Venus has an average surface temperature of 735K. Would it be possible to harness this extreme temperature and use it to generate electricity? After all, that's the principle a radioisotope thermoelectric generator (RTG) runs on. And a large portion of RTG electric power goes to heating the spacecraft to keep its electronics and instruments functional. Solar panels could not be used to power a Venusian lander, because they would get destroyed by the atmosphere, and RTGs are expensive. But if there's a way to convert the atmospheric heat into electricity, we could have a longer Venus mission without having to worry about power.
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3$\begingroup$ The problem is the "differential". Bringing a chunk of ice along would hardly last longer than a charged battery with the same mass. It is just hot, not much of temperature gradients to use. The thick atmosphere evens temperature out on the surface, no eternally shadowed polar craters to use. But some crazy physicists maybe invent a balloon with a wire hanging down through different temperature layers and somehow... $\endgroup$– LocalFluffJul 12, 2016 at 17:42
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2$\begingroup$ @LocalFluff I was about to make the same comment: "What differential?" The lander will quickly heat up to match its environment. $\endgroup$– called2voyage ♦Jul 12, 2016 at 17:43
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$\begingroup$ I see, that makes a lot of sense. I guess I should have thought of that. Or read the Wikipedia page for Thermoelectric Generator more closely. $\endgroup$– PhiterosJul 12, 2016 at 17:44
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$\begingroup$ Thanks, but I wouldn't accept it as an answer! Lets hear from someone who really knows this stuff. I just copycat and compare what others say. I just answer, or comment, when I'm confident enough that others are confident enough that it is nearly true. And sometimes to try to explain stuff other say in a way which is more accessible to a more general public, because I'm on their side. I don't produce, I consume science. $\endgroup$– LocalFluffJul 12, 2016 at 17:50
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$\begingroup$ The simplest answer would be to develop a probe that could operate at 863F. There are ceramics and metals that can exist at this temperature. Vacuum tube technology might be use to create electronics that could operate at this temperature. It's hot enough to use materials that emit electrons directly rather than needing electric powered filaments. It might be necessary to use gold for wiring, because silver would be near melting point. So maybe there is a way to make a probe that can use the ambient temperature for at least part of the energy needed. $\endgroup$– Howard MillerJul 12, 2016 at 21:49
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High temperature in itself is useless. You need a temperature differential, just like the RTG exploits the difference between the temperature of the Plutonium and that of outer space. Kinetic energy (heat) flows from the high temperature region to the low temperature region and can do work on the way.
On Venus, there's no gradient. Everything has the same temperature. You want to keep your spacecraft a lot cooler than the ambient temperature, but that means expending lots of energy to pump the heat out. Running a Peltier element off the difference between the spacecraft and ambient temperature would only decrease the effectiveness of your cooling system, as you'd be letting the heat back in through the Peltier element.
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You might be able to do something like OTEC: https://en.wikipedia.org/wiki/Ocean_thermal_energy_conversion . Say you attach an anchor at the surface, and connect a balloon to it. Let the balloon float at an altitude of 5 km on a tether, which gives you a temperature difference of about 38 C (roughly twice what OTEC could get). The atmospheric pressure is about 70 times that on earth, so you don't need a very big helium balloon to support quite a large weight. On a logarithmic scale, buoyancy in the venerian atmosphere is more similar to buoyancy in water than buoyancy in earth's atmosphere.
The hard part then is the same thing that makes it hard to get OTEC to work. You have to operate a heat engine between heat reservoirs that are separated by a large distance, and the thermodynamic efficiency is going to be fairly low because $\Delta T/T$ is only about 5%. This is probably a far-future, high-tech scenario rather than something that could be exploited for probes in the foreseeable future.
Wind power from surface winds is probably a lot more practical, although just like with wind power on earth it would go away when there was no wind.
