There is abundant $\require{mhchem}\ce{CO2}$ in martian atmosphere, so we could use that in a modified gas turbine. Can this, combined with PV cells (a hybrid power source), be an efficient energy source? It also turns out that supercritical $\ce{CO2}$ gas turbines are smaller than water steam turbines.

  • 1
    $\begingroup$ What would be heating the gas? It seems like that's a much more interesting problem than the medium you're using to transmit power. $\endgroup$ Nov 28, 2015 at 7:22
  • $\begingroup$ oh sorry, I didn't think about it. how about the sabatier reaction that makes rocket fuel for the Mars Ascent Vehicle be also used for heating the gas? please correct me if i am wrong, I maybe thinking about this in an inefficient way . check this link please : [link] (scientificamerican.com/article/…) $\endgroup$
    – Mathav Raj
    Nov 28, 2015 at 8:01

1 Answer 1


Any type of steam cycle of a Stirling engine or a closed-cycle gas turbine would face significant heat exchange challenges on Mars, so what little you saved on size and mass in improved heat transport efficiency for using a higher heat capacity and expansion ratio working fluid (sCO2 thermal efficiency is close to 45%), would be lost to cooling due to operation in extremely tenuous atmosphere and requiring massive heat pipes and large radiators, also adding significant thermal lag to the steam cycle and thus decreasing its capacity and efficiency.

So short of building such regenerative power systems in situ, and for a small settlement of few people, it would be too heavy to transport to the surface of Mars and there's a large number of other regenerative power systems available and in development that are more mass efficient (from heat pipe nuclear reactors coupled with some energy storage to electrolysis of the atmosphere systems).

One other comment tho, that the heating part of such a system that you inquire about would be better off with solar concentrators than photovoltaics, since there's no energy conversion inefficiencies (tho there are still losses, of course).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.