Induction cooker pans?

[Merlin3189]

What sort of pan can be used with an induction hob?
I ask here because this seems electrical rather than electronic, though I think people here will know about it.

My understanding of induction heating is that the heating device produces an alternating magnetic field, probably with a thick coil and a pair of switching FETs, and that this magnetic field induced eddy currents in nearby conductors, such as the base of a pan. In that case I would expect the pan to need to be made of conductive metal like copper, aluminium, brass say. I can see that steel or iron might work, as they are conductive

But I read in WikiP for eg.

In induction cooking, an induction coil in the cook-top heats the iron base of cookware. Copper-bottomed pans, aluminium pans and other non-ferrous pans are generally unsuitable

and similar comments in retailers' web sites (eg. John Lewis).

On the other hand Raymond Blanc sells a set of pans in "heavy gauge aluminium" which is " Suitable for all hobs - Including induction." - just as I would have expected.

Wikip also explains,

Aluminum and copper cookware are more conductive than steel, and the skin depth in these materials is larger since they are non-magnetic. The current flows in a thicker layer in the metal, encounters less resistance and so produces less heat. The induction cooker will not work efficiently with such pots.

But if the resistance is lower, why is the current not greater? They have already explained that the pan base acts as a single turn secondary of a transformer (with the primary being the induction coil), so there should be a certain emf per turn and the power should be (V^2)/R , implying that you get more heating with a good conductor than a bad conductor.

Does anyone here have any real understanding of these devices and can explain how different types of pan fare on them?

 

[duke37]

Iron and ordinary steel are magnetic so will couple to the field much more strongly.
You could try non-magnetic stainless steel to see what difference it makes.

 

[Merlin3189]

Thanks for reply Duke. At the moment I can't try because I have not bought it yet! Just looking into the issues and getting a bit puzzled by some of the advice I've found.

In fact it is not a serious practical problem, as I will treat myself to a new set of cooking pans if necessary - and generally I prefer stainless steel, which seems to be recommended as second best. But I always get intrigued when advice is mixed or conflicts with my intuition, so I'm reading around about it and just thought I'd ask among people who might know what they're talking about.

 

[Arouse1973]

I think the material needs to be feromagnetic as its the eddy currents which create the heat.
Adam

 

[Arouse1973]

Iron and ordinary steel are magnetic so will couple to the field much more strongly.
You could try non-magnetic stainless steel to see what difference it makes.

I think you need adaptor plates for non-magnetic pans so they just act as a normall hob.
Adam

 

[shrtrnd]

I don't KNOW, but my brother had one of those stoves and he SAID the older aluminum pots wouldn't work on it.
With so many of them in use now, I would GUESS that pot & pan manufacturers would have made allowances with aluminum and stainless steel pots, so their wares would work with all types of stoves.(alloys used)
As a side-note. A lot of pots have copper bottoms to distribute heat evenly, the copper bottom would enable a stainless steel pot to heat.
To allow SS pots to work on the induction stoves,it would not surprise me if mfgrs integrated a thin sheet of copper between the inside and outside of a fully stainless steel exterior pot(?).
Again, just so that their wares would work on induction stoves.

 

[Arouse1973]

I don't KNOW, but my brother had one of those stoves and he SAID the older aluminum pots wouldn't work on it.
With so many of them in use now, I would GUESS that pot & pan manufacturers would have made allowances with aluminum and stainless steel pots, so their wares would work with all types of stoves.(alloys used)
As a side-note. A lot of pots have copper bottoms to distribute heat evenly, the copper bottom would enable a stainless steel pot to heat.
To allow SS pots to work on the induction stoves,it would not surprise me if mfgrs integrated a thin sheet of copper between the inside and outside of a fully stainless steel exterior pot(?).
Again, just so that their wares would work on induction stoves.

I think it's going to be the other way around. Copper won't be as good as steel because of the skin depth. To allow copper to heat to the same degree the frequency would have to increase. I guess that's what they are doing for stoves that heat copper and aluminium pans.
Adam

 

[Merlin3189]

I think the material needs to be feromagnetic as its the eddy currents which create the heat.
Adam

Eddy currents is what I thought, but that's why I thought it should be Copper or Aluminum - a good conductor.

