3
\$\begingroup\$

I want to know why an induction heater and an induction ring launcher don't produce the same effect. Both seem to use AC voltage to power a solenoid(which induces a voltage in a pipe) so why doesn't the induction heater launch the pipe inside it, or the induction launcher just heat the "projectile" pipe.

Induction launcher https://www.youtube.com/watch?v=szPwPLNN4IM

Induction heater https://www.youtube.com/watch?v=_v5Hg2zfLjs

\$\endgroup\$
4
  • 1
    \$\begingroup\$ I believe that you can actually get some molten metals to levitate inside the induction heater. I think you also tend to use higher resistance metals with induction heaters rather than lower resistance metals, whereas with a ring launcher you want lower resistance metals. \$\endgroup\$
    – DKNguyen
    Commented Jul 15, 2023 at 15:27
  • \$\begingroup\$ Interesting. Why don't they shoot out though, similar to the induction launcher? In fact, shouldn't it shoot out quicker due to the frequency being higher which would induce a greater current and as a result magnetic field? I have a small induction heater set up using a ZVS driver(50khz) that can heat copper pipes. After watching the video about induction launchers I was wondering why mine doesn't do the same. Thx for the help. \$\endgroup\$
    – Yassin
    Commented Jul 15, 2023 at 15:35
  • 1
    \$\begingroup\$ Higher frequencies may produce a higher rate of change of flux but, the "target's" inductive reactance is also proportionately higher and therefore it circulates the same current even though the frequency is higher @Yassin \$\endgroup\$
    – Andy aka
    Commented Jul 15, 2023 at 15:56
  • \$\begingroup\$ Is there any way to mitigate inductive reactance? \$\endgroup\$
    – Yassin
    Commented Jul 15, 2023 at 16:38

2 Answers 2

4
\$\begingroup\$

Both seem to use AC voltage to power a solenoid(which induces a voltage in a pipe) so why doesn't the induction heater launch the pipe inside it, or the induction launcher just heat the "projectile" pipe.

Both effects happen but, there's a vast swathe of middle ground that reduce one effect and enhance the other. To get a launcher you want losses in the target ring to be very low so that it's reactionary magnetic field is quite high. This causes a launch and very little heating.

In the induction heater, you want losses in the "material" to be quite high so that heating takes place and, accordingly, mechanical reaction is quite low.

\$\endgroup\$
6
  • \$\begingroup\$ Ok. I understand. This seems interesting. I have made an induction heater and want to switch it to an induction launcher. How can I optimise the target ring to launch instead of heat. I have a copper pipe with 0.5mm wall thickness that is 10 mm outer diameter. \$\endgroup\$
    – Yassin
    Commented Jul 15, 2023 at 15:59
  • \$\begingroup\$ I have never made one and, if I were in your position I'd experiment a bit more and try and see what others have done. I notice there is scant information on-line about this so I can't off further advice other than to experiment. I'd be looking to ensure that the pulse of current into the "primary" is over a hundred amps for instance but, because I've never made one it's guesswork. Also make the projectile as light as possible and possibly thicken up the ring wall thickness. \$\endgroup\$
    – Andy aka
    Commented Jul 15, 2023 at 16:11
  • \$\begingroup\$ Ok. Thx for your help. I could increase the current in the coil by altering the turns ratio of the transformer in the ZVS driver. With regards to wall thickness, wouldn't a thinner wall be better due to skin effect at 50khz or am I mistaken? \$\endgroup\$
    – Yassin
    Commented Jul 15, 2023 at 16:42
  • \$\begingroup\$ Losses in the projectile need to be reduced and, the only way to do that is to make the walls thicker. Anyway, the original question is now answered I believe. If you want to make something that fires a projectile my advice is raise a new question. \$\endgroup\$
    – Andy aka
    Commented Jul 15, 2023 at 17:51
  • \$\begingroup\$ Ok, thx a lot for your help. \$\endgroup\$
    – Yassin
    Commented Jul 15, 2023 at 18:08
3
\$\begingroup\$

Three substantial differences:

  1. Induction heaters generally run at higher frequencies, and operate on diverse materials.
  2. Propulsion systems generally run at lower frequencies, and operate on low-resistance materials, optimizing for mechanical efficiency.
  3. As a subset of propulsion systems, launchers of the type shown are pulsed devices, whereas induction heaters and rotating/linear motors are continuous-wave devices.

For a given amount of power delivered to the material (voltage * current induced in it), the flux density, in the magnetic field doing that induction, is inversely proportional to frequency (Faraday's law). Lorentz force (magnetic component) is in turn proportional to flux density: \$\vec{F} = \vec{v} \times \vec{B}\$. Thus, force is inversely proportional to frequency, at a given power level and for given material and geometry.

There are, of course, induction heaters operating at lower frequencies, and at such power levels, that they deliver significant forces, whether to cause material displacement (requiring clamping to constrain the work, or that causes stirring of a melt), or enough to overcome gravity and levitate or launch a workpiece wholesale.

