In the interest of full disclosure, I’m not an HVAC specialist. Nor was thermodynamics one of my stronger college courses. The calculations used here are top-of-napkin at best, using formulas found on the internet or that aforementioned thermo textbook. (Yup, It’s still on my shelf.) So if you happen to find a glaring error in my math or assumptions, please reply to this email and let me know about it. I’ll happily publish a correction in a subsequent edition.
Mars is cold. Really cold.
Hellas Planitia is the largest known impact crater on the Red Planet and in the solar system. At 9 kilometers deep, it’s also the lowest point. In the winter, when the air is coldest and most dense, atmospheric pressure is around 12 Pascals. While this is 12/1000ths Earth’s sea level barometric pressure, it’s still the greatest surface pressure on Mars.
The basin’s floor’s relatively mild climate still drops to -96degrees C ( -140 degrees F!) during a winter night. Keeping a habitat temperate enough for astronaut comfort will be critical to survival and mission success.
How much energy will it take to keep the first astronauts warm on such a night on that first expedition to Mars? We’ll dig into a bit of surprisingly simple math to find out.
But first, let’s review a few assumptions about our equipment. For our first mission, we’ll provide our intrepid explorers a 10-meter diameter dome structure for two months. Both weight and volume will be considerations for the Mars transit, so our habitat should be inflatable. An inflatable structure is lightweight and compressible. A continuous fabric dome will also have fewer seams to generate pressure leaks than a rigid geodesic structure with dozens of triangular panels.
Okay, our dome is standing. How much energy will it take to keep the interior at 20 degrees C (68 degrees F) when it’s – 96 degrees C outside? Here on Earth, the building construction industry relies on a term called the R-value to determine heating system capacity. A few simple formulas using R will provide our answer.
R-value: R = (K)(m2)/W (A value derived experimentally for a given material)
Heat Flow Rate: Φ = 1/R = W/(K)(m2) (The amount of heat energy lost per degree temperature difference per square meter)
Total Heat Flux: F = (Φ)(ΔT)(A) (The amount of heat energy lost through a structure for a given temperature differential)
Where:
ΔT = 96 + 20 =
116K (The temperature spread between our interior and exterior temperatures)
A = (4πr2/2) + (πr2) = (4(3.14159)(5)2/2) + (3.14159)(5)2 = 157.2m2 + 15.7m2 =
172.8m2 (The surface area of a 10m half sphere, our dome, plus the floor area)
All that remains now is to determine what material to look at for our dome. Here on Earth, outdoor adventurers and explorers routinely use an inflatable product that allows them to sleep comfortably on snow, no matter how cold the weather. The Thermarest company manufacturers several inflatable sleeping pads. One model, the NeoAir® Xtherm
TM is a 3-inch thick matt with “Triangular Core Matrix
TM construction and ThermaCapture Technology.” In layman’s terms, they’ve engineered the heck out of this product. We’ll pattern our walls after it.
Granted, I doubt it’s reasonable to use a woven nylon fabric for a Mars habitat. But a similar technology might be durable enough if the inside and outside surfaces were carbon fiber. The Xtherm
TM has an R-value of 7.5. To make things simple, I’m neglecting the heat loss through the airlock.
I concede that the floor won’t be inflatable. Rather, it will need to be an airtight fabric with rigid lightweight panels set on top. For consistency let’s assign an R-value of 7.5 to minimize heat loss through the floor, a reasonable assumption for a rigid foam product. Given the floor is only 15.7 square meters, less than 10 percent of the overall structure, it shouldn’t take up much volume to ship. On to our calculations:
R = 7.5m2K/W
Φ = 1/R = 0.133W/(K)(m2)
F = (0.133 W/Km2)(116K)(172.8m2) =
2,666W
2,666 Watts. That’s a lot of energy. But thanks to our dome’s small size, it’s less than used by an electric water heater in an average American home.
What should we power our habitat heater with? The ISS relies on 2500 square meters of solar panels, over half the area of a football field, to generate the 75 kilowatts (75,000W) specified for its operations.
I’ve had difficulty nailing down energy allocation for the ISS, but a 1988 paper on space station power requirements recommended allocating 45 kilowatts for experiments, leaving the remaining 30kW for oxygen generation, water reclamation, air circulation, thermal regulation, etc. But the size and mass of the required panels, plus the batteries to store power for use at night, precludes solar power on Mars.
NASA has developed a compact 10kW nuclear reactor called a Kilopower Unit. They were designed with the Artemis and Mars programs in mind. While these devices are relatively heavy, they don’t take up much space. A typical unit is a six-foot tall cylinder, with a foldable ten-foot diameter umbrella-shaped heat dissipator. Four or five devices would suffice for a small first mission. More units could be flown in with subsequent missions to grow the base’s power generation capacity over time.
Their drawback is the Uranium-235 core. Fortunately, underground storage of spent cores on Mars is more straightforward than on Earth. The Red Planet lacks the liquid groundwater to dissolve and transport radionuclides like we experience here. Consider the massive on-going effort to clean up the contaminated groundwater at the Hanford Nuclear Reservation in Washington state. On Mars we can dump them in a hole and forget about them.
What about fusion? It’s certainly cleaner. Unfortunately, I firmly believe we’ll reach the Red Planet long prior to developing commercially viable fusion power generation here. Once we finally do, it will be another ten or twenty years before the components are miniaturized enough to ship a plant to Mars. I’m also doubtful about the availability of deuterium and tritium fuel. If that’s the case, fusion may not be viable there until the next century.
In the EPSILON SciFi Thriller Series, I assumed 40kW for Prospector Base, plus an additional 10 kW factor of safety. It seems reasonable, with an eye toward maximizing environmental equipment power efficiency, for our little dome to get by on 40kW.
The wild card could be the power source needed for any rover prepositioned for early missions. If it comes with its own radionuclide battery, it won’t be a drain on the base's energy source. If it runs on hydrogen, utilizing either internal combustion or a fuel cell, the electrolysis of water for fuel will be a tremendous drain and may require additional Kilopower units.
In the meantime, the next time you grouse about throwing an extra blanket on your bed on a cold night, just imagine how much colder it would feel if it were minus 140 degrees F outside!
Happy Reading,
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Want a deeper dive? Check out these sources.
https://www.energy.gov/energysaver/insulation#:~:text=R%2DValues,its%20thickness%2C%20and%20its%20density.
https://www.forbes.com/home-improvement/home/what-is-insulation-r-value/https://en.wikipedia.org/wiki/R-value_(insulation)https://www.nasa.gov/directorates/stmd/tech-demo-missions-program/kilopower-hmqzw/https://www.rei.com/product/217085/therm-a-rest-neoair-xtherm-nxt-max-sleeping-padhttps://ntrs.nasa.gov/api/citations/19880011861/downloads/19880011861.pdfhttps://en.wikipedia.org/wiki/Kilopower#:~:text=The%20prototype%20Kilopower%20uses%20a,to%20electricity%20by%20Stirling%20engines.