I think you need adaptor plates for non-magnetic pans so they just act as a normall hob.
Adam

Yes. I found one of those and it said it was made of Aluminium and Iron, though it didn't say how the metals were arranged.

Copper won't be as good as steel because of the skin depth. To allow copper to heat to the same degree the frequency would have to increase. I guess that's what they are doing for stoves that heat copper and aluminium pans.
Adam

Well this gets right into the can of worms!

Skin depth (a slightly mysterious topic for me, but): if the skin effect effectively increases the resistance of the copper (because the current is confined to a thin layer and can't use the full thickness) then surely this is a good thing for those that say Copper & Al don't work because they're good conductors!?!
The only reason I can see why there would be less skin effect in iron, is because iron is not such a good conductor. So skin effect seems just to help level out the two materials.
Since skin effect is caused by magnetic field, would not iron increase the magnetic field and hence the skin effect?

Frequency: If induced voltage and induced current were fixed, then I can't see what difference frequency would make to copper. It just heats by RxI^2 where I is the RMS amplitude of current - no reference to frequency. Iron on the other hand could well have hysteresis loss which would indeed increase with frequency and cause heating. Is that not the reverse of what you suggest

But if induced V and I are not already determined, what role for frequency?
Well skin effect increases with f usually I think. Good, or bad?
Induced V will be easier at high frequency, which is, I think, why they use it. The secondary is the base of the pan, fixed at one turn. So to get max V you need as few turns as possible on the primary (for any given supply V.) And to get a big voltage across a few turns with a small current requires a high frequency. Now, since copper is a good conductor, we don't actually need such a high induced voltage (P= (V^2)/R), so a low turns ratio is less important and we don't need such a high frequency!?!

My guess so far?
Conductance may be a red herring. We need a reasonable conductor, but its exact conductance may not be critical. I still think more conductance is better, so I want Al or Cu in it, but any metal should work.
If iron has a role, it is in improving the magnetic circuit, so that the maximum flux from the generator coil links through the pan. Copper & Al don't much care one way or another about a static magnetic field (its permeability is much the same as air), but it does shun alternating magnetic fields - why? because eddy currents are generated and their flux opposes the causal flux. But eddy currents are exactly what we want!
And if you use an iron pan to encourage the magnetic flux, then it won't work nearly as well once eddy currents occur. Just as in Cu or Al, they will oppose the flux. So have you gained anything? (Real question, btw)

So my guess for the design of a good induction pan? A base of thick Cu or Al in the form of a ring with an iron core and maybe a sheet of iron on top, but not underneath. Make it look as much like a transformer sawed in half as possible.

Finally: getting it to heat up is not everything.

Cu and Al are also good conductors of heat, so maybe give a more even temperature across the base? I know hotspots are a problem for some non-stick pans, causing local debonding of the Teflon (or whatever it is these days.)

One of the nice features of induction heating is the controllability of the heat. Lowering the input current immediately reduces the heat input. (This was the disadvantage of old-style electric hobs with large resistive elements often encased in metal blocks.) But this is spoilt if the pan itself has a large thermal capacity and low thermal conductivity slowing down the response.

Much as I like Cu & Al electrically, I'm a bit doubtful about having them in contact with all my food. From the looks of it, some of the metal dissolves. Which is why most of my pans are stainless steel. I do worry about the chrome and vanadium a bit, but SS does not show the same tendency to dissolve. I don't mind eating pure iron, but I think the carbides in most ferrous metals can produce some odd compounds when they dissolve. Now what are enamel coatings made of?

Thanks for everyone's interesting comments.

 

[Merlin3189]

Just to add a link about Eddy's currants. WikiP has,

Eddy currents ... are circular electric currents induced within conductors by a changing magnetic field .. ... Eddy currents flow in closed loops within conductors, ... They can be induced within nearby stationary conductors by a time-varying magnetic field created by an AC electromagnet or transformer,... The magnitude of the current in a given loop is proportional to the strength of the magnetic field, the area of the loop, and the rate of change of flux, and inversely proportional to the resistivity of the material....
...Eddy currents are also used to heat objects in induction heating furnaces

(my underlines)

So the emphasis for eddy currents seems to be that you need a good conductor with as low resistivity as possible. (Though what happens with a superconductor - infinite current and no power!?)

