Current nuclear power plants are essentially an enhanced version of a kettle, which seems like a stupidity caused by a lack of other options. We heat the water which turns to steam which rotates the turbine, which is total waste of energy due to the several conversions.

I googled a bit and found that actually there exists the thermoelectric effect which allows for converting heat to electricity directly. Yes, I didn't know about it until today. ;)

Is it possible to turn the heat from the nuclear reactor directly to electricity? Have there been any attempts to do it?

I am not asking why we do not use it currently, my question is about whether it's possible in principle and whether anyone has tried it.

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    $\begingroup$ If you didn't know about thermoelectric coolers until today, you should hold off calling the designers of existing nuclear plants "stupid". Steam turbines are actually very efficient. $\endgroup$ Commented Jun 26, 2015 at 17:28
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    $\begingroup$ ..as are the synchronous alternators connected to them. $\endgroup$ Commented Jun 27, 2015 at 17:59
  • $\begingroup$ @whatsisname Indeed, I think this week I learned a lesson to think twice before treating people as ignorant. 0_o' If only I RTFM more about steam turbines I probably would not ask this question at all... $\endgroup$
    – hijarian
    Commented Jun 28, 2015 at 17:38
  • $\begingroup$ Industrial electric machines (motors, transformers and generators) are paragons of efficiency. Given the amount of power that they deal with, if they aren't, they'd melt. $\endgroup$
    – Nick T
    Commented Jun 30, 2015 at 3:25
  • $\begingroup$ In fact, a nuclear power plant with a thermoelectric power generator WAS built by the russians: BES-5, and used in the RORSAT spy sattelites. The americans built the SNAP-10A. $\endgroup$
    – ntno
    Commented Jul 25, 2015 at 17:35

5 Answers 5


The efficiency of a thermoelectric generator is around 5 - 8%.

The efficiency of a large steam turbine power plant aproaches 40%.

In fact the thermodynamic efficiency of a large steam turbine power plant is over 90%, so it's about as efficient as anything could be. The maximum possible efficiency of a steam driven engine is given by the idealised model called a Carnot engine. The efficiency is ultimately limited by the difference in temperature of the hot and cold ends of the engine, and modern power plants get pretty close to this theoretical maximum.

Thermoelectric generators tend to be used only where other restrictions force their use. For example the Curiousity rover uses a thermoelectric generator with an efficiency of about 6%. The lower efficiency is balanced out by a lack of moving parts, and of course the non-availability of water on Mars from which to make steam.

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    $\begingroup$ So, we use the steam turbines because it's actually the most efficient known method of generating electricity from heat? OK, then, no more silly questions. I thought in the other way. :) $\endgroup$
    – hijarian
    Commented Jun 26, 2015 at 10:51
  • $\begingroup$ The steam cools down so much that water droplets can form in the engine which can have an abrasive effect on the blades. $\endgroup$
    – Name
    Commented Jun 26, 2015 at 17:35
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    $\begingroup$ The Voyager probes also use thermocouples attached to a hunk of plutonium to generate a few hundred watts of electricity at 30 volts. It is a very small amount of power, but it lasts for a long time. $\endgroup$ Commented Jun 26, 2015 at 18:16
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    $\begingroup$ @EricLippert, the plutonium lasts for a long time. However, the radiation causes the thermocouples to wear out after only a few decades. $\endgroup$
    – Mark
    Commented Jun 26, 2015 at 22:31
  • $\begingroup$ A steam turbo-alternator set overall is very efficient, and is unlikely to be superceeded anytime soon. $\endgroup$ Commented Jun 27, 2015 at 17:58

my question is about whether it's possible in principle

The answer is yes.

and whether anyone tried it.

The answer is by all chances, no.

So, how come?

The effect

The thermoelectric effect for electricity generation (called the Seebeck effect) is the phenomenon that a voltage is generated at a temperature different across the ends of a conductor:

$$V=S\Delta T$$

where $S$ is the Seebeck coefficient, a material propety. All we need is a temperature difference - that is, a hot and a cold source. Said in other words, any hot source and any cold source. So sure, if you have a hot source which nuclear certainly is, and water cooling which is a usual method for nuclear power plants, you just need a material that has good properties in this temperature range.

The materials

But here comes the problem. The search for and research in thermoelectric materials is the brake in this field. We simply do not have good enough materials at the moment. The topic still feels new, though discovered some hundreds of years ago, and the best materials at present are still those that were found in the 1950's. We are improving and improving, but the efficiencies are simply too low compared to any other source.


I do not know the typical efficiency of a nuclear power plant. But a diesel engine as an example is at about 40% and is one of the most efficient practical engines existing and in use today.

