Induction heater question

1. Dec 16, 2016

Robertphysics

I know for them for some time and I understand how induction heating works in general , where you have a coil through which a high frequency AC current is fed that creates a magnetic field which changes fast according to the frequency and this excites electrons in the metal introduced and heats it up.
I’ve done some googling and found that some companies offer induction water heaters. If I understand correctly water itself wouldn’t heat up as fast or as good as iron does when inserted in the heater coil but hence water is passed through an iron pipe located in the heater that’s where it heats up.
Now here’s what I wonder , the most common method for electrical water heating is resistive coils. But in those all the electricity is converted to heat so it sounds efficient in terms of conversion but it consumes a lot of energy, is it physically true that heating up say the same amount of water through induction heating would be much more efficient ?
If I understand this then in the induction heating process the only losses are the copper coil circuit losses in the heater itself which are resistive losses and probably some capacitive losses since high frequency AC is employed correct?
But then I don’t understand where would the water or metal or whatever is inserted in the heater coil get its energy ? Because the induction heater is not a transformer with a shorted secondary but an inductor and the heating occurs due to the large eddy losses and other in a solid iron block or pipe which sits there instead of a laminated core as would in normal inductors like those for PFC in SMPS supplies right ?

So this is what I don’t understand , apart from the ordinary circuit copper resistance losses and reactive losses in the induction heater , what is the main source of power loss/transfer from the electrical supply to the heated iron in the middle of the coil , since it’s not a transformer with a shorted or high load secondary but an inductor and as far as I know inductors store applied AC as magnetic field which minus some losses should consume next to no power at all.
Or is it that the induction heater is simple on purpose made an extremely bad and extremely lossy inductor whose main purpose is to cause as much eddy currents as possible instead of eliminating them and by this cause the iron to heat rapidly?
But then I have another question , why do eddy currents cause the inductor to consume power instead of simply storing it as magnetic field and then delivering it back , is it because eddy currents cause the original magnetic field that made them to lose it’s power and so the current making that field is now consumed?

2. Dec 16, 2016

Staff: Mentor

The conductor is not perfect. It has resistance. $I^2R$ works just as well for circulating currents.

3. Dec 16, 2016

Baluncore

The flow of questions you ask have varying answers dependent on the configuration of induction heater you are considering.

The efficiency of water heating often comes from avoiding heat loss from stored hot water while hot water is not being used. Instant hot water heaters eliminate stored hot water losses.
Secondary savings are available if a long pipe does not have to be filled with hot water before a small amount of hot water is needed at a remote location. Heating the water where and when it is needed eliminates that inefficiency.

There are several different ways of applying induction heating to water. Will the water heater provide instant hot water or will you heat a large stored volume slowly? Will it use mains frequency or a higher frequency? If it uses a higher frequency then there will be losses in the frequency converter. If those converter inefficiencies cannot be used to preheat water they will be losses.

Wrong. Consider a metal tube with an insulated copper coil wound around it. When the coil is connected to AC mains frequency, the coil becomes the primary of a transformer, with the internal tube as a single shorted turn secondary. The high secondary current in the tube wall will heat the tube and the water passing through the tube.

The changing magnetic fields induce high currents that cause I2R heating.

Now consider replacing the AC supply and coil with a moving permanent magnet armature, (driven by some other energy than electricity). That may magnetically induce electric currents in the water pipe that will heat the contained water.

4. Dec 22, 2016

Robertphysics

I understand the applications of such a heater and that it would in many ways prove to be more economical etc , my main misunderstanding was about the working principle , I thought of it as an inductor but you are saying me it's a transformer , I kind of thought that a transformer must have a core which isn't a coil and then two coils are magnetically coupled to one another via that core , I didn't imagine that the second coil could be put in the place of the core and serve the same purpose as a coil and the core is eliminated altogether, but whne I come to think of it an ordinary inductor with a single coil and a laminated core is basically like a transformer without a secondary coil so the core simply acts as a magnetic field reservoir which can store large amounts of energy for very short time periods , is this a correct way of looking at it ?

So the induction heating isn't an inductor , it's just a transformer with a very lossy and short circuited secondary coil and without a core ?

is there a difference for the induction heater what kind of metal is used or can all metals or substances that conduct electric and magnetic field be heated up just at different times of heating because of the difference of the EM properties of the substance ?

5. Dec 23, 2016

Baluncore

The laminated magnetic core of a transformer is oriented so as to reduce eddy currents in the laminations. The thinness of laminations makes it possible for the magnetic field to penetrate the full volume of the lamination in half the period of the oscillation. That is a function of the skin effect depth in the core material. The insulation between the laminations allows the magnetic field to reach the full surface of all laminations, and also to prevent eddy currents between laminations.

When a coil is wound around a metal tube, the tube becomes a single shorted turn secondary. The coupling will not be high because there is no obvious core shared by the primary and secondary. But the magnetic wall material of the tube will have to play the part of the core as there is no other material to concentrate the magnetic field. So yes, an iron heated tube can be thought of as a magnetic core with high eddy current losses near the outer surface.

Material and skin depth decide how thick the tube should be for the power frequency used. You must select a material for the tube that is thick enough to handle the pressure while being thermally cycled. If the wall is too thick there will be a long delay between the application of power and the water being heated, which will make it difficult to for a controller to regulate water temperature.

The resistivity of the tube, allowing for skin depth will determine the primary turns count. If you know the secondary resistance R, and specify the wattage, W, then you can calculate the secondary voltage needed as Vs = Sqrt( W * R ). You know the primary voltage, Vp, is the supply voltage, so the number of primary turns will be Np = Vp / Vs.
For example, with a 230 VAC supply, 3kW heating with Rs = 0.001 ohm, Vs = Sqrt( 3 k * 0.001 ) = 1.732 V
Np = 230 / 1.732 = 133 turns.
Now you must find a material that works as a magnetic core and has a high enough resistivity to limit the secondary current.

6. Dec 24, 2016

Robertphysics

What makes me confused is that in a transformer the primary and secondary coils are always parallel to each other and the core that is in the middle of them is always with 90 degrees , so the flux cuts the coils at 90 degrees , but I can see how the flux changes in the pipe that is inserted in the induction coil but I fail to see how that flux can also induce an electrical current in that pipe , I thought it's just magnetic current aka flux, like in the transformer core.
I guess I fail to see how that pipe can form a shorted secondary because a shorted secondary in a transformer is a piece of conductor fully wrapped around a 90 degrees flux core atleast one full time if not more.