BTW, another WikiP on induction heating has,

Heat may also be generated by magnetic hysteresis losses in materials that have significant relative permeability.

So score one for iron. In fact it says,

Iron and its alloys respond best to induction heating, due to their ferromagnetic nature.

agreeing with Adam and most of the manufacturers' advice.
My slight reservations about this article are that they flip between talking about hysteresis or eddy as it suits them, and while there is some maths for eddy currents there is none at all for hysteresis or permeability.
So we seem to be left with people telling us, perhaps based on experience or trials, that iron is best and we just have to take it on trust. I had been hoping for convincing scientific arguments. (Which BTW, is why I always try to pick holes in whatever views are posted. It's just scientific scepticism.)

 

[Arouse1973]

The thickness that copper would need to be to produce the same heat as steel would make the pan impracticable for cooking. The trouble with a lot of the articles on the web about eddy currents is it's all about reducing them. This is because we often want to stop them causing heating issues.

For induction heating we would actually want loads of eddy currents and reduced conductivity to cause heat. We don't want copper, it's too good a conductor. Hysteric heating (Magnetic Friction) adds to the eddy current heating, but will stop at the curie temperature and then all you have is eddy currents.

We need higher permeability because this is what causes the reduced skin depth (Magnetic domains line up). A material that can be easily magnetised reduces skin depth by pushing back on the magnetic field and reducing penetration depth. This in return gives us the higher surface resistance we need for the cooking.

Adam

 

[Merlin3189]

Hi Adam, thanks for the continuing interest. As I said B4, this is a can of worms. There are several factors all of which interact and it's hard to work out which apply or which dominate.

For induction heating we would actually want loads of eddy currents and reduced conductivity to cause heat. We don't want copper, it's too good a conductor.

I suspect you are right here.
Another WikiP article says,
"Skin depth also varies as the inverse square root of the permeability of the conductor. In the case of iron, its conductivity is about 1/7 that of copper. However being ferromagnetic its permeability is about 10,000 times greater. This reduces the skin depth for iron to about 1/38 that of copper, about 220 micrometres at 60 Hz. Iron wire is thus useless for A.C. power lines ...
Iron rods work well for direct-current (DC) welding but it is impossible to use them at frequencies much higher than 60 Hz. At a few kilohertz, the welding rod will glow red hot as current flows through the greatly increased A.C. resistance resulting from the skin effect, with relatively little power remaining for the arc itself. Only non-magnetic rods can be used for high-frequency welding."
So it does appear that iron is the stuff to use because it has a much higher resistance.

However (!) both the examples have the iron between the source and the load, where you don't normally want resistance nor power dissipation, and the iron may not be the principal source of resistance in the circuit. I have been thinking of analogy with a transformer and looking at the secondary winding. To keep it clear, say a torroidal core with completely separate primary and secondary windings and the primary winding being very substantial so that it can handle all the power required. Let's take the ratio to be such that we get a 10V output. Now if the secondary is shorted, how much power is dissipated by resistance in the secondary? If we say the secondary is Cu with a resistance of say 1 Ohm, then we get 10V/1Ohm= 10 Amp and dissipate 10A*10A*1Ohm = 100W. Then if we change the secondary to iron wire (and keep the same length and cross sectional area) then its resistance is about 58 Ohm. So now the current is 10V/58Ohm= 0.17A and the dissipation 0.17*0.17*58= 1.7 W ! If there really is a skin effect here, it will be greater in iron increasing the iron resistance and reducing the dissipation further. (An online calculator gave me skin depth 200 times thinner for Fe than Cu, due to the higher permeability.)

So my conclusion was that we need the lowest resistance possible, because for a given induced voltage, current and power are inversely related to resistance.