Note that there is no threshold between cases, it is a continuum between low and high power levels, low and high resistivity materials, and low and high frequencies. Whether some combination of those variables is enough to cause some given effect, is another matter, but the underlying physics at work do not care about any such threshold.

As for pulsed operation, consider this: a typical implementation might use a capacitor discharge circuit into a coil of wire. The coil and workpiece are positioned such that there is an easy path for them to push apart (whereas a workpiece in the middle of a solenoid coil for example, won't go much of anywhere, if centered properly that is). A typical example is a pancake coil and a plate/ring; another is the cored solenoid in the video, with the ring positioned so that there is more flux density behind it. (Where this position lies, compared to the primary coil itself, depends on the shape and size of the core; it's not obvious from a glance at the video, how the core is shaped inside the base.)

A capacitor discharge circuit can produce massive currents, developing flux densities of several tesla peak even without the help of a magnetic core (indeed, no materials saturate above 2T, so a core is only of modest value at these levels anyway!). Equivalent frequencies can be in the 100s Hz to 10s kHz, which may be on the high side, but this is made up for by the sheer violence that is possible. When many kJ of energy is involved, the peak power level can be near gigawatts, and the forces can be enough to radially pinch a small disc -- a "quarter shrinker" machine for example.

Compare to a typical induction heater circuit: the available current is, at most, either what is available from the AC power supply directly (current-sourcing type), or from the resonant capacitors and Q factor (voltage-sourcing type).

The common ZVS circuit is of the voltage-sourcing type. The push-pull transistors act as a synchronous power mixer ("mixer" in the RF sense), converting supply voltage to AC output voltage. Conversely, this converts AC current flow into DC supply current, so that load resistance manifests as DC resistance at the inverter's supply; thus, DC power is drawn when AC power is dissipated into a workpiece (as must necessarily be the case).

Thus we can conclude, the Lorentz force available from such a circuit, is limited by supply voltage, capacitance, and coil geometry.

To be able to use such a circuit as a launcher, as much power must be drawn from the supply as is delivered to the load -- whether by resistive dissipation or mechanical work. Presumably, several kilowatts are desired, to have a satisfyingly impactful demonstration? This will require quite large transistors, and a power supply capable of at least as much peak power (a regular power supply unit won't do, but a battery perhaps could). The elephant in the room, however, is the capacitors required to reach low enough frequencies to be useful. Which have the effect of increasing the resonant current (the ratio of AC voltage to circulating current is given by \$Z_0 = \sqrt{L/C}\$), but also reducing the Q factor (because the coil has about the same resistance but proportionally less reactance). Eventually you're putting far more power into the coil's resistance, just to make a pitiful magnetic field, than you are into actually pushing the work (and needless to say, you want to avoid heating the work; use copper or aluminum). The coil resistance can be mitigated by adding a magnetic core to the system (which I believe is what the unit in the video clip uses), but at this point you're just making a transformer with extra steps, and you can just plug it into the wall (a ready source of 50/60Hz AC) briefly to do the job.

So, suffice it to say, there isn't much meaningful connection between a toy induction heater module, and something useful for propulsion. While the effects are identical, the circuit and geometry changes are substantial enough that a different engineering solution becomes apparent.

\$\endgroup\$
12
  • \$\begingroup\$ Hi, thx for your response. You mention flux density is inversely proportional to as frequency. Faraday's law is is ε=−N * (dΦ/dt). (dΦ/dt) is the frequency of the alternating magnetic field, so therefore is it not proportional to frequency and inversely proportional to change in time? \$\endgroup\$
    – Yassin
    Commented Jul 15, 2023 at 19:24
  • \$\begingroup\$ @Yassin For some given EMF \$\mathcal{E} = A \sin \omega t\$, and given geometry so that \$B \propto \Phi\$, then \$\Phi\$ and therefore \$B\$ goes as \$\int \mathcal{E} dt = -\frac{A}{\omega} \cos \omega t\$. \$\endgroup\$ Commented Jul 15, 2023 at 22:20
  • \$\begingroup\$ From my knowledge, what this really means is that the rate of change of the magnetic field (or the flux) with respect to time depends on the frequency of the driving EMF. In other words, a higher frequency means the EMF is oscillating more rapidly, which means the magnetic field is also changing more rapidly in response. But this doesn't directly determine the strength or amplitude of the magnetic field. So while ω does appear in the denominator in this context, it's not accurate to say that the magnetic field strength is inversely proportional to the frequency. \$\endgroup\$
    – Yassin
    Commented Jul 16, 2023 at 8:52
  • \$\begingroup\$ @Yassin I defined EMF: "For a given amount of power delivered to the material (voltage * current induced in it)" \$\endgroup\$ Commented Jul 16, 2023 at 12:55
  • \$\begingroup\$ Why would it be inversely proportional though? What is the specific flaw in my reasoning? Your answer seems to contradict the one by @AndyAka, could you explain why his is wrong as well. This should clear any confusion. Thx. \$\endgroup\$
    – Yassin
    Commented Jul 16, 2023 at 14:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.