I’m thinking that, with modern electronics, the most expedient way to heat up a small pottery kiln to a controllable temperature might be induction heating of iron inside the kiln, using coils outside the kiln driven by a simple 300W ultrasonic inverter; this avoids the need for any exotic materials or components.
A small cylindrical space is the kiln proper, as in An electric furnace the size of a sake cup, perhaps 150 mm across and 100 mm tall; it is
enclosed by insulating refractory material. This is divided into a
bottom plate with a raised rim and a sort of inverted bucket that fits
on top of it; lining the upper part of the bucket is a steel or
cast-iron pipe (## in the cutaway diagram below) heated by coils of
copper wire (oo in the diagram) wound around the outside of the
refractory.
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oo| ## ## |oo
oo| ## ## |oo
oo| ## ## |oo
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With this arrangement, when the contents overheat and melt into the bottom of the kiln, they destroy only the bottom plate, which is the easiest part to replace; and, of the electrical components, only the inexpensive steel pipe is exposed to the high temperatures inside the kiln. The copper windings, in particular, are insulated from the pipe by a significant thickness of refractory material.
The raised border around the edge of the plate also serves to lengthen and crookedify the air path between the inside and the outside of the kiln, thus reducing thermal leakage, though probably at the cost of increasing wear when you open and close it.
The power delivered to the pipe can be on the order of 100 to 500 watts, and there isn’t any particular advantage (and maybe a bit of disadvantage) to its tendency to deposit primarily in the outer diameter of the pipe, due to the skin effect, so we can use frequencies as low as is convenient.
(As explained below in “materials and thermodynamics”, the walls and floor probably need to be proportionally thicker than is shown here.)
We can use the type-K thermocouples used for safety cutoffs for gas ovens and pilot lights; these are available at any hardware store and cost about US$3. To avoid creating a low-thermal-resistance path through the kiln wall, the lead can be passed through the wall in a spiral path from the inside to the outside. Accurately measuring the 50-millivolt DC signal from the thermocouple is somewhat challenging, as explored in Inductor thermocouple sensing, but also a well-understood electronics problem with off-the-shelf designs to solve it. We might have to turn the induction heater off when we’re taking temperature measurements.
As described in An electric furnace the size of a sake cup, refractory firebrick has
thermal conductance in the range 0.2–0.5 W/m/K. We can use ghetto
firebrick made by mixing used yerba mate with a ball-clay-based clay
body; as noted in file ceramics-notes, my previous experiments
firing it at 1020° burned out all the yerba cleanly (making a terrible
stench) and left solid ceramic results at ratios of 1:1, 3:2, 1:2,
1:3, 1:4, 1:6, and 1:8 in favor of the yerba; but the 1:6 and 1:8
results were noticeably friable and porous, so 1:2 (probably about
0.8 g/cc) or 1:3 (probably about 0.6 g/cc) is probably a safe bet.
These probably conduct about 0.3 W/m/K.
The surface area of the inner cylinder described above is 100 mm · 2 π 75 mm + 2 π (75 mm)² = 0.082 m², so it will lose 100 W at ΔT = 1000 K if the 0.3 W/m/K refractory averages 246 mm thick; this suggests that 300 W and 82 mm thick is a reasonable goal. You probably want to be able to burst above 300 W in order to hit temperatures at or above 1020 K in a reasonable time.
This gives an outer diameter of 314 mm and an outer height of 264 mm, for a total outer volume of 20.4 liters, 1.8 of which are taken up by the inner chamber; this leaves 18.6 liters of refractory material, about 11.2 kg of fired clay (and 13 kg of used yerba or coffee grounds or sawdust or whatever) if we assume 0.6 g/cc.
It’s probably worthwhile to put 5 mm or so of solid clay on the top of the plate to make it resist impacts and abrasion a little better.
Although you would clearly get better results if you fired the kiln in a larger kiln before using it, I think it will probably be usable even if you don’t do this — the steel pipe will reach a high enough temperature to burn out the yerba in its near vicinity, which will provide enough insulation to allow it to start heating up the inner chamber and burn out the yerba there. The ball clay I was using really needs to be sintered at 1020° to reach full strength, but full strength may not be necessary for this application. The yerba should eventually burn to charcoal at even 250° or 300°, which is probably adequate insulation.
It would be nice to prevent the yerba from stinking before the thing is dry, and maybe a small amount of copper sulfate or something would work as a biocide for this. This depends in part on how tolerant the neighbors are, though.
Nickel-plating the steel pipe might extend its life — steel doesn’t melt until at least 1130° and usually 1300° or more, and pure iron won’t melt until 1492° (and the mild steel we can most easily get will be close to that), but iron or steel in direct contact with air oxidizes rapidly above 200° or so. Nickel-plating isn’t totally trivial (nickel salts are fairly toxic) but can be done successfully with nothing more than salt, vinegar, and electricity.