However(!) what determines the induced voltage? I am assuming the pan base is one turn and the voltage = primary voltage * 1 / (primary turns) . But of course this is for a normal transformer where there is very tight magnetic coupling between the windings - effectively we assume 100% flux linkage I think. With our hob, there is much less flux linkage, so presumably much lower induced voltage. But with an iron pan the higher permeability will mean greater flux linkage and a higher induced voltage. So would the increased permeability of iron increase the voltage enough to outweigh the extra resistance?
Well again there are two factors: iron can be as much as 200,000 times more permeable! but how much of the magnetic circuit is iron? Effectively the only barrier to the magnetic flux is the air (and copper which is about the same as air) and we can ignore the permeability of iron. The effect of iron is simply to shorten the length of the air gap (and calculating the original air gap is very difficult.)
If we guessed the average length of the magnetic circuit as 10cm with no pan present (that is, from the top face of the coil, up through the glass and air, sideways through the air, back down through air and glass, through air back to the opposite face of the coil, assuming the coil is cored) then going vertically through a few mm of Fe pan base would only reduce the magnetic circuit by say 10%. But allowing for the magnetic field to go sideways through the base as well might effectively reduce the air gap to a cm or so, giving 10x the magnetic flux and 10x the induced voltage. That is more than enough to compensate for the 6x difference in resistivity. The (V^2)/R ratio for Fe to Cu would work out to about 10V*10V/6R : 1V*1V/1R = 16.7 : 1 giving the required superiority for FE.
It does all depend on the details of the magnetic circuit and the grade of the iron: there is tremendous variation in the permeability of iron alloys, with some grades of steel hardly better than Cu or Al. My figures are taken from the values for "iron" which is unlikely to be what pans are made of. Provided the permeability is reasonably high, then it mainly depends on the reduction in length of the magnetic circuit versus the increase in resistance. My guess is that iron alloys will generally win overall, but the factor will be rather less than I calculated (break even might be around 50-70% reduction magnetic length.) Without the sort of details we can only get from manufacturers (which they don't divulge) or experiment, I don't suppose we can get much closer.

So, some tangled worms, but now you've made me look closer, they seem to be wriggling towards iron and high permeability materials.

 

[duke37]

Another worm to add to the heap.

Radio frequency coils are often screened to stop the energy escaping. Aluminium or silver is used to reduce the power loss, because of the low resistance. If the resistance is low, the eddy currents will reflect the energy. Every morning I think of this when I admire the image in my mirror.

 

[shrtrnd]

Methinks there is a lot of theoretical over-analysis here.
Mayhaps a pot manufacturer would specify on the outside of a newly purchased sales box, whether or not the pot would work on an induction-type stovetop(?)
Again. With so many of these stoves in common usage. I would think a pot manufacturer would have already analyzed this dilemma, and compensated in alloy use.
But that's just my theory.

 

[Arouse1973]

Hi Adam, thanks for the continuing interest. As I said B4, this is a can of worms. There are several factors all of which interact and it's hard to work out which apply or which dominate.
I suspect you are right here.
Another WikiP article says,
"Skin depth also varies as the inverse square root of the permeability of the conductor. In the case of iron, its conductivity is about 1/7 that of copper. However being ferromagnetic its permeability is about 10,000 times greater. This reduces the skin depth for iron to about 1/38 that of copper, about 220 micrometres at 60 Hz. Iron wire is thus useless for A.C. power lines ...
Iron rods work well for direct-current (DC) welding but it is impossible to use them at frequencies much higher than 60 Hz. At a few kilohertz, the welding rod will glow red hot as current flows through the greatly increased A.C. resistance resulting from the skin effect, with relatively little power remaining for the arc itself. Only non-magnetic rods can be used for high-frequency welding."
So it does appear that iron is the stuff to use because it has a much higher resistance.

