Power distribution above an induction element?

[HowardLi]

Does anybody know the power distribution of an induction element, i.e. how evenly the heat is transferred to the cooking vessel? Is it perfect even, or a series of concentric heating rings, or random, or varying?

[Sam Iam]

My Wolf induction cooktop seems to transfer energy evenly to my Viking pots and pans. Large electromagnets are jizzing the molecules evenly, it seems.

[dcarch]

Does anybody know the power distribution of an induction element, i.e. how evenly the heat is transferred to the cooking vessel? Is it perfect even, or a series of concentric heating rings, or random, or varying?

Induction cook top does not heat up the pot.

The pot heats itself up by converting the magnetic field to heat.

Therefore the design and construction of the pot have a lot to do with it.

dcarch

Edited August 1, 2011 by dcarch (log)

[HowardLi]

Does anybody know the power distribution of an induction element, i.e. how evenly the heat is transferred to the cooking vessel? Is it perfect even, or a series of concentric heating rings, or random, or varying?

Induction cook top does not heat up the pot.

The pot heats itself up by converting the magnetic field to heat.

Therefore the design and construction of the pot have a lot to do with it.

dcarch

Not sure if you're playing with semantics here, but the cooktop does heat up the pot. It runs AC through a big-ass coil which in turn produces a rapidly-changing magnetic field which causes Joule heating in the ferromagnetic (for cooking) material.

So, in effect, what I am asking is if you had a flat sheet of iron on the cooktop, turned on the cooktop, and looked at the top with thermo-vision, what would it look like? No need to lecture me about how conductive certain materials are, this isn't about that.

[gfweb]

Somewhere on eG there is a post about the hot spots on induction pans. I can't find it.

Essentially it is hottest over the coil, typically a several inch ring. A thin pan will burn in that region. A thicker pan will distribute better. You could put a layer of flour in a pan and see easily where the hot spots are in a few minutes.

[andiesenji]

I've put my cast iron griddle (12 1/2 inch diameter) on the induction burner and checked the temperature at various spots with my ThermoWorks Combo IR thermometer.

(Incidentally this is currently on special for $69., $20. off the regular price. I've had mine for a few years and it works great in both modes.)

The temperature of the griddle does not vary more than 5 degrees (F) from the center to within an inch of the outer edge. The far outer edge is cooler by 15 degrees but as I'm not cooking on that rim, it is of no concern to me.

Edited August 1, 2011 by andiesenji (log)

[dcarch]

Does anybody know the power distribution of an induction element, i.e. how evenly the heat is transferred to the cooking vessel? Is it perfect even, or a series of concentric heating rings, or random, or varying?

Induction cook top does not heat up the pot.

The pot heats itself up by converting the magnetic field to heat.

Therefore the design and construction of the pot have a lot to do with it.

dcarch

Not sure if you're playing with semantics here, but the cooktop does heat up the pot. It runs AC through a big-ass coil which in turn produces a rapidly-changing magnetic field which causes Joule heating in the ferromagnetic (for cooking) material.

So, in effect, what I am asking is if you had a flat sheet of iron on the cooktop, turned on the cooktop, and looked at the top with thermo-vision, what would it look like? No need to lecture me about how conductive certain materials are, this isn't about that.

Not at all playing with words. I am sorry if you feel that I am lecturing.

Induction cooking is very unique that it really does not heat up the cookware. The cookware, by the effects of hysteresis energy and eddy current generated (two of the most undesirable effects in transformer design are being taken advantage in induction cook tops), heats itself up.

How the heating distribution menifests, as I understand, depends a great deal on the design of the cookware and not the cook top.

dcarch

[HowardLi]

Does anybody know the power distribution of an induction element, i.e. how evenly the heat is transferred to the cooking vessel? Is it perfect even, or a series of concentric heating rings, or random, or varying?

Induction cook top does not heat up the pot.

The pot heats itself up by converting the magnetic field to heat.

Therefore the design and construction of the pot have a lot to do with it.

dcarch

Not sure if you're playing with semantics here, but the cooktop does heat up the pot. It runs AC through a big-ass coil which in turn produces a rapidly-changing magnetic field which causes Joule heating in the ferromagnetic (for cooking) material.

