-1
$\begingroup$

Within the CO$_2$ deposit near the south pole of Mars, according to this article 3 subunits have been found with one having a top CO$_2$ layer of about 300 meters thick covered by a water ice layer with a thickness of up to 20 meters (Figure 4).

The Russian OK-900A nuclear fission reactor has been used to power icebreakers. It has a fuel load of about 151 kg and a power production of about 171 megawatts with a total energy production of over 2000 GWh !

Dropping a similar reactor on the water ice layer covering the CO$_2$ subunit would sublimate the water ice beneath the reactor and it would sink down to the CO$_2$ layer, leaving a hole above it.

Phase diagram of carbon dioxide and water Phase diagram of carbon dioxide and water

Looking at the CO$_2$ phase diagram above, the CO$_2$ solid heated by the reactor would turn into gas, leaving the shaft above, and the reactor would sink relatively fast to the bottom of the deposit, probably within days.

And what would happen next ?

Would not the lifting, cooling gas in the shaft deposit on the inner surface near the opening, eventually closing the shaft ?

Sitting at the bottom of the deposit the reactor would sublimate more and more solid CO$_2$, creating a big room around itself, with ever increasing pressure within. Would not that eventually lead to a huge explosion of pressurized gas ?

But if the shaft would stay open, will the initially very cold gas circulate enough to prevent the reactor from overheating ?

$\endgroup$

closed as primarily opinion-based by Jan Doggen, peterh, Rob, Hohmannfan Sep 22 '18 at 20:25

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Industrial nuclear reactors need a strong cooling, it happens typically with water. It is unclear, how could it happen on the Mars. They are also very sensitive things, and contrary the common misconception, problems cause them to stop, and not to explode. The popular belief is that the nuclear reactors can easily explode. The fact is that they can easily go under, but can't be exploded (old, Chernobyl-type reactors could be exploded but it was hard even in their case). $\endgroup$ – peterh Sep 22 '18 at 13:13
  • $\begingroup$ @peterh Could not an adjusted pressurized water reactor stop itself at 70$^0$ C and start itself again at 50$^0$ C for instance ? $\endgroup$ – Conelisinspace Sep 23 '18 at 8:23
  • $\begingroup$ It would be practically unusable for power production, the theoretical maximum of its effectivity would be $\approx$ 6%. Industrial reactors are using pressurized water which boils only at $\approx$ 300 ${}^\circ C$. That reactor would be essentially an air-cooled reactor with a water primer circle. But 0.02atm CO2 doesn't make a very good coolant. I would be surprised if its maximal power could be over some kW. $\endgroup$ – peterh Sep 23 '18 at 12:59
  • $\begingroup$ @peterh Effectivity will be not important, this is just about heat production. Initially the CO2 will be at least minus 100$^0$ C when coming from the solid and with rapid sublimation the pressure could rise considerable with high speed circulation. I think it's a very complex situation to judge. $\endgroup$ – Conelisinspace Sep 23 '18 at 13:53
5
$\begingroup$

OK-151 is a pressurized water reactor. This has a primary cooling loop with water under high pressure, this loop must be cooled in a heat exchanger by a secondary loop. The OK-151 is designed to be used on a ship, with a constant supply of fresh water for the secondary loop.

In your situation, the water and CO2 would evaporate, so you can't fill the secondary loop. What happens then depends on details of the reactor design not present in the Wikipedia article. The primary loop will heat up until steam is formed, but by then you've probably tripped pressure relief valves and you're losing primary coolant. When the reactor boils dry, the reaction has slowed, but maybe not enough to prevent a meltdown.

So you can't just take an OK-151 and drop it in. It has to be modified with a primary cooling loop that heats the outside surface of the reactor module. That surface has to be pretty big to reach equilibrium with ~300-500 MW of heat feeding into it.

at about 100 meters below the surface the pressure would increase such that the heated solid would turn into liquid CO2.

Where do you get that idea? The reactor won't seal the hole it's melting out, gas will escape along the sides of the reactor.

Would not the lifting, cooling gas in the shaft deposit on the inner surface near the opening, eventually closing the shaft ?

You may get some deposits, but I don't think the shaft will close.

  1. The ambient temperature is right around the temp where CO2 freezes. So you only need to add a little heat for the CO2 to sublimate. In fact, once the CO2 is exposed, it'll probably start sublimating on its own during the day.

  2. The reactor produces so much heat in a small package that it can easily heat the column of CO2 gas to a point where it can't freeze again in the time it has to rise to the surface.

  3. The CO2 gas will escape quickly: it expands by a huge amount during sublimation, so you'll get a high-speed jet of gas.

$\endgroup$
  • $\begingroup$ Thank you for the answer ! I guess that at 100 meters below the surface the pressure will be about 10 bar. Looking at the phase diagram of CO2 at that pressure the solid would turn first into liquid so the wall of the hole above the reactor could be liquid, but you're right, it won't seal the hole. But what about the moderator ? en.wikipedia.org/wiki/Pressurized_water_reactor#Moderator Will it not prevent losing coolant ? $\endgroup$ – Conelisinspace Sep 21 '18 at 16:12
  • 1
    $\begingroup$ 10 bar at 100 m is about right, but that only applies if there's a column of liquid water above you. On Mars, when you thaw water it goes straight to the gas phase because the atmospheric pressure is too low for a liquid to exist. $\endgroup$ – Hobbes Sep 21 '18 at 17:07
  • $\begingroup$ The coolant is the moderator, it can't prevent its own loss. $\endgroup$ – Hobbes Sep 21 '18 at 17:08
  • $\begingroup$ I don't understand, in the "Moderator paragraph" nothing is said about "losses". The coolant is already moderating before any losses could occur. $\endgroup$ – Conelisinspace Sep 21 '18 at 17:38
  • $\begingroup$ That's the question (and one I don't have an answer for): is the moderation due to steam pockets enough to lower the power output to a safe level, when the secondary coolant loop is empty? $\endgroup$ – Hobbes Sep 21 '18 at 18:31

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