear over with me.
I only want to air my next attempt to get to a number a bit more precise than substantial amount.
If I look at e.g.
https://www.homemade-circuits.com/simple-induction-heater-circuit-hot/#The_Tank_Circuit
Who tests this tiny IH which makes his screwdriver glow in an instant (~600° C)
/>Where he uses 12 V and 5 A to make it happen.
I assume I do not need to heat the cooking vessel that much to boil 18 liters of water.

I also assume (time will show) that providing enough power to heat the vessel to 150° C in 30 min. Would do the trick. Of course the vessel will never reach that temperature, since the water will absorb all that heat until it boils.

The amount of μT used in the video I calculated being 283 μT. To heat the screwdriver inside the coil to 600°. Linear thinking gives me 283/4=71 μT to heat to150°. The volume of the heated screwdriver is 126 mm³, the volume of the bottom of the vessel is 67.280 mm³ which is 534 times greater.
Then, would the answer be, we need 71*534=37.914 μT? That is at least substantial :-)

In the meantime I've learned neither a pancake coil nor a solenoid coil is ultimate. As usual something in the middle.

I'm now making a speadsheet calculator to test various coils to get to the desired result.

Looking forward to your comments.

Title: Re: Induction cooker without electronics
Post by: klugesmith on December 05, 2022, 04:52:22 PM
>>The amount of μT used in the video I calculated being 283 μT.

Please show your work.   What's the current in coil? Do you think it increases when workpiece is inserted?  Is the field strength in air with no workpiece, or air near workpiece, or inside the skin of workpiece?

Computers aren't needed.   Engineers used slide rules when huge induction heaters proliferated in metalworking factories in the 1930's.   EM physics theory was identical to that of today.
Title: Re: Induction cooker without electronics
Post by: Solhi on December 06, 2022, 06:50:17 AM


>>The amount of μT used in the video I calculated being 283 μT.<<

Please show your work.   What's the current in coil? Do you think it increases when workpiece is inserted?  Is the field strength in air with no workpiece, or air near workpiece, or inside the skin of workpiece?

My damned zeros again? The result was 28.2743 Gauss, which should be 2827 μT.

The guy in the video used a rewired V meter as Amp meter, saying you have to halve the shown values. Without load it shows ca. 1.2 A. See pict. My Sim shows 900 mA. Further he shows that the amperage rises to 5A when inserting the load in the center of coil.

My 283 μT comes from 5(A) x 9(T) x 0.01(R to center coil, I missed a commaplace)
But maybe I should have used the 1.2 A in the  formula (Without load)?

In that case the formula shows 628 μT.

So to recompute my last post.
2827/4 = 707 x 534 = 377404 μT
Substantial indeed. What I did not include is the time of 30 min. Would that mean another devision of 30?
In that case we would get a result of 12580 μT
 
Title: Re: Induction cooker without electronics
Post by: klugesmith on December 06, 2022, 07:54:48 AM
You have a few pieces of the puzzle right-side-up and correctly put together.

One piece from wrong puzzle is your idea that microteslas (in center of empty work coil) can be scaled to heating power density in steel (no matter how large).
It's not that simple. To begin with, heating power in any configuration will go up as the _square_ of current, voltage, and magnetic field amplitudes.

The guy in video is measuring the DC current from 12 volt power supply, which is NOT the same as the oscillating current in work coil and tank capacitors.   I bet the latter is much higher, especially when unloaded. I have not yet studied the circuit or searched for analysis at this forum.   Just ordered a couple of modules like the one in video, to play with and measure properly.  Dipping toes in the water before starting any serious IH project. The "1000 watt" label, in video caption and online sellers' description, far exceeds the DC input specs.

To model the induction heating process without finite element analysis, a more advanced student might
start with an infinitely long conductive rod inside an infinitely long solenoid. 
Follow a good fields-and-waves textbook, skin effect chapter, to figure the penetration and phase shift of induced voltage, current, and magnetic field.
The planar geometry in textbook should adapt easily when skin depth is very small compared to rod diameter.
We might even get values for electric circuit model (transformer with resistive load on secondary) per unit length.

[pedantic] The editors of excellent hyperphysics page, that Solhi cited, mistakenly capitalized unit names like ampere and gauss.   The official _symbols_ are capitalized (A, Gs, Pa) when the unit is named for a person.  Personally I say 10 gausses or 10 torrs, just like 10 amperes or 10 kilometers, in the name of regularity.
Title: Re: Induction cooker without electronics
Post by: Solhi on December 06, 2022, 12:36:10 PM

One piece from wrong puzzle is your idea that microteslas (in center of empty work coil) can be scaled to heating power density in steel (no matter how large).
It's not that simple. To begin with, heating power in any configuration will go up as the _square_ of current, voltage, and magnetic field amplitudes.

The scaling no matter how large is done in industry, but they use teslas instead of μT

My challenge is that the vessel is outside the coil center, so bigger loss.

I found now what you said about going up to the square is haunting me.

If my assumption about needing ± 12000 μT is somewhat right, (!klugesmith?) I can get that with a bifilar Brooks coil with 220 V 6.7 A producing + 13000 μT, but still learning and searching.

That is then 1,4 KW to do a 3.5 KW job.

Quote
The guy in video is measuring the DC current from 12 volt power supply, which is NOT the same as the oscillating current in work coil and tank capacitors.   I bet the latter is much higher, especially when unloaded.

The circuit is a ZVS converter producing AC sin wave of 12 V. If I remember right the coil has a freq of round 240 MHz. I do not know how much the capacitors boost the current.
Edit: looks the circuit has 2 x 330 nF  caps.

Quote
The "1000 watt" label, in video caption and online sellers' description, far exceeds the DC input specs.

He said raising the input to 48 V 24 A would melt a 10 mm nail. That is more than 1000 W

Quote
To model the induction heating process without finite element analysis, a more advanced student might
start with an infinitely long conductive rod inside an infinitely long solenoid.
Follow a good fields-and-waves textbook, skin effect chapter, to figure the penetration and phase shift of induced voltage, current, and magnetic field.
The planar geometry in textbook should adapt easily when skin depth is very small compared to rod diameter.
We might even get values for electric circuit model (transformer with resistive load on secondary) per unit length.

Haha no way I can go that way. No intention to become a electro engineer and besides my age tells me to sit on the porch in my swing chair or go fishing.

Quote
[pedantic] The editors of excellent hyperphysics page, that Solhi cited, mistakenly capitalized unit names like ampere and gauss.

Yes, is confusing. Leaving that place now. Need more advanced calculations, so the spreadsheet will hopefully do a better job
Title: Re: Induction cooker without electronics
Post by: petespaco on December 06, 2022, 05:59:38 PM
Regarding the little ZVS heater a few posts back:
  If is NOT a 1000 watt heater. It is sold as a 100 (sometimes 120 ) watt ZVS induction heater.
Don't be too impressed by the heating of the nail.  Most cameras make iron objects look a LOT hotter than they really are.
Also notice how much the current dropped when the nail reached the Curie point.
They are sold as 5 volt to 12 volt input devices.  I think that, at 48 volts, you would see a lot of fireworks in  a short period of time.
I just saw one for sale on Aliexpress for USD $1.52.  Maybe you should get one and power it with a car battery to get a bit of actual experience as you move forward.
Title: Re: Induction cooker without electronics
Post by: Solhi on December 07, 2022, 05:53:40 AM
I just saw one for sale on Aliexpress for USD $1.52.  Maybe you should get one and power it with a car battery to get a bit of actual experience as you move forward.

Ordered one a long time ago. Never know if it arrives here.  The price including shipping was $ 6.50
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