There's a related question: In space can the difference in temperature between the inside and outside, provide useful energy?

The problem with that one is that it relies on the concept of "hot inside" and using space as the heat sink - while there are sources (like the Sun) which would pretty much heat up the "outside" more than the "inside", and generally a range of other problems.

Now, there should be a much simpler way of exploiting heat transfer: Let's take a pad of Peltier cells, with the side exposed to the Sun, with highly absorbent coating, and an array of radiators of high emmissivity on the "dark side".

The basic problem on Earth is the "ambient temperature" - about as much heat as is radiated out of the "cooled side" is returned to it by radiating Earth, and balanced out by convection+conduction of heat from air. In space, there's Cosmic Microwave Background (CMB) which, if I understand correctly, is fairly weak. There is no convection but radiation is fairly efficient. That means considerable amounts of heat could be transferred from the Sun side to the "outer space" side, at least as long as we remain relatively near to the Sun.

Of course such heat transfer could be exploited through any energy source that exploits heat transfer - like a Peltier cell, or even if you're willing to go that far, through a Stirling engine.

Now, how would such systems fare against "tried old" photovoltaic cells? Could someone throw some numbers at me, comparing what we could expect from the same surface/mass/price of one versus the other?

  • $\begingroup$ Suggest doing some research, at least on the Wikipedia. Newer generations of PVs are available. The key word is efficiency. $\endgroup$ Commented Jul 29, 2013 at 9:00
  • $\begingroup$ @DeerHunter: Note the solar cell efficiencies are only for single band or a strictly limited number of bands in multi-band cell. All the other frequencies are wasted entirely. Meanwhile, with thermal transfer, you can apply a dumb black paint over a layer of sheet metal, which will absorb and turn into heat everything starting far towards X and ending somewhere beyond microwaves or further. That's a lot of heat and even losses of order of 91% may not be enough to outweigh the gain of simply absorbing everything that can be absorbed. $\endgroup$
    – SF.
    Commented Jul 29, 2013 at 9:36
  • $\begingroup$ the problem is that you have to reject all that heat afterwards. $\endgroup$ Commented Jul 29, 2013 at 10:06
  • $\begingroup$ @DeerHunter: The higher the temperature the stronger the radiation. At certain point the radiators would catch up and we'd have a rather high difference between the two surfaces (unless the whole thing burns and melts first). Now, what difference would it be? What kind of heat transfer? $\endgroup$
    – SF.
    Commented Jul 29, 2013 at 10:11

1 Answer 1


This post doesn't include any real world construction problems.

if we guess that the temperature on the cold side is -30.0 degrees F (243 K) and the temperature at the warm side is 250 F (394 K) (source), the Carnot cycle predicts an efficiency of 0.383.

Multiplying by the maximal efficiency of a peltier element of 15% (Thermoelectric cooling), the net efficiency is 0.057, or 5.7%.

A 1 square meter surface lit with an energy of 120 W (source), at 5.7%, yields only 6.9 W.

This is much lower that the energy you get with typical solar panels (in the rage of 20% source).

  • $\begingroup$ edited - efficiency much lower than solar panels. The efficiency per m^2 is much higher than with solar sails - but then m^2 of solar sail costs a fraction of a penny; their efficiency comes from enormous surface. $\endgroup$
    – SF.
    Commented Aug 7, 2013 at 0:07
  • $\begingroup$ @Quonux But what about the efficiency per kg? That would be much more interesting. $\endgroup$
    – Philipp
    Commented Aug 7, 2013 at 13:18
  • $\begingroup$ @Philipp: Currently, Peltier cells are somewhat heavier than solar cells. On the other hand, nobody ever bothered to build Peltier cells that are very lightweight. Most of their weight comes from heavy copper elements distributing the heat, the thermoelectric elements comprising a relatively small fraction of the volume, so there's a lot of room for improvement. $\endgroup$
    – SF.
    Commented Apr 27, 2014 at 12:57
  • $\begingroup$ Can you give a derivation or reference for the starting assumption that the cold side would be a mere -30 F? $\endgroup$ Commented Aug 28, 2015 at 2:52
  • $\begingroup$ @NathanTuggy just was a guess, if you want to calculate it exactly you have to calculate the incomming heat/energy from the sun, the heat conduction of the material and the temperature on the dark side is then the steady state solution of this (partial?) differential equation. The Stefan-Boltzmann law is used to calculate the (black body) heat radiation of the illuminated/shadow side. $\endgroup$
    – Quonux
    Commented Aug 30, 2015 at 15:41

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