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I have an inductive stove, and some very large pots that don't work on the stove. It seems that the material is invisible to the induction coil, and if I were to place a steel disk inside the pot, then that steel disk would act as a heating element. Would this work?

This would be for liquids, boiling large amounts of mostly water (bone broths, beer, etc.), so the cooking surface isn't really a factor.

edit: The stainless pots I'm using are quite thin. I have tried placing them on a cast iron skillet similar to the adapter idea, but that doesn't work well at all. I am hoping the disk would heat, and being inside the pot, would be much more efficient. I could also test with some generic bits of steel I can find. I thought I'd ask first.

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    If you have a cast iron pan that will fit the pots you want, put the pot in the pan and use the induction stove. Used this trick in a Tokyo AirBNB to use my aluminum Bialetti coffee maker. – Chris Cudmore Mar 9 at 14:11
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It would not work

As Benjamin Kuykendall stated, you may place magnetic iron or steel under the pan. I will tell you why placing it in the pan won't work.

Induction

As the name states, induction cooking works through induction: The stove runs an alternating current through a coil, producing a changing magnetic field. This changing magnetic field then induces a current in anything conductive placed in it.

Conduction

What happens when you place an aluminium or copper pan on top of the stove? A current will be induced in them, but they are very good conductors, and pans are much thicker than the copper in the coils, so not a whole lot would happen. The stove would likely notice that little energy is being absorbed from the magnetic field, complain and turn the stove off.

Impedance

So what is the difference with iron? Well, iron is not as good a conductor as copper or aluminium, but it's not that bad. The thing is though, iron is magnetic, and resists the magnetic field from the stove. This means that all the current is confined to the very surface of the iron, greatly increasing the current density, and therefore the power dissipated.

You can tell that this is the reason by finding a non-magnetic stainless steel pan and trying it. It won't work. (it's the nickel in the alloy that changes the crystal structure, making it non-ferromagnetic)

So what happens?

When you place the steel under the pan, the steel acts as usual, confining the current and heating up.

When you place the steel in the pan, the entire pan between the steel and the coil will happily conduct the current, and heat up less than the coil under the stovetop.

Other tricks

So, the reason that iron works is that the current is confined to a small thickness of the metal. Can we use this in other ways?

Foil

Aluminium foil is very thin, so it doesn't matter whether the material conducts well. A strategically placed piece of aluminium foil can absorb all that energy, melt, and totally ruin your induction stove. Don't do that.

Possibly you may be able to improvise a pan with glass cookware containing aluminium foil and water, but I wouldn't recommend it

High frequencies

There's another trick that engineers can use. At high frequencies, alternating currents will confine themselves to increasingly thinner thicknesses of metals. This means that if you use a high enough frequency, any conductive metal will work. There are stoves that use this principle, but they are rare and new.

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  • I don't understand your explanation. You agree that the iron disk will heat up as usual when placed under the pan. You agree that the pan itself will not act as a barrier. So why would the disk not heat up when placed inside the pan, just like it heats up when placed on top of some non-pan object? – rumtscho Mar 10 at 21:35
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    @rumtscho the pan IS a barrier. A sheet of glass and a sheet of aluminum will not have the same effect on a magnetic field. Don't take my word, try it for yourself.There's some sophisticated engineering involved in inductive cooking, but fundamentally it all comes down to rapidly changing electromagnetic fields generating currents in metal. Ferrous metals are better for generating heat from electrical resistance because their lower conductivity means electricity converts to heat more efficiently. That's why heating elements are made of alloys of iron and its kin, like nichrome. – barbecue Mar 10 at 23:04
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    @jpa: en.wikipedia.org/wiki/Induction_cooking#Cookware is pretty good; it explains that skin-depth as well as resistivity is a key property, and that a layer of highly conductive metal below the iron will shield it. Note that you're not shielding against a fixed magnetic field, you're merely shielding against inductive effects from alternating current like the wall of a Faraday cage. A Faraday cage doesn't need to be ferromagnetic. – Peter Cordes Mar 11 at 21:42
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    @PeterCordes excellent analogy. Aluminum is often used for RF shielding because it's both light and inexpensive. – barbecue Mar 12 at 0:35
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    @PeterCordes Hmm, I'm still not convinced! But I posted the question on Physics: physics.stackexchange.com/questions/535951/… – jpa Mar 12 at 7:33
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The usual approach would be to put the steel disk under the pot. Such disks are even produced commercially for this purpose: look for an induction interface disk. The basic mechanism is straight forward. The disk heats due to electromagnetic induction; then heat moves from the disk to the cooking vessel via conduction. As long as your cooking vessel is a good conductor and the bottom of the pan is in direct contact with the disk, it will heat up relatively efficiently.

Putting the disk inside the cooking vessel seems iffy. If your cooking vessel has a thick bottom, then the disk will be further from the stove-top. Because the magnetic field strength decreases with distance, induction would be less efficient. Also, if the disk is inside the vessel, it will come into contact with the food and probably need to be washed. Putting it under the pot leaves one fewer item to wash.