So, in effect, what I am asking is if you had a flat sheet of iron on the cooktop, turned on the cooktop, and looked at the top with thermo-vision, what would it look like? No need to lecture me about how conductive certain materials are, this isn't about that.

Not at all playing with words. I am sorry if you feel that I am lecturing.

Induction cooking is very unique that it really does not heat up the cookware. The cookware, by the effects of hysteresis energy and eddy current generated (two of the most undesirable effects in transformer design are being taken advantage in induction cook tops), heats itself up.

How the heating distribution menifests, as I understand, depends a great deal on the design of the cookware and not the cook top.

dcarch

Yes, I suppose if you could that you take away the ferromagnetic substance above the cooktop, then yes, nothing is happening. However, in the non-trivial situation where there is a vessel containing ferromagnetic material above the cooktop, the power goes into the cooktop and is transferred somewhat efficiently into the vessel material. You even admitted that eddy currents are necessary within the material; where are the eddy currents coming from, if not the source of the energy?

Electrical energy -> magnetic energy -> electrical energy -> thermal energy

Are you trying to make the point that heat energy is not conducted through the cooktop? Everybody knows that.

Anyway, enough on this point - we are speaking of words and words alone.

[dcarch]

I am puzzled why you seem to object to my answer to your original question, which is how even an induction cook top can heat up the cookware.

The point I am trying to make is that a lot has to do with the cookware, not just the cook top because the induction cook top transfers no thermal energy to heat up the cookware.

dcarch

[andiesenji]

I am puzzled why you seem to object to my answer to your original question, which is how even an induction cook top can heat up the cookware.

The point I am trying to make is that a lot has to do with the cookware, not just the cook top because the induction cook top transfers no thermal energy to heat up the cookware.

dcarch

I'm with you on this. I've been using induction burners since they became available for home use - I had a propane portable stove explode and wanted no more of that - back when I was catering.

I went to an appliance store that demonstrated it for me. They used a skillet (steel - non-stick), that had been cut in half. He turned the burner on and when the skillet was hot, broke an egg so it was half on the skillet and half on the bare burner.

The half on the skillet cooked, the other half remained raw.

That was a demonstration that was easily understood and which convinced me. It was a Mr. Induction, was a bit pricey and more bulky than the ones available today.

Later I got a Fagor, then a Supentown and more recently a Max Burton 6500 with the interface disc that allows me to use my copper cookware. (When I bought mine the disc was included.)

Frankly, I don't care why it works. It is enough for me that it actually works the way it is supposed to. The technical minutiae is not something that holds any interest for me.

Edited August 2, 2011 by andiesenji (log)

[HowardLi]

I am puzzled why you seem to object to my answer to your original question, which is how even an induction cook top can heat up the cookware.

The point I am trying to make is that a lot has to do with the cookware, not just the cook top because the induction cook top transfers no thermal energy to heat up the cookware.

dcarch

Correct. No thermal energy is being transferred... which, at least, is a change from "the cooktop does not heat up the pot".

Please describe how and which characteristics of a cooking vessel affect how evenly itself is heated.

Edited August 3, 2011 by HowardLi (log)

[Anna N]

.....

Frankly, I don't care why it works. It is enough for me that it actually works the way it is supposed to. The technical minutiae is not something that holds any interest for me.

Ditto! I cannot imagine going back to any other stove-top cooking method. I have found that most magnetic pans work just fine. Like using gas or electricity you learn the characteristics of each pan and adjust accordingly. It is important though that the pans be "flat" i.e., free of warps. I use cast-iron, enameled cast iron, non-stick s/s, s/s, carbon steel (which heats superfast)and one of these days I will try the interface disk to see how it fares with a clay pot.

[dcarch]

"----Please describe how and which characteristics of a cooking vessel affect how evenly itself is heated. "

If I look inside the guts of a typical induction cooker, this is what I see:

The circuit on the top right is the power board. Normal household power is converted to low voltage/high current by ferrite toroidal inductors. The converted power is connected to the high frenquency converter board on the left which also converts the AC power to DC power to power the electronics and the DC brushless cooling fan you see at the lower right corner.

The power at around 25khz (cycles per second) powers up the high current coil (orange colored round coil). At this point, the induction cooker becomes the primary side of an electric transformer. When you place a ferromagnetic cookware on the cooker, the cookware become the secondary side of the transformer. Because the cookware behaves as a short circuited single turn secondary coil, an even lower voltage and much higher electric current is generated, and that will produce a lot of heat (at 1 watt = 3.5 BTUs).

