Super Flywheels… Some Performance Figures

My recent Tweet on super-strong carbon nanotube fibres for use in energy storing flywheels – as well as the underlying News story – has inspired some imaginative posts on Brian Wang’s “Next Big Future” news-blog, here and here. Brian likes to extrapolate hard and has a devoted audience of fact checkers, so read the comments and watch out for “Goat Guy” who always brings hard figures into the discussion. Educational discussions are often had at “Next Big Future”.

Let’s look at the underlying physics and make a comparison with some other energy storage options.

Firstly some basic Rotational physics. Imagine the flywheel as a solid cylinder of even composition, so its density \(\rho\) is constant. For any rotating object the important figures are the Moment of Inertia \(I\) and the angular velocity \(\omega\). For a cylinder of mass \(m\), length \(l\) and circular radius \(r\) the Moment of Inertia is:

$$I = \frac12 mr^2 = \frac12 \rho\pi r^2lr^2 = \frac12 \rho\pi r^4l$$

Rotational Kinetic Energy is:

$$E = \frac12 I\omega^2 = \frac14 mr^2\omega^2$$

To constrain the possible we have to look at material strength and how it affects the flywheel rotor’s performance. The relevant factor is the stress in the flywheel’s material – that is, the centrifugal force that is trying to throw it apart. The material’s molecular structure must counter that force with its tensile strength, the intra-molecular force that the material exerts on itself holding it together. In the case of the new nanotube fibre it’s internal strength is measured as upwards of 80 billion pascals or 80 gigapascals (GPa). Most steel alloys have between 0.25-0.5 GPa tensile strength, so the new material is up to 320 times stronger.

In a spinning rotor the maximum stress to be countered is on its maximum radius and is computed by:

$$\sigma_t = \rho r^2 \omega^2$$

The maximum the rotor can spin is when the maximum stress equals its tensile strength. Past that point and the material will eventually ‘fail’, pulling itself apart violently due to all its kinetic energy. To operate safely a rotor should run at a maximum of some fraction of that limit. A factor of 50% is considered reasonable, allowing wiggle room for fluctuations. Thus the maximum operating stress should be about 2/3 of 80 GPa or ~54 GPa.

Notice that the stress and the rotational kinetic energy look very similar. In fact the relationship is simply

$$E = \frac{m\sigma_t}{4\rho}$$

Engineers love to express a material’s properties in terms of ‘figures of merit’ – in energy storage applications it’s usually Specific Energy density which is the stored energy per unit mass, thus we can rearrange the equation as:

$$\frac Em = \frac{\sigma_t}{4\rho}$$

For carbon nanotube material, with a density of about 1,300 kg/m3, and an operating maximum stress of 54 GPa, that means a specific energy density of 10.4 MJ/kg. For comparison a kilogram of natural gas represents about 50 MJ of chemical energy, if fully burned. However that comparison neglects the efficiency that the chemical energy can be turned into useful work. Flywheels convert their stored energy into electricity at very high efficiency – 95% or so. Running a gas powered generator turns maybe 20% of the gas’s chemical energy into electrical energy. The rest is lost as heat, via both friction and the Carnot Limit. That’s still about 10 MJ/kg, so why use a flywheel? The other assumption is that one can neglect the mass of the oxygen in the air that burns the fuel. If the application we’re using the flywheel for is in space, then the oxygen has to be added in.

Natural gas, methane, burns with oxygen via the following reaction

$$CH_4 + 2 O_2 = CO_2 + 2 H_2O$$

Every molecule of methane masses 16 atomic mass units (16 amu), while every molecule of oxygen (O2) masses 32 amu. For every kilogram of methane burned, one needs 4 kilograms of oxygen to burn it with. Thus the useful energy content is 20% of the 10 MJ/kg we calculated above. In that case flywheel storage is better than chemical energy.

Every rocket, since rockets *must* carry their fuel and oxidiser, is subject to this limitation, so how do flywheels compare to chemical fuels? Rocket engines are far more efficient than reciprocating internal combustion engines, as they’re much simpler and don’t operate according to a Thermodynamic Cycle. The Space Shuttle Main Engines, for example, converted about 92% of the chemical energy they liberated into the kinetic energy of their jets. However no chemical rocket can burn fuel perfectly – typically there’s an excess of fuel to oxidiser to ensure maximum burn-up – so the theoretical chemical efficiency is higher than the attainable maximum.

Rocket jets have the figure of merit known as the Specific Impulse (Isp) – the “gee-force” (weight-force) produced by reacting a unit mass of whatever propellant mixture is being used. The SpaceX “BFR” promises an Isp of 375 seconds for the natural gas + liquid oxygen Raptor engines it will use. That’s an effective jet velocity of 3,677 m/s – equivalently that means every kilogram of propellant it vents represents 6.76 megajoules of useful kinetic energy. So in that sense, the flywheel’s energy density is better than a rocket. The trick is applying it to lifting a vehicle against gravity into empty space. The usual alternative in that application is the Space Elevator. But how do Space Elevators really compare? Can super-strong Carbon Nanotubes and Super-Flywheels truly make the Space Elevator a possibility? Part II

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