Normally we think of steam engines as being fairly slow, needing minutes to hours to get up a head of steam. But what if our heating elements and the spaces between them are really thin? They could have the fractal-heat-exchanger structure I described in Dercuano, where a very large and very rumpled surface pierced with many thin, short “capillaries” permits the transfer of a great deal of thermal energy into water or another working fluid very quickly.
Suppose you have 1-mm-diameter heating elements made of copper pierced or separated with 1-mm-diameter water channels. What’s the time constant of the relaxation of this thermal system when the copper is much hotter than the water?
Copper has a thermal conductivity of 401 W/m/K, liquid water of 0.5918 W/m/K, and steam around 0.01 W/m/K. My first thought is that a crude approximation is that the water in the middle of the passages is insulated from the heat from the copper by about 200 microns of water plus an insignificant amount of copper, which would give you about 3kW/m^2/K. Each passage, if circular, has a circumference of about 3 mm, so that’s 9 watts per millimeter of passage per kelvin. A millimeter of passage has on the order of 1 mg of water in it, and 9 watts would heat a milligram of 4.184 kJ/kg/K water at about 2000 kelvins per second, so under these assumptions the characteristic relaxation time is about half a millisecond.
That is, if you have a temperature difference of 1000 K, you’ll have 9000 W per mm of passage, which will be heating the mg of water in that mm by 2 megakelvins per second, so every microsecond the water closes 1/500 of the remaining temperature gap.
However, if you have Leidenfrost stuff going on, the water will start to be insulated from the walls by a layer of steam, which will slow the process down by a factor of 60 or so, up to timescales of 30 ms or so. On the other hand, if you have turbulent flow that recirculates the mass flow of water or steam from the center of the passage to the walls and back on submillisecond timescales, the process will accelerate further. On the gripping hand, hot steam condensing onto cold water also accelerates heat transfer, which is how nucleate boiling transfers heat.
So I think millimeter galleries being about a millisecond is probably about right. Maybe in reality it’s a tenth of a millimeter or something because it sure takes a lot more than 10 milliseconds for hot water flowing through a 10-millimeter pipe to cool down to the temperature around the pipe, but maybe that’s mostly because the concrete around the pipe is less conductive than copper.
Copper has some advantages for this kind of boiler thing, although it’s not as strong as some other metals. It has great conductivity and tends not to corrode until well above boiling.
PV = nRT, and the molar mass of water is like 18 grams. At one atmosphere and 0° a mole should be 1 mol RT/P = 22.4 liters and 1/18 mol should be 1.25 liters. At 100° it’s 1.70 liters, so 1 cc of water boils into 1.7 liters at that temperature, but it’s not doing any work at that point, since it’s at 0 gauge pressure. At 250° at 1 atm it’s 2.38 liters.
Suppose the steam is pushing a piston 10 cm in a cylinder of 20 mm diameter, thus 314 mm². That’s 31.4 milliliters of steam volume. For it to do 1000 J of work over that distance, it needs 10 kN, which means averaging 31 MPa, 314 atmospheres. This would need to be supercritical steam; water’s critical point is about 22 MPa at 650 K (377°). At 250° its vapor pressure is only about 3 MPa, so it would only do about 100 J of work, which is still pretty okay.
At 250° at 3 MPa the ideal gas law gives us a volume of 80 ml for 1 g of water. PV/RT is about 390 mg, so if we have more water than this, some of it will remain liquid. Say 500 mg.
Heating 500 mg of water to 250° should cost about 0.5 g × 4.184 J/g/K × 230° = 480 J, though I guess the specific heat goes down a little at higher temperature, and then boiling it (normally 44 kJ/mol) only takes about 32 kJ/mol or 1.8 kJ/g, so about another 900 J, for a total of about 1.4 kJ. This is not a very efficient steam engine, under 10%.
If we want the 1.4 kJ to be lost in the sensible heat of some copper as it drops from 300° to 250°, well, copper’s specific heat is 0.385 J/g/K (at room temperature anyway), so that’s about 73 g of copper. This is an unreasonably large amount of copper to put in contact with 500 mg of water, so a better approach may be to maintain the temperature of the copper at 300° (to keep it from oxidizing) by running electricity through it as the water boils. Alternatively, we could use a smaller amount of copper at a higher temperature and just sacrifice it; copper boils at 2562°, and so we’d only need 1.6 g of copper at that temperature to boil the water, though at that temperature the specific heat might be significantly lower. And of course the engine won’t work for many iterations.
Dumping 1.4 kJ into the copper electrically during the millisecond of boiling would require 1.4 MW. If we use 2000 V, above which we start to encounter special problems, we need 700 A, requiring 2.86 ohms or less to avoid needing even higher voltages. If we dump this in from a capacitor, the capacitor needs to have an ESR that’s not too large compared to those 3 ohms.
This is pretty challenging. AVX’s “BestCap” low-ESR supercaps include, say, the BZ01CB153Z_B, which handles 12 volts, 15 millifarads, and 420 milliohms (at 1 kHz, which is a good speed for this). This is 28 mm × 17 mm × 6 mm, more or less, and an energy capacity of about 1.1 J. You’d need 1300 such caps to hold the 1400 J, totaling 3.7 liters; a bit awkward.
It gets worse, though. That’s a time constant of 6 milliseconds, so in 1 ms you can only get about 15% of the energy out of it. Other supercaps are similar. Aluminum electrolytics are faster but even bulkier.
No, resistance heating is not the way to go. The right way to flash-boil the water is with a packed bed of little balls of something inert, like quartz or aluminum oxide or porcelain. You can heat it to the requisite temperature by blowing hot air over a Kanthal or Nichrome filament and then over the packed bed, then pump in the water once the packed bed is hot.
Granite’s specific heat is 0.79 J/g/K, fused silica 0.703, crystalline quartz sand 0.835. Alumina (thermal conductivity 30 W/m/K, still far better than the water) is 0.96 J/g/K, I think, and doesn’t melt until 2277°. At 1100° the 1400 kJ might need 1.3 grams. 1-mm balls of fused silica might withstand the thermal shock better than the stronger but more expansive aluminum oxide, though, even though alumina conducts heat better: granite’s thermal conductivity is about 1.8-3.8 W/m/K, fused silica around 1.4 W/m/K, sapphire closer to 27. Embedding copper wires below the surface might help.
Such oxides could perhaps also improve the efficiency and compactness of the engine by withstanding higher temperatures without corroding. If one gram of steam is at 600° instead of 250°, then nRT/P at one atmosphere would be 4 liters instead of 2.38 liters; at 3 MPa it’s 134 ml. If allowed to expand to only 31.4 ml, nRT/V gives us 12.8 MPa; if this pressure were constant throughout this expansion, because the steam is generated exactly as fast as it expands, it does 400 J of work. If the pressure is higher at first, because steam generation finishes earlier, it could do more.
I haven’t calculated here the energy needed to heat the water and then steam to this temperature, but the 1.4 kJ above was to heat half this amount of water to 250° and then boil it off at that temperature, so probably it’d be around 3 kJ.
How do you power the resistance heater? Suppose you have four 18650s (weighing 250 g or so) and you use the 1800-mAh 15C types sold for quadcopters. Each can provide nominally 27 A at 3.7 V, which is 100 W. So all four together can provide 400 W, thus providing these 3 kJ of heat over 7.5 seconds. So they could activate this piston every few seconds.