I recently learned about how an n-type and p-type "sandwich" with different thermal properties can create a potential difference in the presence of a temperature gradient. The subsequent current can power basic electronics (in the dozens-hundreds of watts range), as long as a decent heat source exists such as a radioactive isotope in an RTG.

Since this is fully solid-state, it seems preferable to other heat-to-electricity means from a maintenance POV. But why are they so inefficient compared to nuclear power plants that use turbines? What specific mechanisms yield an overall RTG inefficiency of 3-7% vs nuclear power plant efficiency of, say, 45%? What's the fundamental limitation on RTG efficiency?

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    $\begingroup$ This question might be better suited to physics.SE (or some other SE site) rather than specifically space exploration. $\endgroup$ – Starfish Prime Feb 24 at 18:14
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    $\begingroup$ They use thermoelectric generators. Those are inefficient for reasons discussed here: physics.stackexchange.com/q/481303/223601 $\endgroup$ – ikrase Feb 24 at 19:47
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    $\begingroup$ @StarfishPrime all questions about engine efficiency are about physics also, and yet we answer those here because some users like to answer them and do a good job. "Better suited" is never a close reason. This is on-topic here and so should not be closed. $\endgroup$ – uhoh Feb 24 at 22:11

Here's a brief answer since some people are trying to close the question and prevent answers:

As @ikrase points out answers to the Physics SE question Why is the Peltier / Seebeck Effect's efficiency so low in practical devices? are helpful here.

Briefly, there are two main parts to an RTG's conversion efficiency

Thermodynamics limit

The fraction of the thermal power that can be theoretically converted to electrical power is called the Carnot efficiency. It is given by $\eta = (T_{hot} - T_{cold})/T_{hot}$. For example, the MHW-RTG has hot and cold temperatures of 1273 K and 573 K, with $\eta$ of 0.55.

Now why on a cold planet or in space is the "cold" side so hot? Radiation efficiency which scales as $T^4$. It's hard to radiate heat if you are not hot yourself!

Material properties limit

The thermoelectric materials limitations as discussed in depth in the answers to the linked Physics SE question are currently the biggest source of inefficiency.

The need to have low thermal conductivity and simultaneously high electrical conductivity; it needs to be a good thermal insulator and simultaneously a good electrical conductor. Semiconductors can fall into this general category, but unfortunately useful materials don't fall far enough into this category to be highly efficient.

A little extension from @Uwe:

Unfortunately thermal and electrical conductivity is related, both depend on the movement of free electrons. So aluminum and coper are good conductors for both electrical and thermal flow. If there is a material with good thermal isolation, the movement of free electrons must be very low, so this material could not be a good electrical conductor.


Related here in Space SE:

  • $\begingroup$ I flagged it when you had your placeholder answer text! I would absolutely say that your answer now is great. However, whether the question itself is better here or on the physics stack exchange, I would say it is up for debate. $\endgroup$ – Mark Omo Feb 24 at 23:09
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    $\begingroup$ Actually, upon further reflection, I agree that this question is on-topic for this stack exchange. I retracted my close vote. $\endgroup$ – Mark Omo Feb 24 at 23:14
  • $\begingroup$ @MarkOmo thanks! ya I know, but there were 4 close votes and a fifth would have shut down answer posting, so I added a placeholder. It took 45 minutes, longer than I'd expected. I have heard (looking for the info now) that there is some kind of 4 hour grace period for adding answers to just-closed or just-held questions, but I've never tried it. $\endgroup$ – uhoh Feb 24 at 23:16
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    $\begingroup$ Unfortunately thermal and electrical conductivity is related, both depend on the movement of free electrons. So aluminum and coper are good conductors for both. A semiconductor with better electrical conductivity is not likely a better thermal insulator $\endgroup$ – Uwe Feb 25 at 11:01
  • $\begingroup$ @Uwe that's a very important point; please feel free to add a new answer or edit that into this answer, thanks! $\endgroup$ – uhoh Feb 25 at 14:56

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