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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.

Brian Michalk
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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 09 '20 at 14:11

3 Answers3

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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.

Benjamin Kuykendall
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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 09 '20 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 09 '20 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 '20 at 08:07
  • @rumtscho I've used both, it's not that different. – JimmyJames Mar 10 '20 at 13:42
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    @rumtscho Or rather, I find it pretty dubious that having a plate inside a pot that is heated is going to be more like heating the pot directly than heating it from the outside. If anything, I would expect it to me more different. The pot would be a heat sink, not a heat source. It's an interesting concept though if you extend upon it. I wonder if ball bearings could work. – JimmyJames Mar 10 '20 at 14:04
  • @JimmyJames The pot is a heat sink regardless, it's not where you're ultimately trying to bring the heat. But if you put the plate on the inside, the heat would be generated where you already want it, and all the outside would be the "sink". If you put it below, you'd still have the heat sink of the pot, but you'd also be directly heating further away from where you actually want the heat to be. You should get better heat transfer with your heater directly in the place you're trying to heat. You would have to weigh that against the decrease in inductive heating due to distance though. – JMac Mar 10 '20 at 14:20
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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 '20 at 14:27
  • I just tried this for myself. It does not work very well. – barbecue Mar 10 '20 at 15:00
  • @JMac "The pot is a heat sink regardless." In both (normal) induction cooking and conventional, the pot is heated and it's what is heating the food. In order to heat the food, the pot must be warmer than the food. In an immersion situation such as this, you now are heating from the contents and it's heating the pot and the pot will never be warmer than the contents. The only caveat with that is that the plate might be in direct contact in some part of the bottom. Immersion heating is interesting but to suggest that it's more like using an inductive pan doesn't make much sense to me. – JimmyJames Mar 10 '20 at 15:17
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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 '20 at 16:00
  • @JimmyJames Yes; the pot heats and it is heating the food; but it's only an intermediary at best. The goal is heating the food. As the pot heats, it loses a lot of that heat to the surroundings instead of heating the food. The pot only has to be warmer than the food if the heat for the food is coming from outside the pot. If the heat is coming from inside the pot, the contents of the pot could be higher temperature than the pot itself; and this would actually help reduce how much heat is lost to surroundings. – JMac Mar 10 '20 at 16:27
  • @JMac I'm not sure why you are repeating what I wrote back to me. – JimmyJames Mar 10 '20 at 18:30
  • @JimmyJames I totally misread your comments, my bad. I read back through your comments and I think I do agree with what you are trying to say. That said the whole "source/sink" distinction does get pretty blurry with the pots. With external heating the pot would act as a source and a sink, so there are definitely advantages if you can minimize pot heating by heating the pot internally; for some reason thought you were arguing against that. I see now that it doesn't seem like you were. – JMac Mar 10 '20 at 18:55
  • @JMac Cool, I'm glad we worked that out. There are definitely advantages to immersion cooking. I'm planning to see if I can get a ball bearing (or some other ferrous object) to heat up water in pyrex with my cheap counter-top induction burner. – JimmyJames Mar 10 '20 at 19:17
  • @JimmyJames It was re-reading the ball-bearing comment that made me realize that we were probably on the same page. It would be interesting to see if that works/to what extent. – JMac Mar 10 '20 at 19:21
  • @JMac Being farther away might be an issue. IIRC, my burner won't turn on if there's not a ferrous pan there. It might not 'recognize' or detect it. I'll let you know – JimmyJames Mar 10 '20 at 19:38
  • @barbecue yes, the pot heats the food. But there is a huge difference in how you heat the pot. In resistive, you need the heat to get transferred from the outside to the outer surface of the bottom, and then get conducted through the pot to the inner surface of the bottom. This is inefficient and reacts very slowly to changes in energy. For induction, you heat the inner surface of the pot bottom, which is very efficient (gives you much more heat per energy amount used) and reacts immediately to changes. Especially the quick reaction makes the cooking experience different. – rumtscho Mar 10 '20 at 21:27
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    @rumtscho Everything you just said is true, but it does not affect how the heat transfers from the pot to the food. Electricity->Element->Pot->Food vs Electricity-Pot->Food may be more energy efficient but it does not change the rate at which the heat transfers from the pot to the food, that depends on the heat content of the pot and the food, not how the heat got there. Heat does not remember where it came from. – barbecue Mar 10 '20 at 22:17
  • @barbecue Where it comes from has an effect on where it goes though. Applying the same amount of heat right next to the food is different than applying that same heat below the pot and the food. When it's applied further from the food, more of that heat will go into things that aren't the food (like the pot and the air surrounding the the pot/element). This slows down the rate of cooking and the efficiency, assuming the same amount of net heat is sent via electricity. Finding out if it does send the same amount of heat is what would need to be tested. – JMac Mar 10 '20 at 22:26
  • @JMac I agree that induction is more efficient at heating the pan than traditional burners, there's no question about that. But the efficiency of converting electricity to heat doesn't affect the cooking. The rest of the room might get hotter, but as long as the pan is hot the food heats the same. That's all I'm saying – barbecue Mar 11 '20 at 00:03
  • @barbecue But if the same amount of energy heats the pan/what's in the pan more, then it does affect the cooking. It takes less heat, and therefore less time, to cook the same thing, if you assume the same amount of heat being used. – JMac Mar 11 '20 at 00:35
  • @barbecue your statement that it "doesn't change the rate of heat transfer" assumes a constant heated state of the pot. This is not correct (and not desirable) in cooking. The cook changes the heat frequently, on purpose. The inductive stove reacts quickly, while the resistive stove reacts very slowly. This ignores the additional effect (which is not necessarily positive) given by an induction stove using time modulation for reducing its heat output (turning the coil on and off every few seconds) as opposed to the changed amount of current running constantly through resistive. – rumtscho Mar 11 '20 at 06:58
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    @rumtscho you seem to be under the mistaken impression that I'm saying something I'm not saying. I've never disputed that induction is more energy efficient. – barbecue Mar 11 '20 at 18:11
  • @barbecue You are arguing the same point as I was earlier but with more detail. The only difference from cooking results perspective is the time to heat the pan. This makes we wonder if I would get better results using my induction burner for searing instead of my gas range. – JimmyJames Mar 11 '20 at 20:02
  • @JimmyJames I'd say probably yes, your cast iron pan will get hot faster, and recover faster with even a low-power induction hob than with anything but the biggest commercial gas stoves. A heavy duty induction hob of 4000 watts or more will beat anything gas for speed. The biggest limitation of induction is the requirement for ferrous metal. There are induction units that can work with aluminum pans, but they're less efficient and in my experience less reliable. – barbecue Mar 12 '20 at 00:20
  • @barbecue I am making two points, independent of each other. One is the energy efficiency. The other is the speed of reaction. Both are excellent in induction, and both get lost when there is a disk placed under the pot.Your answer reads as if placing the disk underneath is the only right way to go, without mentioning that there are downsides which will be avoided if the "disk inside" solution works. That's why I find it misleading. – rumtscho Mar 12 '20 at 10:53
  • @rumtscho That's not what I said. IMO, the only right way to go is to put a ferrous pan directly on the induction cooktop. Disk-in-pot and disk-under-pot are both inferior options. Disk-under-pot basically converts your induction cooktop to a traditional burner. Disk-in-pot either won't work at all (induction cooktop won't "see" the disk through the aluminum) or won't get hot enough. Don't take my word, go try it yourself. If you want the lighter weight and faster heat distribution of aluminum, just use an aluminum pan with a built-in steel base. Greenpan makes some nice ones. – barbecue Mar 12 '20 at 13:18
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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.