Now, the maximum efficiency of thermoelectric devices is given as:

$$\mu_{max} =\frac{T_C-T_H}{T_H}\cdot \frac{\sqrt{1+ZT}-1}{\sqrt{1+ZT}+T_H/T_C}=\mu_{Carnot}\cdot\mu_{r}$$

where $T_H$ and $T_C$ are hot and cold end temperatures, and the Carnot efficiency is $\mu_{Carnot}=\frac{T_C-T_H}{T_H}$. The $\mu_r$, called the reduced efficiency or conversion efficiency, is maybe realisticly at around $10\%$, while the Carnot could be at maybe $60\%$ - multiply them together, and the maximum efficiency $\mu_{max}$ is simply too low.

Figure of merit

It all comes down to the so called figure of merit $ZT$, given as:


For present state-of-the-art materials it is only at around 1 to 1½. I am myself working with the alloy Bismuth Telluride, the best material at the moment in the lower temperature range around room temperature to 100-200 degress C, of which the higest $zT$ achieved is $zT \sim 1.45$ It is usually stated that it needs to reach around 3-4 for a material to be usable in industry. See the graph of the $\mu_r$ below showing the value for different $ZT$s at varying temperature.

enter image description here

Source: Rowe, D. M.: ”Termoelectrics Handbook - Macro to nano”, Taylor & Francis Group, 2006.

The problem is mainly the issue of reducing the thermal conductivity $\kappa$. This is mainly material science and a material problem, we need to overcome - but I fully agree that this physical phenomenon must have a huge potential at some point.

So, I do not know if anyone ever tried putting it into a nuclear power plant. But I really don't think so. Whenever better materials are found, it will take years for them to be integrated into largescale plants.

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    $\begingroup$ They are used in space probes/rovers where reliability is more important than efficiency. $\endgroup$
    – Eph
    Commented Jun 26, 2015 at 12:38
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    $\begingroup$ @hijarian Yep and in the case of power plants energy efficiency is vastly more important than maintenance/reliability, so I agree that its use specifically in earth based power plants is very unlikely. I figured the phrasing "tried it" could be in the sense of making small scale prototypes, which I think the power generators would qualify as. $\endgroup$
    – Eph
    Commented Jun 26, 2015 at 13:15
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    $\begingroup$ @hijarian - So the radioactive decay of plutonium is not nuclear power? That seems to be a pretty fine limit on just what a nuclear power plant is, when you are asking about unconventional ones to begin with! $\endgroup$
    – Jon Custer
    Commented Jun 26, 2015 at 13:28
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    $\begingroup$ @hijarian - I'm OK with the limit! But, it is worth contemplating the engineering scaling issues between the two as part of the 'why we don't do that'... $\endgroup$
    – Jon Custer
    Commented Jun 26, 2015 at 13:39
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    $\begingroup$ The RTG's used on space probes etc. are not usually classed as 'nuclear reactors'. That term is usually only applied to fission reactors and so RTG's are outside the scope of the OP question. $\endgroup$ Commented Jun 27, 2015 at 18:01

In principle, the drop in the Gibbs energy when the uranium gets converted to the fission products is available for doing useful work. While a steam engine will not come close to the maximum possible efficiency attainable (which is very close to 100%), a thermoelectric device will have much worse performance, as pointed out in detail in the other answers. The only way you could really get close to the almost 100% theoretical efficiency is to utilize the kinetic energy of the fission products directly, instead of capturing the energy that results from that kinetic energy being dissipated into heat first.

  • $\begingroup$ You know, this answer is actually contains _exactly_ what I wanted to know when asking the question and no more, no less... Amazing insight. 0_o' $\endgroup$
    – hijarian
    Commented Jun 28, 2015 at 9:58
  • $\begingroup$ @hijarian - if you feel that way about this answer, you should feel free to accept it rather than the one with the most votes. It's your question... you pick the answer that helped you most. But John's answer, and Steeven's, are pretty good too... $\endgroup$
    – Floris
    Commented Jun 29, 2015 at 5:39
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    $\begingroup$ @hijarian And to add to that, if we could capture the kinetic energy of the fission products directly, we'd be able to make perpetual motion machines - it's actually quite impossible. It's the same kind of technology that would allow you to move heat from colder object to a warmer object. There's a few suggestions on more direct energy conversion in fusion power plants (unlike fission products, some of the fusion products are electrically charged), but steam is still probably going to win out - possibly less efficient, but it captures more of the total energy released to offset this. $\endgroup$
    – Luaan
    Commented Jun 29, 2015 at 8:27
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    $\begingroup$ One can only capture some fraction if the kinetic energy, in case of fission, that fraction will be very close to 100% though, because the energy released is far larger than the thermal energy. The farther you are from thermal equilibrium, the more of the energy is available to perform work. $\endgroup$ Commented Jun 30, 2015 at 16:57

Yes, people use thermoelectrics as part of very-small-scale nuclear power generation systems, mainly in spacecraft: See Radioisotope Thermoelectric Generator.

People do not normally call these things "nuclear power plants", but they are definitely a type of nuclear power generation.


Lockheed Martin makes an LH3 and LH8 that are thermoelectric nuclear power generators. You can by new or used. You have to get the fuel from DoE or the Russians.


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