However (!) both the examples have the iron between the source and the load, where you don't normally want resistance nor power dissipation, and the iron may not be the principal source of resistance in the circuit. I have been thinking of analogy with a transformer and looking at the secondary winding. To keep it clear, say a torroidal core with completely separate primary and secondary windings and the primary winding being very substantial so that it can handle all the power required. Let's take the ratio to be such that we get a 10V output. Now if the secondary is shorted, how much power is dissipated by resistance in the secondary? If we say the secondary is Cu with a resistance of say 1 Ohm, then we get 10V/1Ohm= 10 Amp and dissipate 10A*10A*1Ohm = 100W. Then if we change the secondary to iron wire (and keep the same length and cross sectional area) then its resistance is about 58 Ohm. So now the current is 10V/58Ohm= 0.17A and the dissipation 0.17*0.17*58= 1.7 W ! If there really is a skin effect here, it will be greater in iron increasing the iron resistance and reducing the dissipation further. (An online calculator gave me skin depth 200 times thinner for Fe than Cu, due to the higher permeability.)

So my conclusion was that we need the lowest resistance possible, because for a given induced voltage, current and power are inversely related to resistance.

However(!) what determines the induced voltage? I am assuming the pan base is one turn and the voltage = primary voltage * 1 / (primary turns) . But of course this is for a normal transformer where there is very tight magnetic coupling between the windings - effectively we assume 100% flux linkage I think. With our hob, there is much less flux linkage, so presumably much lower induced voltage. But with an iron pan the higher permeability will mean greater flux linkage and a higher induced voltage. So would the increased permeability of iron increase the voltage enough to outweigh the extra resistance?
Well again there are two factors: iron can be as much as 200,000 times more permeable! but how much of the magnetic circuit is iron? Effectively the only barrier to the magnetic flux is the air (and copper which is about the same as air) and we can ignore the permeability of iron. The effect of iron is simply to shorten the length of the air gap (and calculating the original air gap is very difficult.)
If we guessed the average length of the magnetic circuit as 10cm with no pan present (that is, from the top face of the coil, up through the glass and air, sideways through the air, back down through air and glass, through air back to the opposite face of the coil, assuming the coil is cored) then going vertically through a few mm of Fe pan base would only reduce the magnetic circuit by say 10%. But allowing for the magnetic field to go sideways through the base as well might effectively reduce the air gap to a cm or so, giving 10x the magnetic flux and 10x the induced voltage. That is more than enough to compensate for the 6x difference in resistivity. The (V^2)/R ratio for Fe to Cu would work out to about 10V*10V/6R : 1V*1V/1R = 16.7 : 1 giving the required superiority for FE.
It does all depend on the details of the magnetic circuit and the grade of the iron: there is tremendous variation in the permeability of iron alloys, with some grades of steel hardly better than Cu or Al. My figures are taken from the values for "iron" which is unlikely to be what pans are made of. Provided the permeability is reasonably high, then it mainly depends on the reduction in length of the magnetic circuit versus the increase in resistance. My guess is that iron alloys will generally win overall, but the factor will be rather less than I calculated (break even might be around 50-70% reduction magnetic length.) Without the sort of details we can only get from manufacturers (which they don't divulge) or experiment, I don't suppose we can get much closer.

So, some tangled worms, but now you've made me look closer, they seem to be wriggling towards iron and high permeability materials.

I think the reason for this anomaly is the thermal conductivity of the two metals. Wattage is irrelevant unless you know how hot the material is going to get with a certain amount of wattage. Materials in general that are not good conductors of current, won't conduct heat away very well either.

This is generally down to the amount of electrons in the conduction band and because they have very little mass they heat up easily but also transfer heat easily. so if the material does not have many free electrons it can't get rid of the heat as well.

So copper has a TC of 401W/m-deg C and Iron is 80W/m-deg C. This shows that iron can't dissipate heat anywhere near as well as copper so it will require less energy to heat up.

Adam

 

[Merlin3189]

Duke - Yes. Shielding + good conductors go together, though electric field shielding may also be a reason. I know you're right about the coils in Al cans - I've seen it. But my recollection was the can was quite large. I suspect you don't want too much eddy currents around an RF coil, it would lower the Q.

Shrtrnd - Oh yes! Lots of over-analysis! I hope I made clear, my interest is in understanding, not just in getting the right pans. Perhaps some pot manufacturers have worked this out and surely some induction hob makers must have. But do I believe what they tell me? Not really. If they explained it, ok, but just saying, "our pans work on all hobs" doesn't convince me. As I mentioned at first, one hob maker said, only iron & steel, but Raymond Blanc says his aluminium pans are "suitable for all hobs, including induction." Perhaps some manufacturers are honest and just don't tell us anything they don't think we need to know, but I think most manufacturers just tell us what they want us to believe.