Dennis Williamson
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AI0867
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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 '20 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 '20 at 23:04
  • @rumtscho as barbecue says, the pan *is* a barrier. If you place the disk *under* the pan, it will be 'closer' to the coil than the pan, current will be induced in it, and it will heat up. If you place the disk *in* the pan (that is, on top of the bottom of the pan), it will not, because the current will be induced in the pan itself, and not generate any heat there. Viewed differently, the magnetic disk is a barrier to the magnetic field, limiting the induced current to the disk. Aluminium will conduct the current, but does not form a barrier, allowing the current to spread out. – AI0867 Mar 11 '20 at 16:28
  • @jpa the distance is not much of an issue - you can easily place a cutting board on an induction stove and your pot on top of it and it works (I have tested this empirically, with a cheap consumer grade portable stove, so can vouch for it). It may heat up less at the same energy output, but it does heat up sufficiently. As for the barrier thing, I admit I am a bit confused, the physics of electric fields are not my strong suit. Maybe I should try to dig my old portable induction stove out and simply test. – rumtscho Mar 11 '20 at 17:37
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    @jpa: https://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 '20 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 '20 at 00:35
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    @PeterCordes Hmm, I'm still not convinced! But I posted the question on Physics: https://physics.stackexchange.com/questions/535951/does-aluminum-in-between-stop-induction-cooktop-from-warming-iron – jpa Mar 12 '20 at 07:33
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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.

cbeleites unhappy with SX
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  • https://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 '20 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 '20 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 '20 at 00:12