What all this means is this: the fact that the entire bottom surface of the cookware is one single “turn”, I believe the electric current flow probably would be very uniform and producing even heating, assuming the construction (both material and geometry)of the cookware is even and flat, otherwise hot/cold spot will likely happen.

I don’t know if I have given you any help at all.

dcarch

[HowardLi]

I believe the electric current flow probably would be very uniform and producing even heating, assuming the construction (both material and geometry)of the cookware is even and flat

Yeah, this is pretty much what I was looking for, though I didn't think it would take 20 posts to achieve it.

[Blether]

"All I need is co-ordination"

I'll admit to a dilettante's interest in the shape of the field. If the magnetic field round a bar magnet is like a(n American)-football-shape around the bar, with wheatsheaf sprouts from either end, what does the field that that coil induces look like ?

From what you say, dcarch, it's doughnut-shaped from above, but what's the cross-section ? Elliptical, like a squashed doughnut ? Does the requirement for flat cookware mean that it's a very shallow field ? How wide is the doughnut ? Just the area of the coil, viewed from above ? Extending beyond it ? Or not even reaching the edges ?

The effective area of induction will influence the degree to which the cookware has to rely on thermal conduction to heat evenly, and so the thickness of metal that is ideal.

Edited August 3, 2011 by Blether (log)

[dcarch]

Good questions Blether.

I believe the shape of the induction cooktop induction coil produces magnetic field in a very limited area. If you look at the photo I posted, you can see there are many components which are senitive to magnetic field. If there is an alternating magnetic field, they can be damaged easily. For instance, I think the fan motor rotor is a circular magnet, which can be de-magnetized if it is near another magnetic field.

I am not sure of the cross sectional shape of the induction field, however, as far as the cookware is concerned, the field will be essentially uniform because the cookware is right next to the "Primary coil".

Although the heating of the cookware is only on the thin layer of metal next to the coil(That's why only ferromagnetic cookware works) the thickness helps to even out the distribution of the heat.

dcarch

[Mjx]

I cannot find the post that mentions it, but unless I'm completely mistaken, I recall a member mentioning an induction cooktop that has no burners: You just put your pot down wherever, regardless of its size, which suggests that more than one coil configuration exists (a diagram should be available from the manufacturer).

[Ashen]

Because the cookware behaves as a short circuited single turn secondary coil, an even lower voltage and much higher electric current is generated, and that will produce a lot of heat (at 1 watt = 3.5 BTUs).

What all this means is this: the fact that the entire bottom surface of the cookware is one single “turn”, I believe the electric current flow probably would be very uniform and producing even heating, assuming the construction (both material and geometry)of the cookware is even and flat, otherwise hot/cold spot will likely happen.

I don’t know if I have given you any help at all.

dcarch

That is cool. As a single turn it would only be able to push 1 volt right? But the higher frequency against the resistance of the shorted turn produces the heat. I had wondered how it didn't also produce enough electricty in the pans to shock people using it .

[dcarch]

Mjx, there are many designs using multiple induction coils. I think I have seen some models which has the entire cooking surface active, doesn't matter what size pot and where you put the pot. Each coil has a sensor to turn on when it senses the pot.

Ashen, actually voltage does not produce heat, current does. When you comb you hair with a plastic comb, you can be generating a million volts of electricty.

The voltage generated by the pot follows the transformer design principle. (step-up/step-down) If the primary coil (the induction coil) has a 100 turns and powered by 10vac, the pot, being a single turn, will have 1/100 of the voltage, ie 0.1 vac.

No, at this voltage, you will not be shocked electrically.

dcarch

[dougal]

...

Please describe how and which characteristics of a cooking vessel affect how evenly itself is heated.

Base flatness does matter to evenness of heating.

On my (old) deDietrich you only had to raise the pan a fraction of an inch (maybe 1/8 inch?) to effectively stop the energy input.

But its not mega-critical; most pans are quite happy cooking "through" a sheet of paper. This was advocated by some as a deliberate technique to catch the fat spatter during deep frying -- but, because it doesn't burn on to the cooker, its very easy to clean anyway.

The microstructure of the metal matters, but it should be fairly consistent across the pan base.

