Induction cooker without electronics

[Solhi]

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.

[Solhi]

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?

[Solhi]

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.

[Solhi]

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:

"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.

[klugesmith]

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.

[Solhi]

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).

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.

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.

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

[klugesmith]

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?
   

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.

[Solhi]

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.
   

~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

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

[klugesmith]

>> 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.

[Solhi]

>> 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 :/

[Twospoons]

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.

[Solhi]

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.

[Solhi]

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.

[klugesmith]

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.

[Twospoons]

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.

• Consider the problem as an N:1 transformer - you have rough dimensions for your coil, so if you model your "pot" as a ring of similar inner and outer diameter you can calculate the resistance of the ring, since you know the thickness and material.
• From there you know you want 3kW, so you can calculate the current that must be circulating in the ring.
• Using the N:1 turns ratio you can estimate the primary current required, assuming a reasonable coupling factor.  For two large flat coils in close proximity I would guess somewhere in the region of 0.8 for the coupling, maybe higher
• Knowing the primary current you can work out the power loss in the primary coil (primary self-heating)
• Last you need to figure out how to generate that primary current

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.

[klugesmith]

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?

[Solhi]

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.

[klugesmith]

>>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.

[Solhi]

>>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
 

[klugesmith]

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.