Rudnev et al.’s Handbook of Induction Heating, 2nd Ed. explains that it’s possible to heat pretty much any metal (and also graphite) by induction heating, but there are significant differences; in particular, ferromagnetic metals have a much thinner skin depth δ in which the heat is deposited than non-ferromagnetic metals. For good heat transfer, the metal must be thick enough to be “electromagnetically thick” (p. 65), at least six times the skin depth δ.
This is substantially complicated when you are heating steel, because its skin depth increases by about a factor of 15 when it passes its Curie point (p. 64), which is around 720–770°, and also because the skin depth in ferromagnetic materials depends on the the magnetic permeability of the material, which is not constant but varies with the magnetic field intensity — perhaps 300μ₀ at room temperature and low fields, it may drop to 150μ₀ at room temperature and 20% of the saturation field intensity (Fig. 3.10, p. 60).
Now, in theory, we might not have to worry about this, because we don’t really care where the heat gets deposited inside the kiln; it will eventually conduct to where it’s needed. But at 60 Hz the skin depth in SAE 1040 steel is already 2.5 mm at low field intensity (Table 3.5 on p. 64). Six times that (to be “electromagnetically thick” and absorb nearly all the field) would be 15 mm, and 15 times that (when we pass 700°) would be 225 mm. Since I think it’s going to be hard to find or work with steel pipes that are more than 5 mm thick, it’s probably worthwhile to use a higher frequency; if we want a skin depth of a 90th of 5 mm, we need 100 kHz (0.06 mm in SAE 1040 at 10 A/mm and 21°, according to the same table).
(However, the penetration depth scales as the square root of the frequency, and the power delivered scales as a negative exponential of the number of skin depths, so even two or three skin depths would already be pretty good, so lower frequencies might be adequate. Probably not powerline 50 Hz, though.)
On p. 97 et seq. we have expressions for the electrical efficiency ηₑₗ and related quantities. It points out that you have losses from the resistance of the coil itself and losses from heating up nearby random metal objects, and it gives this approximate formula for heating a solid cylinder in an electromagnetically long solenoid coil made from electromagnetically thick copper tubing:
ηₑₗ = 1/(1 + (D'₁/D'₂)√(ρ₁/(μᵣρ₂)))
Here D'₁ and D'₂ are the inside diameter of the coil and the outside diameter of the cylinder, offset by their respective skin thicknesses, and ρ₁ and ρ₂ are their respective resistivities.
Now, the pipe isn’t a solid cylinder, and the coil here isn’t electromagnetically long or electromagnetically thick, but let's assume that the situation is close enough. We have the major disadvantage of having 82 mm of refractory between the coil and the pipe, so D'₁/D'₂ ≈ 2.1, which would give us ηₑₗ ≈ 0.3 absent other considerations; but our resistivity ratio ρ₁/ρ₂ is about 0.017μΩm/0.16μΩm ≈ 0.1 (Table 3.1, p. 54), and, at least at first, the μᵣ of the steel pipe ≈ 300, so our √(ρ₁/(μᵣρ₂)) ≈ √(0.1/300) = √.00033 = 0.018, so ηₑₗ ≈ 1/(1 + 2.1 · 0.018) ≈ 96%, though it might decline to 83% as the steel heats up and stops being ferromagnetic. On the other hand, the hotter steel will be even more resistive, so the situation might improve.
An interesting thing about this equation is that it doesn’t include any explicit dependence on the number of turns or the current, except indirectly through the lower μᵣ that comes with a higher field. I don’t yet understand what the implications of more or less turns are, except that more turns means more inductance and thus more impedance — easier to drive at low frequencies, harder to drive at high frequencies.
Suppose we decide on 20 kHz and 50 V as being relatively friendly numbers. To get 300 W out of 50 V we need 6 A, and thus about 8 Ω of impedance; with a 50% duty cycle we’d need 4 Ω. A MOSFET like the IRF540N (see My attempt to learn about jellybean electronic components) should work well for this if we have a source for the 50 V, but it can handle 33 amps; maybe a more interesting approach is to drive the low-impedance induction coil through a step-up transformer, perhaps stepping up voltage 4:1 from 12 V to 48 V and stepping current down from 24 A to 6 A. Perhaps even a higher induction coil voltage could work.
Okay, but what does that imply about our coil? Presumably we want its inductance to be high enough that, if the pipe weren’t conductive (maybe we replace it with laminated steel that blocks the eddy currents), it has a much higher impedance than 4 Ω at 20 kHz, maybe 40 Ω, and we want its dc resistance to be much lower than 4 Ω, maybe 0.2 Ω. AWG20 wire can reasonably carry 5 A and is 0.812 mm in diameter and 33 mΩ/m, so you could reasonably use 6 m of it to get 0.2 Ω, which would be just about 6 turns around the outside of the kiln; this would make it a sort of 6× stepdown transformer to the steel pipe, so you’d see 36× the resistance of the (skin depth of the) steel at the induction-coil terminals.