The old deDietrich could produce a 'doughnut' of nucleating bubbles in the base of a Le Creuset saucepan when bringing water to the boil quickly. ISTR that it was noticeable on the smaller rings, at high heat, with a wider pan ... it does still matter to match the diameters of pan and ring!

Some current high-end deDietrichs have a large 'heating zone' (which might be what Mjx is recalling) - but I have no idea how even the heating might be (or why people might want to have plural pans without independent control).

They had a "Continuum Zone', but now they have gone further ... http://www.dedietrich.co.uk/93cm-zoneless-induction-hob-the-piano-p-10001716.html

Suffice it to say that I think that generalising globally across brands and models wouldn't seem like a terribly wise idea!

[dougal]

Just as a PS, a very important feature of most (all?) induction controllers is that they give dimmer-like control - the power distribution in time is steady, like a gas flame, rather than 'lumpy' like most ceramic/glass hobs/cooktops.

[Paul Kierstead]

The old deDietrich could produce a 'doughnut' of nucleating bubbles in the base of a Le Creuset saucepan when bringing water to the boil quickly. ISTR that it was noticeable on the smaller rings, at high heat, with a wider pan ... it does still matter to match the diameters of pan and ring!

My portable induction burner does that with my LC dutch oven and, as far as I can tell, it is far worse with the dutch oven then any other vessel. With the LC, you are in serious danger of burning things in that zone. It works well with most other pots, but the "ring" does show some in other pots as well, but not nearly so strong.

Just as a PS, a very important feature of most (all?) induction controllers is that they give dimmer-like control - the power distribution in time is steady, like a gas flame, rather than 'lumpy' like most ceramic/glass hobs/cooktops.

Again, on my portable unit, it definitely cycles, and to the point where if you are very near a boil, it will boil on and off some. Might be just that unit.

Edited August 4, 2011 by Paul Kierstead (log)

[Ashen]

Ashen, actually voltage does not produce heat, current does. When you comb you hair with a plastic comb, you can be generating a million volts of electricty.

The voltage generated by the pot follows the transformer design principle. (step-up/step-down) If the primary coil (the induction coil) has a 100 turns and powered by 10vac, the pot, being a single turn, will have 1/100 of the voltage, ie 0.1 vac.

No, at this voltage, you will not be shocked electrically.

dcarch

I didn't mean the voltage was generating the heat. I beleive that is from the frequency of the oscillation of the magnetic field, vs the resistance of the atoms in the metal in the pot/pan. I still wonder how they make the whole thing work without both the "windings" being around an iron core though.

[dcarch]

Ashen, actually voltage does not produce heat, current does. When you comb you hair with a plastic comb, you can be generating a million volts of electricty.

The voltage generated by the pot follows the transformer design principle. (step-up/step-down) If the primary coil (the induction coil) has a 100 turns and powered by 10vac, the pot, being a single turn, will have 1/100 of the voltage, ie 0.1 vac.

No, at this voltage, you will not be shocked electrically.

dcarch

I didn't mean the voltage was generating the heat. I beleive that is from the frequency of the oscillation of the magnetic field, vs the resistance of the atoms in the metal in the pot/pan. I still wonder how they make the whole thing work without both the "windings" being around an iron core though.

You don't need an iron core for the "transformer" effect to happen. A coil will generate a magnetic field without iron core and induces a voltage/current in another coil nearby. However, for the current/voltage to be induced in another coil the magnetic field must be changing in strength or polarity, hence the primary coil (the cooker) must be driven by high frequency current.

dcarch

[slkinsey]

The old deDietrich could produce a 'doughnut' of nucleating bubbles in the base of a Le Creuset saucepan when bringing water to the boil quickly. ISTR that it was noticeable on the smaller rings, at high heat, with a wider pan ... it does still matter to match the diameters of pan and ring!

My portable induction burner does that with my LC dutch oven and, as far as I can tell, it is far worse with the dutch oven then any other vessel. With the LC, you are in serious danger of burning things in that zone. It works well with most other pots, but the "ring" does show some in other pots as well, but not nearly so strong.

This seems ridiculous to me! What's the point of having induction if the heat isn't even across the burner?! Is there some technical reason why these have to be an empty ring and couldn't be absolutely even across the surface of the burner, or is it cheapness on the part of the manufacturers?