Cost of energy on Mars

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Solar array with a Mars settlement in the background

The cost of energy on Mars is one of the prime parameters required to analyse martian settlement scenarios. To go beyond generalizations and actually evaluate one scenario against another, the cost of energy and the cost of transportation to and from Mars need to be known, or at least estimated to compare on a common basis.

There can be no absolute answer to this question, as the cost of energy will vary depending on the source and the level of development of the colony. If all the energy producing equipment comes from Earth, the cost will be higher than if it is produced in-situ. As the settlement grows larger, economies of scale will come into play to reduce energy costs. As automation increases, and productivity increases accordingly, the cost of energy may go down significantly. Self replication of production equipment may eventually bring down the cost of energy to very small values. This will allow the realization of projects using energy on a scale that will be, literally, cosmic.

But before such a grand time comes, we can estimate a preliminary value for the cost of solar energy on Mars for a growing settlement of a few hundred to a few thousand settlers.

Solar energy

Design

For this estimation we will use a solar panel array built on Earth with the following characteristics:

Solar array characteristics
Characterisitic Value References
Efficiency 30% This would require fairly expensive multijunction production cells. Large volumes of production would keep the cost low.
Fill factor 80% This is a conventional average for good quality cells.
Mass 3,5 kg/m2 This is the NASA 2018 BIG challenge target of 3 kg/m2, plus 0,5 kg/m2 for ancillary equipment. This corresponds to about 30 We/kg.
Cost on Earth 800 $/m2 An estimate based on an average cost of 500$/m2 for conventional panels on Earth

Transported by SpaceX Starship with other cargo and unfolded on Mars semi-autonomously. For the Cost of Transportation to Mars, we could simply use the cost proposed by SpaceX for their transportation system, 140 $kg to Mars. However, 500$ per kg to Mars may be a reasonable conservative choice, in particular if we apply it to all technologies.

The design is a tracking array. This allows for more energy production with the same peak power characteristics. The added complexity is estimated to be a small cost compared to the transportation costs from Earth. The moving mechanism would be fitted with elements allowing for a self cleaning phase for dust removal. The panels are mounted on a single rotating and extending boom, that allows for tight packing at transportation and can be unfolded using traction from a vehicle that can also serve to transport and set down the array, eliminating the need and mass of self extension mechanisms.

Solar panel cost on Mars

The cost would be 800$/m2 + 3.5 kg/m2 x 500$/kg = 2550 $/m2 of solar panel on Mars. Transportation is by far the largest part of the cost.

Note that 30% is not current but a projection. 22-24% would be closer to the performance of current cells.

Solar panel energy production

The solar constant is estimated at 590 W/m2 over the whole year. Atmospheric dust losses are set at 20%. An additional surface dust factor of 10% is included. This lowers the solar illumination to 413 W/m2. Using solar tracking for 10 hours per day, the available energy per day is calculated as 413 W/m2 x 10h x 3600 s/h = 14 868 000 J.

The actual energy production should be 14 868 000 x 30% x 80% = 3 568 320 joules per day.

The installed power, based on the 10 hours of operation, should be 413 W/m2 x 30% x 80% = 99 W/m2. The same cells on Earth would have a power of about 99 x 1300/413 = 300 W/m2. The mass ratio metric is then 300 W/m2 / 3 kg/m2 = 100 W/kg. this seems safely conservative compared to recent (2017) NASA projections[1]

The availability of the solar panels is estimated at 95%, to account for repairs, tracking failures and other problems.

So every year a m2 of panel would produce about 3 568 320 x 365 x 0,95 /1e9 = 1,23 GJ/m2.

With an estimated life time of 20 years the total energy produced would be 20 x 1,23 = 24,7 GJ/m2. When divided by the cost of 2 550$ the bare cost of the energy would be 103 $/GJ. Majoring in financing, profit and uncertainties, we could expect a multiplying factor of 1.5, for a final cost estimate of 150 $/GJ for martian solar power. In other units this is 0,15 $/MJ, or 0,55$ per kWh. This is about twice the cost of electricity on the Islands of Hawaii. For 22% efficient cells, the cost might be increased to 200 $/GJ

This is not necessarily the selling price, as there might be taxes added or other costs, such as the distribution system costs. The price for energy might also be modulated; for example, the cost of night time energy, requiring the installation of energy storage systems, might be significantly higher. Or the cost of these storage systems could be factored into the base cost, if the settlement determined this was a useful policy.

This price also has a very broad range of uncertainty, as the cost of solar has been going down steadily, and the cost of SpaceX to Mars is pretty speculative.

Nuclear energy

Design

Nuclear reactor designs for Mars might be based on the Kilopower[2] 10 kW nuclear reactor. It used Highly Enriched Uranium(HEU), presumably at 20% concentration.

Characteristic Value References
Power 10 kW electric

43 kW thermal

Solid core, sodium heat pipe cooled reactor. Stirling engines.
Efficiency 23%
Mass of reactor 7 We / kg A 10 kWe reactor would include 43.7 kg of 235U and mass about 1430 kg.[2]
Cost on Earth 400 000$ 5500 $/kg of HEU (240 000$)[3]

100$ per kg for fabrication of the rest (140 000$).

Nuclear power station cost on Mars

Using 500$ per kg for transportation of 1430 kg and 400 000$ for construction, the cost is about 1 100 000$.

Nuclear energy production

The reactor, operating at 95% availability for 10 years will produce 10 000J x 10 x 365 x 24 x3 600 x0,95 = 3 e12J or 3000 GJ

When divided by the cost of 1 100 000$ the bare cost would be about 300 $/GJ. If we have the same multiplication factor of 1,5 then the final cost would be 500 $/GJ. this is about 3 times more expensive than solar. If the nuclear reactor could be refurbished and refueled for another ten years, for about 500 x 43,7 + 240 000 = 260 000$ then the cost would be 1 360 000 / 3000 = 226$/GJ x 1,5 = 339$/GJ or about twice the cost of solar.

Larger nuclear reactors might be more mass efficient and

References