I Heating/cooling moon tunnels, why and how?

  • Thread starter Fred Bobo
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Summary
Heating/cooling moon tunnels, why and how?
Read a story in the Analog, April 1990, issue.

The author describes a cooling system for moon tunnels using water that is cooled via large fins on the surface.

An explanation is given that rock is a poor conductor of heat and eventually equalizes. And therefore the need for cooling.

Why would it not be better to use heat-pump tech?

How deep does the heat/cold penetrate the moon?

What is the temp below that?
 

berkeman

Mentor
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Um, why would you need to cool tunnels on the Moon?
 
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Was that the one with a small nuclear power plant, like the US had in their underground Greenland base ??

At least it was underground, against the threat of dust, solar flares, cosmic rays and micro-meteorites.

Against that, those micro-meteorites could make the cooling circuit rather vulnerable...
 
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No.
And yeah.
 
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Why would it not be better to use heat-pump tech?
Because heat pumps transfer heat on both sides by convection (blowing air across a heat exchanger), and there is no convection in vacuum on the moon's surface.
 
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Because heat pumps transfer heat on both sides by convection (blowing air across a heat exchanger), and there is no convection in vacuum on the moon's surface.
heat pump lines toward the center, not the surface.
 
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Active Heat pumps designed for such may move heat either way.
IIRC, passive 'heat-pipes' are inherently one-way only.

If you have a small nuclear reactor, disposing of its waste heat may present a problem. Can't send to river, sea, air or ground-water. External 'radiators' are strictly radiative, may struggle through the fortnight of lunar 'daylight' . Look at ISS for tech issues...
 
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Because heat pumps transfer heat on both sides by convection (blowing air across a heat exchanger), and there is no convection in vacuum on the moon's surface.
That is a problem for any type of cooling system - not just for heat pumps - and it is solved with radiators. They are more effective when used with heat pumps instead of a passive system. But a passive cooling system is less complex and requires no additional energy.

Heat pumps can't pump heat into or out of rock; you can't blow rock across a heat exchanger with a fan.
Hot dry rock geothermal energy exists. By reversing this process heat could also be pumped into the rock. But that is not suitable on the moon because it would be difficult or even impossible to create a system of flow paths with a sufficiently large surface inside the rock that is isolated from the vacuum.
 
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I suppose that the waste heat of a moon base inside the tunnel will heat it and cooling will be the problem.
A lot of waste heat has to be dumped by the NASA Space Shuttle and the ISS.
 

jrmichler

Science Advisor
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A quick search with search terms moon subsurface temperature indicates a temperature of about -6 deg F at one meter down. The effect of the one month day/night cycle disappears at that depth. The core of the Moon is hot, but the temperature gradient is low, so the temperature at any reasonable depth should be close to -6 deg F.

Maintaining a comfortable temperature should be easy. Just adjust the amount of insulation until the waste heat gives the desired temperature.
 

stefan r

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...External 'radiators' are strictly radiative, may struggle through the fortnight of lunar 'daylight' . Look at ISS for tech issues...
The radiator fin would be perpendicular to sunlight. If you feel the need you could also put up a shade.

I would expect a nuclear reactor to down scale operations during daylight. There are some very good solar options since clouds are scarce on the moon. The solar panel could be a shade for the radiator.

Some locations like, for example, the south wall inside of Copernicus crater will receive much less sunlight than nearby terrain.

The mean temperature on the moon is 220K at the equator. Negative fifties is like Alaska in the winter. You would want a lot of the heat stored in rocks. I picture lots of HESCO barriers stacked inside of the air tight seal inside of the tunnel. Could also use gravel water treatment systems to double as heat storage.
 

jrmichler

Science Advisor
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Just for grins and giggles, I did some ball park calculations to estimate how much insulation would be needed. Starting with some assumptions:

1) A moon base would be 100 feet underground so that the weight of the overlying rock would be greater than the air pressure inside. The tunnels would need only air sealing, with no need to be designed as pressure vessels.

2) Life support will be a closed biological system. Composting toilets plus light equals food.

3) One person needs gets 100 square feet of living space plus 100 square feet of work and common space plus 500 square feet of life support space.

4) One person has an average metabolic rate that results in total heat loss of 400 BTUH. This heat comes from food.

5) Photosynthetic efficiency is 2%, and LED grow lights at 30% efficiency.

6) Power is from a nuclear generator with 30% thermal efficiency.

Now for the ball park calculations:

7) Growing 400 BTUH of food requires 400 / 0.02 = 20,000 BTUH of light. If that light comes from LED grow lights at 30% efficiency, then those lights need 20,000 BTUH / 0.30 / 3412 BTU/Kwh = 20 Kw of electricity. Add another 10 Kw for other needs.

8) The nuclear power plant will generate 30 Kw / 0.30 = 100 Kw of heat, which equals 340,000 BTUH per person. That heat will conduct out through the walls, floor, and ceiling. After coming to equilibrium, the heat out the floor will decrease to zero because the interior of the Moon is at a higher temperature. After coming to equilibrium, the heat out the walls will be zero because of adjacent tunnels. The only heat loss will be through the ceiling.

9) Since the net ceiling area will be 700 square feet person, and net heat generated will be 340,000 BTUH, the heat loss is 500 BTUH per square foot of ceiling. An approximate heat transfer coefficient for the air film adjacent a bare rock ceiling is 1.0 BTU/ (hr-ft^2-deg F), so conducting that amount of heat would require a temperature difference of 500 deg F. No insulation needed, and some sort of heat removal system would be necessary.

10) Assume the rock has average density of 150 lbs/ft^3 and specific heat of 0.2. Then 100 feet of rock has total heat capacity of 100 X 150 X 0.2 = 3000 BTU per deg F. The average heat loss of 500 BTUH/ft^2 would heat the rock above at an average rate of 500 / 3000 = 0.17 deg F per hour, or 4 deg F per day, or 120 deg F in one month. Similar calculations apply to heating up the rock walls and floor.

11) These are over simplified ball park calculations. They show that a Moon base requires careful attention to thermal design. And that energy conservation is important, even if there is an unlimited source of power.
 

sophiecentaur

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Hot dry rock geothermal energy exists.
Afaik, the heat (from Geothermal rocks) is transported using pumped water. Not ideal for the Moon.

Moon base requires careful attention to thermal design
Pretty much the same as life in any other vacuum conditions.
 

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