Author Topic: Induction cooker without electronics  (Read 1584 times)

Offline Solhi

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Re: Induction cooker without electronics
« Reply #20 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.
« Last Edit: November 25, 2022, 07:41:20 AM by Solhi »

Offline Solhi

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Re: Induction cooker without electronics
« Reply #21 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?
« Last Edit: November 24, 2022, 08:10:55 AM by Solhi »

Offline Solhi

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Re: Induction cooker without electronics
« Reply #22 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.

Offline Solhi

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Re: Induction cooker without electronics
« Reply #23 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.
« Last Edit: November 26, 2022, 03:03:09 PM by Solhi »

Offline klugesmith

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Re: Induction cooker without electronics
« Reply #24 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.

Offline Solhi

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Re: Induction cooker without electronics
« Reply #25 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

« Last Edit: November 27, 2022, 03:08:46 PM by Solhi »

Offline klugesmith

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Re: Induction cooker without electronics
« Reply #26 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.
« Last Edit: November 27, 2022, 06:17:37 PM by klugesmith »

Offline Solhi

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Re: Induction cooker without electronics
« Reply #27 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

« Last Edit: November 28, 2022, 10:01:42 AM by Solhi »

Offline klugesmith

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Re: Induction cooker without electronics
« Reply #28 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.

Offline Solhi

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Re: Induction cooker without electronics
« Reply #29 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 :/

Offline Twospoons

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Re: Induction cooker without electronics
« Reply #30 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.

Offline Solhi

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Re: Induction cooker without electronics
« Reply #31 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.

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Re: Induction cooker without electronics
« Reply #31 on: November 29, 2022, 01:25:43 AM »

 


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[Light, Lasers and Optics]
klugesmith
November 18, 2022, 10:05:29 AM
post Re: Induction cooker without electronics
[Electronic Circuits]
klugesmith
November 18, 2022, 06:54:55 AM
post Re: Induction cooker without electronics
[Electronic Circuits]
Twospoons
November 18, 2022, 06:29:23 AM
post Buck converter question
[Dual Resonant Solid State Tesla coils (DRSSTC)]
dru
November 18, 2022, 06:10:52 AM
post Re: TV flyback number of primary turns rule of thumb
[Transformer (Ferrite Core)]
davekni
November 18, 2022, 05:53:15 AM
post Re: Induction cooker without electronics
[Electronic Circuits]
Solhi
November 18, 2022, 05:21:52 AM
post Re: Vacuum pump
[Laboratories, Equipment and Tools]
davekni
November 18, 2022, 04:00:04 AM
post Re: Vacuum pump
[Laboratories, Equipment and Tools]
alan sailer
November 17, 2022, 10:24:35 PM
post Re: TV flyback number of primary turns rule of thumb
[Transformer (Ferrite Core)]
John123
November 17, 2022, 08:30:00 PM
post Re: TV flyback number of primary turns rule of thumb
[Transformer (Ferrite Core)]
klugesmith
November 17, 2022, 07:28:16 PM
post Re: TV flyback number of primary turns rule of thumb
[Transformer (Ferrite Core)]
John123
November 17, 2022, 03:38:00 PM
post Re: TV flyback number of primary turns rule of thumb
[Transformer (Ferrite Core)]
davekni
November 17, 2022, 04:03:51 AM
post Re: what type of diode is better for make high voltage diode?
[Voltage Multipliers]
davekni
November 17, 2022, 03:57:40 AM

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