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.
