4 ms·
The Carnot efficiency of a heat engine is 1 - T_low/T_high. When operating between 305K and 300K (room temperature(ish)), for example, this is 1 - 300/305, or a
by theothermkn 8y ago
The Carnot efficiency of a heat engine is 1 - T_low/T_high. When operating between 305K and 300K (room temperature(ish)), for example, this is 1 - 300/305, or about 1.6%. In other words, the absolute most amount of energy a device operating over a 5 K temperature difference at room temperature is pretty dismal. Just because their band gap (for lack of a better term) is small does not make this device "useful."
For all stories about heat engines running between smaller temperature differences, the smaller the temperature difference, the less excited you should be. For example, I have a device that reaches the Carnot efficiency across temperature gaps of 0K! HINT: It's a rock. Or a paper clip. Or a wet tissue. Or...
- beambot 8y agoWet tissue usually has a greater temperature delta due to evaporative cooling... But that's just pedantic; everything else was spot on. ;)
- mrfusion 8y agoIt could still be really useful for remote sensors and who knows what else. A lot of places a little power is all you need.
- JackFr 8y agoYeah -- interstellar space, we're talking 67.5% Carnot efficiency.
- foota 8y agoI don't think heat engines would work in space generally due to the low mass of stuff around you that can absorb heat.
- Baeocystin 8y agoThey work great as long as you have a source of heat and a good radiator. You don't need mass as a heat sink.
- foota 8y agoInteresting, I wasn't thinking about that. Thanks
- theothermkn 8y agoAt a practical level, your initial instinct was right. In space, nobody can feel you convect. Essentially all of the waste heat has to be disposed of via radiation, and this is orders of magnitude less effective than convection at the temperatures that any known material can withstand. Worse, you don't have to radiate "just" the waste heat. Eventually, all of your electric energy becomes heat, too. In other words, all of the energy you extract from your decay source will end up as heat that you have to get rid of via radiation. (Well, you could wind a giant spring, I suppose. But, a spring that large would be a better radiator than energy storage device.)
- woodandsteel 8y ago>In space, nobody can feel you convect. I know I'm on HN, as opposed to one of the political sites I read, when I run across that sort of remark.
- Baeocystin 8y agoAll that being said (and true!), you also have a very cold sink to reject waste heat to, whether you want to or not. The end result is that heating is as much an issue as cooling. Thus spacecraft usually wind up with blankets and radiators, both. :) Here's a detailed look at the thermal control system for the ISS, for the curious: https://www.nasa.gov/pdf/473486main_iss_atcs_overview.pdf https://www.nasa.gov/pdf/473486main_iss_atcs_overview.pdf
- JackFr 8y agohttps://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_ge...
- acqq 8y agoFrom there: “RTGs use thermoelectric generators to convert heat from the radioactive material into electricity. Thermoelectric modules, though very reliable and long-lasting, are very inefficient; efficiencies above 10% have never been achieved and most RTGs have efficiencies between 3–7%” Also interesting, the price of RTGs: "Plutonium 238, which is used to power Radioactive ThermoGenerators such as the ones that power the Voyagers, Galileo, and Cassini probes has not been produced in the US for 25 years, but NASA and DOE have budgeted $50 million to restart production of about 2 Kg per year for 5 years…an average cost of about $5 million per Kg. That is expensive stuff!" One SNAP-27 unit, the type used in Apollo missions, providing only around 70 W, had around 4 Kg, which means its Pu price alone would be at least around 20 million, if it would be produced now.
- namibj 8y agoSo, a kind of thermal solar panel that could possibly scale down/cheap better than a solar panel?
- chiph 8y agoOne of the batteries that Dad worked on back in the 1950's was for the Union Pacific railroad. It was intended to be loaded into an underground vault along the tracks in remote areas, and would provide power to run signals through most of a winter. If modern LED power draw is low enough, and the new thermocouple is efficient enough, they could run the signals off the temperature difference between the mountain air and the temperature of the earth about 10 meters down.
- windows_tips 8y agoRequiring a 10C difference to generate power isn't useful when you only have a 5C difference.
- vpribish 8y agoparent was implying that a 5C difference is practically useless, if technically possible, so don't bother popping open the champagne when it's announced.
- namibj 8y agoIt's not useless. There are not that few IOT style applications where one could extract significant energy from e.g. atmospheric temperature cycling during the day relative to a thermal mass. This would not require sunshine, only that the temperature over the day changes about 10K between minimum and maximum. Or with a large enough thermal mass it could use this cycling over multiple days, which might be more reliable. It also has the great benefit of not requiring something fragile exposed to the sun. This makes the feasible reliability potentially much higher, depending on how good this thermoelectric generator lives.
- aidenn0 8y agoThere are many devices for which power consumption is measured in the microwatts. This could certainly be useful for them.
- mikeash 8y agoGenerating useful energy from large temperature differences is old hat, though. 1.6% efficiency could be plenty for some tiny device that needs very little power.
- namibj 8y agoErr, it's 1.6% of heat flux gets turned to electricity in a perfect device. This device is not perfect, and you are likely lucky to turn 1% of heat flux into electricity.
- nanomonkey 8y agoThis is not a heat engine, as there is no cyclical nature and it's not turning the heat into mechanical energy, therefore the Carnot efficiency need not apply.
- Scaevolus 8y agoCarnot efficiency still applies, even if the thermoelectric effect doesn't have a mechanically cyclical nature. http://www.pnas.org/content/112/27/8205 http://www.pnas.org/content/112/27/8205
- nanomonkey 8y agoThe article you linked doesn't follow your statement, in fact it suggests a different theoretical limit, and only suggests a engineering formula for high temp TEGs to mesh up with the observed results due to the Thomson effect on heat transfer.
- Scaevolus 8y agoThey suggest a new formula because the "the conventional efficiency formula (Eq. 1) often misleads and gives rise to an impractically high efficiency prediction". > The [thermoelectric] conversion efficiency by Eq. 1 is the product of the Carnot efficiency (ΔT/Th) and a reduction factor as a function of the material’s figure of merit Z = S^2 ρ^−1^κ−1, where S, ρ, and κ are the Seebeck coefficient, electrical resistivity, and thermal conductivity, respectively.
- theothermkn 8y agoInterestingly, while the Carnot Limit is derived using heat engines, it applies to any "device" that extracts energy from a temperature difference. If it were not so, then one could place, for example, a thermoelectric device between two energy reserves at different temperatures and use it to power a Carnot heat pump to drive the temperature difference higher, or to sustain the temperature difference and extract work. This is a perpetual motion machine, violating the thermodynamic laws of which the Carnot Limit is a manifestation. (It is a general result, obtained from hypothetical devices.) One can sort of "feel" this result by imagining the voltage pushing a cloud of electrons around a circuit, producing shaft work in an electric motor, and/or heat if there is any electrical resistance.