Another question has just arisen about this cooker. What size circuit do I need to fit it to? You might think that this essential piece of info would be provided by every stove maker - in the form of a power consumption or a max current rating - but no, not a single one, that I've looked at, gives this data. And so far only one has replied to my queries about it. Oddly enough they all quote an energy consumption, such as "0.79 kW Hr", but that is totally meaningless to me. (NB. not kW per hour, nor kW Hr per day, nor kW Hr per chicken - though one quotes "kW Hr per standard load" as if it were a washing machine!) So much for manufacturers knowing or telling people what they need to know.

Adam - interesting points. Ok the thermal/electrical conductivity tie up and the free electrons.
Thermal conductivity does matter for evenness of heating. The only criticism I've seen in user reviews of induction hobs, is that some pans heat very unevenly and the hotspots cause problems. Some makers are using Fe&(Al or Cu) sandwiches because they want Fe to heat up and Al or Cu to spread it around. That was common for stainless steel pans even before induction hobs.

So copper has a TC of 401W/m-deg C and Iron is 80W/m-deg C. This shows that iron can't dissipate heat anywhere near as well as copper so it will require less energy to heat up.

I'm not sure I quite agree with this. If I've looked it up right, yes the thermal conductivity is right, but I find Cu needs 0.34 kJ/kg degC and Fe 0.45 kJ/kg degC ( or just in case, Cu 3487 kJ/m^3 degC & Fe 3533 kJ/m^3 degC)
But I don't think the thermal capacity of the pan is important. One wants to heat the contents and we want the minimum amount of pan to contains it and converts the mag field to heat.

In case anyone has access to IEEE journals, this sounds like a serious treatment of the subject which is supposed to deal with this issue.
W. C.Moreland, The Induction Range: Its Performance and Its Development Problems, IEEE Transactions on Industry Applications, vol. TA-9, no. 1, January/February 1973 pages 81–86

 

[Arouse1973]

Ok so this is the only other thing I can offer. After that it becomes a much more difficult problem for me and I would have to do some more research.

When we think about the current in a wire and the heat generated due to the Joules/s losses we tend to think about voltage current and resistance. With a radiated field another two parameters come into play. That is permeability and reluctance, reluctance is considered as the resistance to magnetic flux.

So although copper has a low resistance to conventional current it offers a very high reluctance to magnetic flux. On the other hand Iron is the opposite it has a very low reluctance to magnetic flux. Now as the magnetic flux is responsible for the eddy currents and the eddy current are responsible for heat generation it's clear to see who's the winner here.

So when the transport mechanism for the energy is from alternating magnetic flux and not electrostatic potential as in conventional current the reluctance of the material is the most important aspect.

Adam

 

[Merlin3189]

Yes, that's the conclusion I'm coming to.
Fine on the reluctance, the reciprocal of permeability, which I've used. Like talking about conductance and resistance.
So we're happy that we need low reluctance, and I'd say low resistance as well, with the low reluctance being most important.
Looking at the construction of these things (there are several YouTube vids), in some it does not seem to be an iron cored coil, but does have several rectangular slabs of ferrite positioned radially underneath. Presumably although these have very low reluctance, they do not get hot, because ferrite has very high resistance.
The coils also appear to have not many turns - one looked to be about 50. But I don't know what the primary voltage would be, so I can't say what the induced emf would be. One tech article I found seemed to think that the primary circuit was resonant. I don't know whether it's resonant unloaded or with the pan on, but I don't see how they get both, unless they change the frequency, because an iron pan must make a big change in the coil's inductance.
Another interesting comment they made was that induction hobs are made for the Far East market that do work with copper pans (because they are traditional there). I think they use higher frequency oscillations.

 

[Arouse1973]

Great conclusion. Yes higher frequecy must be it for copper. I think I mentioned that earlier. Put to bed I think.
Good discusion Merlin thankyou.
Adam