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2$\begingroup$ Also there's the problem of making a balloon capable of surviving in Venus's atmosphere. $\endgroup$– PhiterosJul 12, 2016 at 21:12
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$\begingroup$ @Phiteros: I suppose you're visualizing a mylar or rubber balloon. Buoyant forces on venus are 70 times greater than on earth, so a coffee can full of helium would float. $\endgroup$– user687Jul 13, 2016 at 13:35
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$\begingroup$ No, I understand that it would be a metal balloon. But the combination of sulfuric acid, intense heat, and pressure would mean that such a balloon, as well as the wire connecting it to the lander, would have to be extremely hardy. It's difficult enough to build a lander that can survive on Venus's surface. $\endgroup$– PhiterosJul 13, 2016 at 15:11
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I'm recently on "specific energy" checking spree, so let's test Ice Water Energy Storage for Venus application.
Let's see how much energy can we store as temperature gradient on Venus.
Surprisingly, Water is a material with about the highest specific heat across a range of temperatures. You might bump it by half a percent or so by additions, but let's take pure water for easier calculation.
Ice water: 2.108 kJ per kilogram per Kelvin Latent heat of melting: 334 KJ/kg @ 273.15K Liquid water: 4.184 KJ per kilogram per Kelvin Latent heat of vaporization: 2264.76 KJ/kg @ 373.15 Water steam: 1.996 kJ per kilogram per Kelvin
We cool the block of ice down to 77.2 K while using cold gas thrusters - Liquid Nitrogen, performing the reentry burn. That's the initial temperature.
Venus surface temperature: 735 K
First, heating the ice up to 0. 273.15-77.2 = 195.95K. Times 2.108 = 413KJ. Then melting it: 334 KJ. Then heating water: 4.184 times 100K = 418.4KJ Then evaporating: 2264.7KJ Then heating the steam: 735-373.15 = 361.85 times 1.996 = 722.2KJ.
4.152MJ per kilogram.
That's not bad. About the same energy density as thermite, and about 8 times more than Li-Ion batteries.
Unfortunately, that places it below Li-Ion batteries as electric energy source.
Thermocouples, though very reliable and long-lasting, are very inefficient; efficiencies above 10% have never been achieved and most RTGs have efficiencies between 3–7%. source
So, with generous 10%, we're at 0.4MJ - specific energy of consumer level alkaline batteries. Poor!
BUT! The rest of energy isn't wasted. It's used to cool the craft, like a sublimator. So the idea isn't entirely bad - it could buy the probe a good couple hours of life! It's just not good as power source alone.
...and of course my calculations are waaay off due to Venus pressure. The actual performance will be worse. But I can't really find the data for heats of water at 90bar, and what we have should give you a clue what we're dealing with.
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$\begingroup$ A thermocouple or thermopile is a type of heat engine, so its efficiency is governed by the 2nd law of thermodynamics. The efficiency is going to depend on the temperature difference. Your figures quoted from WP are for RTGs on spacecraft. For the Cassini RTG, the temperature difference is about 1500 K. In your example, the temperature difference starts out at about 700 K and goes down to zero. So if the average temperature difference is about 350 K, we would expect the efficiency to be about 1/4 as good as Cassini's. $\endgroup$– user687Jul 13, 2016 at 13:30
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$\begingroup$ You've estimated the energy density as energy per unit mass, but it would be interesting to see what it is per unit volume. Even at 90 atmospheres, a kilogram of steam takes up a lot of space. $\endgroup$– user687Jul 13, 2016 at 13:39
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$\begingroup$ @BenCrowell: As it expands, it would be vented obviously... still, it could be vented only after it achieves the target 700K, which would either mean excessive volume, or excessive pressure. $\endgroup$– SF.Jul 13, 2016 at 13:41
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$\begingroup$ Well, that increased volume could be used to drive a turbine or something. $\endgroup$– PhiterosJul 13, 2016 at 15:10
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$\begingroup$ @Phiteros: That's probably more viable than thermocouples, if reasonably light turbines can survive the spaceflight and reentry. Steam density at Venus conditions is 31kg/m^3, so 3% of density of water. That's a pretty good expansion coefficient that can surely be harnessed. $\endgroup$– SF.Jul 13, 2016 at 15:18