Could Wind Power a Mars Colony?

I started reading Red Mars by Kim Stanley Robinson last week and was struck by Robinson’s description of nuclear energy on Mars. The first reactor on Mars is a “Rickover” reactor. These reactors are a type of pressurized-water reactor used in nuclear powered submarines. This choice makes sense to me, yet there was one character in the book that claimed

[They] were not reassured by Arkady sending radio messages down from Phobos insisting that they did not need such a dangerous technology, that they could get all the power they needed from wind generation. – Kim Stanley Robinson, Red Mars

Of course I stopped reading to figure out if this could actually work. First, I needed to answer some basic questions:

How can we calculate the power produced by a wind turbine?
What is the air density on Mars?
What is the average wind speed on Mars?

By answering these questions, we can figure out if a Mars colony could reasonably be powered by wind energy.

Deriving the Function for Wind Powering

Wind turbines essentially convert the kinetic energy of air (wind) into electricity. The equation for kinetic energy is simply,

\[KE = \frac{1}{2}mv^2 \text{ [J]},\]

where m is the mass of air flowing and v is the velocity of that air. But we want to know how much power this kinetic energy will produce over a certain period of time. To do that we simply change m into \(\dot{m}\) to indicate the flow of mass through the turbine blades.

\[P_{turbine} = \frac{1}{2}\dot{m}v^2 \text{ [J/s]}.\]

We’re not quite done yet, though. Without measuring the total mass of air passing through the blades at each moment, we can’t know how much power it produces. Fortunately, we can rewrite mass in terms of density and a volume of flow, \(\dot{V}\). We can further rewrite this volume flow as a cross-section area times a flow velocity, v. So we have

\[P_{turbine} = \frac{1}{2}\rho \dot{V} v^2 \text{ [J/s]},\]

and finally,

\[P_{turbine} = \frac{1}{2}\rho \pi L^2 v^3 \text{ [J/s]}.\]

Where L is the length of the turbine blade. From this we can see that wind power is proportional to the cube of wind speed and only linearly related to the density of the air. This is potentially good news since Mars, despite having 1/100th the atmosphere of Earth, is known for its dust storms. Does this mean it’s windy enough to sustain a Mars colony?

Best Case Scenario

It turns out that wind is not much faster on Mars than it is on Earth, with average speeds around 10 m/s. The fastest recorded wind speed on Mars is 30 m/s, during a dust storm. The average wind speed in the United States is around 5 m/s. This is good news for Mars, but there is a competing effect from Mars’ thin atmosphere. The greatest density for Mars’ atmosphere is roughly equal to the density of Earth’s atmosphere at 35 km. That density is approximately \(\rho_{Mars} \approx \rho_{Earth}|_{35km} \approx 0.0112\), calculated with linear interpolation of data.

Now we can calculate the ratio of wind power on Mars to wind power on Earth, and we can plug in some numbers (\(\rho_{E} \approx 1.225 kg/m^3\) at sea level)

\[\frac{P_{M}}{P_{E}} = \frac{\frac{1}{2}\rho_{M} \pi L^2 v_{M}^3}{\frac{1}{2}\rho_{E} \pi L^2 v_{E}^3}\] \[\frac{P_{M}}{P_{E}} = \frac{\rho_{M} v_{M}^3}{\rho_{E} v_{E}^3} = \frac{(0.0112)(30)^3}{(1.225)(5)^3} \approx 1.975\]

So, at best placing a wind turbine on Mars will generate twice as much power as an identical turbine on Earth (roughly speaking). But that was calculated using the highest recorded wind speed on mars. What if we use the average wind speeds for both?

Average Windspeeds

The average wind speed on mars ranges from 2 m/s to 7 m/s in the Martian summer and 5 m/s to 10 m/s in the Martian fall.

Let’s take the highest value, once again.

\[\frac{P_{M}}{P_{E}} = \frac{\rho_{M} v_{M}^3}{\rho_{E} v_{E}^3} = \frac{(0.0112)(10)^3}{(1.225)(5)^3} \approx 0.073\]

Even if the wind speed averaged 10 m/s year round, at best a wind turbine on Mars would generate just 7% of the energy an identical turbine on Earth would produce. I also neglected that fact that wind turbines have an ideal operating range and a “cut off” speed. The turbine produces the same power throughout its ideal range and the turbine is stopped beyond its cut off speed for safety reasons. If we applied the ideal range of velocities to our above equation, air density becomes the dominant factor.

The lesson here is that 1/100th of the air density means 1/100th of the power output.

Final Thoughts

Not only do wind turbines produce far less energy on Mars than they do on Earth, but they also have an exceptionally low power density. This means it would be impossibly expensive to launch all of the necessary materials to build wind turbines on Mars.

The answer in Red Mars is that nuclear power is required. “Shikata na gai, there is no other choice.”