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    Your answer reads as if there would never be a reason to place it on the inside. But it has one large advantage you are not mentioning: the heat is generated in the disk and is thus in contact with the food, while putting it underneath requires the pot to heat up first, basically making the induction oven work like a resistive oven. – rumtscho Mar 9 at 18:32
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    @rumtscho "while putting it underneath requires the pot to heat up first" which is pretty much what happens with a pot that works directly with an inductive 'burner'. – JimmyJames Mar 9 at 21:09
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    @JimmyJames no, that's not how inductive burners work. There, the current creates heat directly on the surface of the pot, including its inner surface. On a resistive stove, you have a hot burner, which makes imperfect contact with the cold metal pot, and you have to wait for heat conduction through the whole thickness of the bottom until the heat reaches the food. This makes for very different behavior during cooking. – rumtscho Mar 10 at 8:07
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    @rumtscho In both conduction and induction cooking, you heat the pot and the pot heats the food. How you heat the pot is different, but how the pot heats the food is the same. – barbecue Mar 10 at 14:27
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    Note that some induction tops says that warranty is void when using interface disks, because it makes surface hotter than designed - imperfect transfer from disk to pot means disk is not cooled by the food as efficiently as the induction pot would. – Mołot Mar 10 at 16:00
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Let me add to the other answers - this is mostly a too long comment trying to clarify some of their points.

We have 3 physical effects that are important for heating the food inside the pot

  • heat conduction: heat will flow from a hotter material to a colder material - the power (energy per time) depends crucially on the contact surface area.

    • this contact area is very good between a hot pot and its liquid contents
    • the contact area of a pot on a hot disc (whether resistive stove or induction disc between pot and induction stove) can be anywhere from basically the whole surface for nicely plane pot & disc to only a few small spots for a pot with warped bottom.

    => conclusion 1: with an induction disc you'll need a good pot like for use with a ceramic or metal top resistive stove. A thin-bottom pot (as e.g. for gas cooking) won't work well because you don't get the heat efficiently from induction disc to pot bottom.

    • (As induction cooking (with induction pot) avoids this heat transfer, induction pots don't need as plane bottom surfaces.)
  • induced current having Ohmic losses (aka Joule heating): Here, the pot bottom or induction disc works as receiving antenna of the electromagnetic field "sent" by the induction stove. As @AI0867 explained, the (alternating) magnetic field of the induction stove induces a corresponding (eddy) current in any electrically conducting material. Since the pot is not superconducting, this current has Ohmic losses: electric energy is converted into heat. This works with any pot material that conducts electric current: in the aluminum foil video linked by @AI0867, water in the "pot" (foil) would have been heated, the foil or an alu pot would not melt as long as there's water inside, see also general induction heating.
    To clarify: we don't get as much power transfered to copper or aluminum as to a ferromagnetic iron or steel: Ohmic losses are proportional to the resistance (and to the square of the current), and that (AC) resistance (= impedance, see @AI0867's answer and skin effect) is (much) lower in non-ferromagnetric conductors. So we have less dampening (absorption, see below) of the electromagnetic field. But of course for practical cooking purposes, already transfering, say, only 10 % of the power compared to ferromagnetic material (copper or aluminum pot would be below that) means that this is practically useless: It's not only that you'd have to wait 10 times as long to transfer the energy you need - the losses from your heated pot don't get lower, so you may not get it to boiling temperature at all.

  • Hysteresis losses: when a ferromagnetic (or ferrimagnetic) material is magnetized back and forth, a certain part of the energy becomes heat (this is the area inside the hysteresis curve of the material), the so-called hysteresis loss. This heat is used in induction cooking, according to it amounts to https://en.wikipedia.org/wiki/Induction_heating 1/3 of the heating power.
    We miss out on this on "non-induction" metallic pots.



  • You can also consider using a glass "pot" and then put your steel disk (or maybe something like an iron fish - here it would help for once...) inside - glassware doesn't disturb the field. Steel is fine as long as it is (ferro)magnetic. The spacing may make the whole setup a bit less efficient, but it may very well be more efficient than an outside induction disc and suboptimal heat conduction.
    Some early patents on induction stoves used concepts with glassware and a disc inside, btw.

Please excuse the sources being in German, the English Wiki page on induction stoves doesn't have that information and I did not find an English-language equivalent of the forum explanation of the pot detection.

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  • en.wikipedia.org/wiki/Induction_cooking#Cookware explains that skin depth and resistivity are key to having enough resistance in the pot to have it dissipate most of the energy as heat: copper and aluminum have a much deeper skin depth (and somewhat lower bulk resistivity) so the induced current doesn't generate enough heat. Aluminum foil may be thin enough, but a pot isn't. Stoves that work with all metals have to run at much higher frequencies than the normal 24kHz, further reducing the skin depth and thus being able to transfer enough power into the pot. – Peter Cordes Mar 11 at 22:17
  • @PeterCordes: yes, there's not so much attenuation since the lower resistance doesn't attenuate the eddy currents. But the part of the EM field that is attenuated does become heat. In practice, the induction stove is probably detecting "insufficient attenuation -> switch off". And efficiency would be abysmal as ohmic losses in the sending coil aren't small compared to the transferred energy any more. – cbeleites unhappy with SX Mar 11 at 23:51
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    @PeterCordes: for the aluminum foil, also the "show effect" is good because you don't actually need that much energy to melt abit of alu foil - a quick back-of-the-envelope guesstimate indicates that maybe 200 J could melt a 3 cm x 1 cm x 24 μm piece of foil (not counting heat loss due to air convection, ...). So if a nominally 2 kW induction stove gets only 10 % of the power transferred to the aluminum foil (compared to proper induction pot), it would still be done in a second. Heating 50 ml of Water about 1 °C is the same energy, but much less showy... – cbeleites unhappy with SX Mar 12 at 0:12

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