I’m going to hazard a guess, though, that even with 6000 mA going through it at 20 kHz, six turns of AWG20 wire isn’t going to be able to induce detectable heating in a 5-mm-thick steel pipe, especially once it passes its Curie temperature. Let’s suppose we’re starting up and so our skin depth in the steel is 0.14 mm, and take that as a reasonable approximation of how much steel is actually on our one-turn transformer secondary dissipating the eddy currents (to avoid having to do the integral). So we have a strip of steel 140 μm thick, maybe 50 mm wide (half the height of the inner chamber), and 470 mm in circumference, with steel’s resistivity of 0.16 μΩm. This comes out to 10.7 milliohms, so with only a 36× boost from the 6:1 turns ratio, we’re still at only 0.39 Ω. So if we want to use only 50 volts ac on the inductor to heat up the steel, we are going to need more turns.
Not as many more turns as you’d think, though — only about 19 turns, 18.7 m of wire, for which we need less than 20 mΩ/m to hit 0.2 Ω, thus AWG17 or bigger. But that’s for heating up the cold steel. Once it warms up and the skin depth increases 15×, we’ll need more turns to hit the same resistance — though at 700° carbon steel’s resistivity has climbed from 0.16 μΩm to about 1.1 μΩm, so only 2½× more turns. These extra turns would be detrimental at lower temperatures if we’re using a constant-voltage power supply, since the higher impedance means lower current and thus lower power, but maybe it’s best to optimize for low power at low temperatures, where heat leakage is low, and high power at high temperature, so 40 or 60 turns might be worthwhile. 60 turns is 59 m of wire, so to keep the resistance below 0.2 Ω, we need 3.4 mΩ/m, so AWG10 (2.6-mm diameter) or less.
(Alternatively you could complicate things and use two separate coils or a variable frequency.)
So at startup we have 10.7 mΩ in the skin layer of the steel, which looks like a 39 Ω resistance in the 60 turns of the winding at ac, although their dc resistance is 0.2 Ω. We’re putting 50 volts across it, although it’s really only 25 volts rms (a square wave has the same p-t-p and rms), so we drive 640 mA rms through the winding, producing 16 W, 99.5% of which gets dissipated in the steel inside the kiln. So maybe 60 turns is going a bit overboard with favoring the higher temperatures, or maybe I calculated something wrong; the power is low by a factor of 20.
Let’s try 30 turns: now we see 9.6 Ω in the winding at ac, so we drive 2.6 A rms through the winding — 65 W, which is probably reasonable. If the skin depth increases so that we’re using all 5 mm of thickness of the steel, its resistance would drop to 0.3 mΩ, except that the compensating resistivity increase gets us to 2 mΩ or so, which looks like 1.8 Ω in the winding at ac, thus 28 A rms (!!) and 1400 W (!!!) of which 10% is dissipated in the winding itself if it’s still 0.2 Ω (!!!!).
(Hmm, I think I’m still calculating something wrong, because the eddy-current EMF should be 30·50 V = 1500 V both when it’s cold and when it’s hot, so V²/R should only increase by a factor of 5.3, not 22.)
So we probably need to take some kind of measure against that. Just using thinner steel might help some, but increasing the frequency is another possibility. Also, you could adjust the duty cycle.
The leakage inductance of the coil presumably also imposes significant high-frequency reactance, which will limit the current you can push into it.
(Because of skin effect, it might be necessary to use thinner wire, perhaps two or more parallel windings.)
(15× may be understating it — this skin depth is atypically shallow because we’re using an atypically weak field and thus have atypically high permeability, so we will have an atypically extreme increase in skin depth when we smash into Curie.)
What scaling laws govern the dimensions of such a kiln?
Well, at a fixed inner volume, the necessary wall thickness is inversely proportional to the power. So if we could increase the power from 300 W to 600 W, we could use 41 mm of refractory instead of 82 mm of refractory, reducing the weight of the thing by more than half and its volume by a third. Voltages and currents would increase. Lower powers would require using proportionally thicker walls (and, perhaps, proportionally smaller inner chambers) which would eventually make the pipe too short to pick up much current.
The inner volume is proportional to the cube of the characteristic dimension, and the leakage surface area to its square. So, for example, a 10% increase in width, height, and depth would increase leakage surface by 21% and volume by 33%. Maintaining the same power would thus require increasing the wall thickness by that same 21%, which very quickly gets us to absurd dimensions. It might be more reasonable to increase the wall thickness proportionally (by 10%) and the power proportionally (by 10%) as well.
Electrical efficiency ηₑₗ stays the same as long as the wall thickness and inner diameter increase proportionally. Using the proportionally thicker metal we could accommodate more easily in a larger chamber would permit proportionally lower frequencies due to proportionally thicker skin depths.
So a small prototype, like the sake-cup-sized thing suggested in An electric furnace the size of a sake cup, might be a good thing to start with.