Presentation¶

Some notices and reflections to build an induction furnace.

You can view the docs on readthedocs and clone the sources from github

Getting started¶

In a bash-terminal:

git clone https://github.com/charlyoleg/induction_furnace.git
cd induction_furnace
npm install
npm run install_py
npm run the_docs

Presentation¶

The metal is an interesting material because of its rigidity. This property let us make mechanical parts that are closed to an ideal solid, which shapes and dimensions do not vary regardless of the applied strength and temperatures.

The high rigidity of the metal is also an issue when it comes to form the shape of the parts. Direct strength become inefficient. Two methods let us bypass the rigidity property of the metal:

• the electro-erosion takes advantage of the electrical conductivity of the metal
• the fusion (i.e. foundry) takes advantage of the relative small quantity of energy required to melt a metal

For years, the difficulty was to bring the thermal energy to the metal at high temperature. But nowadays, thanks to the induction technology, it is quiet easy to bring this energy as the heat is generated within the metal.

Induction overview¶

An LC-circuit generates an alternative magnet field, which generates Foucault current (a.k.a. Eddy current), which generates heat thanks to the Ohmic law.

The Ohmic law:

Wth : the thermal power generated by a current

The Foucault current are limited to the area closed to the surface of the metal if the frequency is too high. This is the skin effect

Thickness of the skin effect:

To get enough thickness for the Foucault current, we work with a frequency below 1kHz. So we can melt all metal. Also, at high temperature, the iron loses its high magnetic permeability, which makes it behave closer to the copper regarding the skin effect.

LC circuit¶

To create an alternative magnetic field, we need an oscillating circuit like the LC-circuit.

Formula:

Inductance sizing¶

The volume of 1kg of iron is 127000 mm^3. In chip, it may requires the double volume, that means 250000 mm^3 or 50*50*3.14*30, a cylinder of radius 50 mm and height 30 mm.

Cylindrical inductance:

Considering a cylindrical inductance of radius 80 mm (diameter: 160 mm) and a height of 30 mm. If the wire loop are distance of 5 mm, we get 6 loops for one way. For both ways, we get 12 loops:

Capacitor sizing¶

To reach a resonance frequency of 1kHz, we need a capacitor of:

Power transfer¶

We want to transfer the energy from electricity to heat up to 1kW i.e. 1kJ/s. We have 1000 oscillations per second.

If each oscillation transfers completely its energy in one oscillation, we need 1J in the LC-oscillation:

If 10% of the energy is transfer per oscillation, we need 10J oscillating in the LC-circuit:

u = 307 (V)
i = 408 (A)

If 1% of the energy is transfer per oscillation, we need 100J oscillating in the LC-circuit:

u =  973 (V)
i = 1291 (A)

Presentation¶

When looking for a rigid substance, metal is a good choice compare to stone, wood, glass, ceramic, plastic. The rigidity is relevant when designing parts with the solid mechanics. In this matter, shapes and dimensions are considered invariant, regardless of the strength, pressure, collision and temperature conditions.

Metal overview¶

An iron-part with a width of 10 cm gets longer of 11.8 um when the temperature increases of 10 °C.

An iron-part under strength with a width of 10 cm gets longer of 200 um before getting a plastic deformation.

Energy for fusion¶

At 101325 Pa (normal atmospheric pressure)

Material	Cp [solid]	Fusion Tc	Fusion heat	Cp [liquid]	Boiling Tc	Vaporization heat	Cp [gaz]
Units	J/(kg.K)	°C	J/kg	J/(kg.K)	°C	J/kg	J/(kg.K)
Water (H2O)	2108	0	334000	4187	100	2264705	1996
Nitrogen (N2)		-210	25700		-196	200000	1025
Iron (Fe)	444	1538	247134	820	2861	6214000
Aluminium (Al)	897	660.3	396590	1180	2519	10859000
Copper (Cu)	385	1084.6	206170	490	2562	4721000
Zinc (Zn)	380	419.5	112402		907	1819000

Energy to melt 1 kg of metal:

Water-steam   :  100 * 4187 + 2264705 = 2683405 J  <=> 0.745 kW/h
Iron          : 1538 *  444 +  247134 =  930006 J  <=> 0.258 kW/h
Aluminium     :  669 *  897 +  396590 =  996683 J  <=> 0.276 kW/h
Copper        : 1084 *  385 +  206170 =  623510 J  <=> 0.173 kW/h
Zinc          :  419 *  380 +  112402 =  271622 J  <=> 0.075 kW/h

Reminder:

\[1 bar = 100000 Pa\]

\[1 Pa = 1 Nm^{-2} = 1 J m^{-3}\]

\[1 cal = 4.1855 J\]

\[1 mol = 6.022 * 10^{23}\]

Alumina capacitor¶

Commercial ultracapacitors¶

Conclusion¶

The naive approach with alumina insulator provides poor result compare to the current state of the art of the industry.

• Specific energy ratio:

• Energy density ratio: