(See also Millikiln.)
There are lots of ways to heat things up to, for example, fire ceramics. Setting things on fire is probably the most fun way, but it’s inefficient and hard to control. You can also concentrate sunlight, absorb microwaves, or run electricity through refractory wires.
Suppose we reduce the kiln problem to make it easier: we have no convection, and we just want to heat the kiln to 660°, the melting point of aluminum, using one of the three non-fire methods. And let’s say we’re satisfied with a kiln about the size of a sake cup: 50mm across, 60mm tall, 118mℓ. How much insulation do we need?
(Note that in the sunlight case, you also need a sufficient solar concentration factor, which works out to be 42 suns, 4.2 million lux.)
Well, of course, that depends on our power budget. At 2500 watts the problem is easy; at 2 watts the problem is hard. As a reasonable medium, let’s take 500 watts, which is 2 amps here in Argentina, less than a toaster, hair dryer, or microwave oven, and about half a square meter of sunlight.
With such a small kiln, we aren’t going to get a very good approximation by pretending the thermal gradient is constant across our insulating wall, but let’s do it anyway.
Our top and bottom walls are 2·π·(25mm)² ≈ 4000 mm²; our outer wall is 2π·25mm·60mm ≈ 9400 mm²; total is 13400 mm². Our total thermal flux is 500 W/(11400 mm²) ≈ 37 kW/m². Over a difference of 640 kelvins, that’s about 58 W/m²/K. Expressing this as an insulation value (R-value) instead, we have 0.017 m² K/W. Loose-fill vermiculite is 17 m·K/W (according to Regenerator gas kiln, though I don’t know where it got that figure) so this requires about 1.2mm of vermiculite. Other insulators are in the ballpark.
So actually the constant-thermal-gradient approximation is going to be pretty good. I’d go with making it 10 mm or 20 mm of insulation anyway, just in case, and maybe doubling the linear size of the kiln.
500 watts into 118 mℓ works out to a bit over 4 W/mℓ, which is adequate for a few kelvins per second of temperature rise, if somehow the whole microkiln is full of solid and yet thermally homogeneous. Since that would reach full firing temperature in a few minutes, it seems like a more than adequate specific power.
In the sunlight case, say you go with a solar concentration factor of, say, 100, just to be safe, and you also go for a whole square meter, to get a kilowatt. Now you are faced with the problem that to squeeze the whole square meter into the kiln aperture, which ideally faces directly downwards, you must exceed this solar concentration factor; the kiln aperture can’t be more than a 500th of a square meter at that kiln size. This is another reason to want to increase the size of the kiln, and by more than a factor of 2.
So, revised design parameters: internal volume is a cylinder 150 mm across and 100 mm tall. Now we have 1.8 ℓ of volume, 0.018 m² of surface area on the bottom to cut a sun hole in, and 0.059 m² of total internal surface area to lose heat through by conduction. We use 20mm of insulating refractory for the walls and floor, which is kind of crappy as insulators go, so maybe its insulation value is only 8 m·K/W; this works out to 0.37 W/K, which means that at 640 K of inside-outside temperature difference we need 236 W of power input to maintain temperature. But we actually have 400 W of power, a comfortable safety margin, which works out to 230 mW/mℓ, which would heat water by about four kelvins a minute if the thing were full of water, but in the more reasonable case where it contains like 200 grams of stuff, we get 40 kelvins a minute, which gets us to full firing temperature (disregarding heat loss through the insulation!) in 15 minutes.
(Hmm, Henan Sinocean says their high-alumina insulating firebrick conducts 0.18 to 0.5 W/m/K, which I guess is insulating at 2 to 5.6 m·K/W, at a density of 0.3 g/cc for the best insulation up to 1 g/cc for the worst. These numbers are very high for insulating substances, barely better than ordinary bricks, but they are only a little worse than the Sheffield Pottery insulating firebrick numbers in Wikipedia and the BNZ Materials insulating firebrick brochure, and I probably can’t do better, though I might do worse. So I might need 40mm of insulation instead of 20mm.)
400 watts of sunlight is 0.4 m²; a solar concentration factor of 100 (theoretically capable of reaching 900°) then requires a hole of 0.004 m², which is nearly a quarter of the floor area, which is still probably okay. A solar concentration factor of 100 could also be thought of, in imaging optics, as 10× magnification, or, in antenna design, as 20dBi. Both of these are eminently achievable. You could even reach that level with a trough reflector, though it wouldn’t be the easiest way.
Fusión Refractarios in Avellaneda offers imported 0.81 g/cc 0.3 W/m/K 1427° firebricks for $87, which are 229×114×63mm; at $87/(2·229·114 mm²) at 31½mm thick, the cost is $1670 per square meter. I could maybe get two or four of these firebricks and try to stick them together.
Ten meters of Kanthal costs AR$70, whether it’s 0.2 mm, 0.3, 0.4, or 0.5, and it’s good to 1250° — not ideal for a pottery kiln, but it would do. It’s not clear which Kanthal this is; there are various grades of Kanthal; one is Kanthal AF, which is good to 1300° and has a resistivity of 1.39 Ω mm²/m, but doesn’t come in such narrow gauges.
How do I figure out how much wire I need? In theory you can get an arbitrarily large amount of power out of a fixed voltage source by putting an arbitrarily small resistance across it, since P = E²/R. However, in this case the crucial fact is not really the resistivity of the wire — as I thought it was — but its surface area! It needs enough surface area to emit the desired amount of power at its safe temperature — at these temperatures, almost entirely as radiant heat.
That’s why the Kanthal data page gives its fully-oxidized emissivity: 0.7. So how much wire surface area do I need to emit 400 watts with emissivity 0.7 at, say, 1100°? 0.7 σ T⁴ = 141kW/m², so I need 2800 mm² (0.0028 m²), about 5% of the total inner wall area of the microkiln.
That’s somewhat alarming! But, it turns out, not fatal.
Ten meters of 0.5 mm diameter Kanthal has almost 7900 mm² of surface area, and so thus, emitting at 1100°, it emits 1108 watts; at 1200°, it emits 1470 W. So really four meters of it (AR$28) should be adequate. (I’m a little skeptical that the price is actually correct, since all the other prices seem to be much higher.)
(Is it bad that I’m ignoring the radiant heat absorbed by the wire? I think it’s okay, since it’s hypothetically so much hotter than everything else.)
There is the engineering question of making the resistance low enough to get enough power out, without making it so low you draw too much power. In this case we want about 2 amps at 220 volts, which requires 110Ω. Four meters of 0.5 mm wire should be about 10 · 1.39 Ω / (π (¼mm)²), which is close — it’s 28 ohms. The easy thing to do is just go ahead and use a longer wire: at 7Ω/m, you need 15.5 meters. Or a narrower one: ten meters of the 0.4 mm wire should be just right, and has 12566 mm² of surface area.
At 150 mm diameter, that’s about 20 turns of a spiral around, which puts the turns about 5 mm apart up the walls if I don’t coil them. Or I could coil them and use, say, four turns.
The volume of refractory is, say, the difference between a 230mm diameter, 180mm height cylinder, and the 150mm diameter, 100 mm cylinder space within: 7478 mℓ - 1767 mℓ = 5711 mℓ. If we take a middle-of-the-road figure of 0.7 g/cc for the refractory, it’s 4 kg of clay.
To control the temperature, we need a thermocouple, probably AR$200, probably type K.
What would this microkiln design look like if I wanted to include safety margins to ensure that it would heat up and wouldn’t burn out? After all, many unexpected things will no doubt occur — perhaps the brick walls will leak, the Kanthal may be a counterfeit that burns out at 1100° or has the wrong resistivity, the bricks may conduct more than expected, and so forth.
I’m not sure what I would do to give it a temperature safety margin so that it wouldn't burn up if it hit 1400°, which is a temperature I’m extremely interested in. Except to make it purely solar, I guess, or arc-driven. The issue is the maximum service temperature of the wire. Kanthal A-1 is supposedly good to 1400°, and Kanthal APM is supposedly good to 1425°. Kanthal A-1 is available on MercadoLibre, but at a 15× higher price.