Domestic induction stoves can't be used with thick aluminium or copper pans, but when I try to heat aluminium tape it does get hot.
Can someone explain why this is the case?
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\$\begingroup\$ Please link the article you read and please explain precisely what model of stove it is. \$\endgroup\$– Andy akaCommented Jan 16 at 8:10
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1 Answer
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Induction stoves induce eddy currents in objects placed on top of their varying magnetic field. The heating effect size is determined by the conductivity of the object. Too low (say a ceramic pan) there will be virtually no current so no heating. Too high (thick aluminium or copper) the current will be there, but the resistance will be so low that little heat will be generated.
For domestic stoves, steel & iron hits the sweet-spot and allows effective heating. Aluminium pans without steel inserts don't.
Very thin aluminium foil/tape will heat as the thin section means the resistance is much higher than with a thick pan. However, you are unlikely to get the full power.
As noted in comments by Tim Williams, at a more detailed level, the optimum pan material is related to impedance matching of the load to the drive circuit magnetic leakage impedance. Very low resistance pans will be able to sustain eddy current sufficient to oppose the drive field with very little power dissipation in the pan. (The ultimate example would be a superconducting pan, which will just exclude the field with absolutely no heating.) However, the key to the heating of the foil is still the higher resistance due to thin section.
Note, higher frequency drive can heat more conductive metals due to the impact of skin-effect localizing the eddy currents in the surface layer (similar to the impact of the tape), but the electronics tend to be more expensive and less efficient, and the surface heating of less conductive metals can lead to overheating. Hence the choice of frequency for the domestic market.
It is also the case that ferro magnetic materials (iron and many steels) have a much larger skin effect below the curie point, which increases their effective resistance (and delivered power) in the domestic case.
(I have provided a more detailed answer in an earlier question on induction hobs here.)
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\$\begingroup\$ This explanation sounds fishy to me. Isn't it basically a transformer? When you reduce impedance across the secondary, the power goes up. There has to be a current limit built into the stove, that causes it to have an optimum load impedance \$\endgroup\$– tobaltCommented Jan 16 at 20:16
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\$\begingroup\$ @tobalt It's an impedance divider against the leakage inductance between coil and load. When the load is highly conductive but thin, it can have the same equivalent impedance and therefore power dissipation as a better-suited solid material. \$\endgroup\$ Commented Jan 16 at 20:30
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\$\begingroup\$ @tobalt thanks for highlighting I brushed over the details of why target resistance matters. I've added more detailed text on the role of impedance matching load resistance and leakage impedance. However, to a first order the reason al foil works but pans don't is still increased resistance due to thin section. \$\endgroup\$– colintdCommented Jan 16 at 22:16
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1\$\begingroup\$ @TimWilliams thanks for comments. I've added more detail in my answer. \$\endgroup\$– colintdCommented Jan 16 at 22:17
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1\$\begingroup\$ The other factor is that iron is ferromagnetic which causes the skin depth to be much shorter than other metals, resulting in current localizing to the surface and corresponding high resistance. Some steel pans work poorly with induction stoves if they're not ferromagnetic. \$\endgroup\$ Commented Jan 16 at 22:54