Flywheels
Introduction and Description
One interesting form of energy storage useful for space applications is by using a flywheel. Flywheels have good energy density, charge/discharge rates, and high efficiency. They could also double as reaction wheels.
Given certain material properties, this calculator will calculate the best-case properties of a flywheel made out of it.
Calculator
Material
Strength^{[1]}: | |
Density: | |
Specific Strength^{[2]}: |
Example Materials:
Geometry
Mass-Efficiency Factor \(k\)^{[3]}: (typical ~\(0.6\) to ~\(0.9\); ideal maximum \(1.0\))
Performance
Specific Energy:
Energy Stored: | |
Mass Required: |
^{[1]} Ultimate tensile strength, although for ductile materials yield strength may be preferred for engineering reasons. See discussion of strength below.
^{[2]} Strength divided by density. Note if using yield strength that usual engineering numbers are quoted for ultimate tensile strength.
^{[3]} How closely the flywheel approximates an ideal flywheel. See discussion of geometry below.
"Strength"
The main limit on flywheel design is the flywheel material's "strength", but we need to be careful when defining that. When spun up, the flywheel expands somewhat according to the relationship of stress and strain. Here, stress is the force per area induced by the flywheel's motion while strain is the change in radius caused by that stress.
- In brittle materials (like glass, concrete, or carbon fiber), the strain increases nonlinearly until the material suddenly fails at the material's ultimate tensile strength.
- For ductile materials (like most metals and alloys), the strain is roughly linear up until the material's yield strength. The material then starts deforming plastically (permanently) and work hardens, becoming able to take more stress. The material fails at the ultimate tensile strength, which is somewhat-higher than the yield strength.
Hence, one must consider the radius of the flywheel when it is spun up, and the strength should be no more than the ultimate tensile strength (and for ductile materials, probably only the yield strength, as work-hardening the flywheel may prove too difficult or hazardous to do practically).
For ultimate tensile strength, the ratio of strength to density is called the specific strength or, colloquially, the strength-to-weight ratio.
Geometry and the \(k\)-Factor
It is not immediately obvious how to design an optimal (greatest energy stored per unit mass) flywheel. A thin-walled cylinder has the highest moment of inertia, but can only hold itself together by tensile strength around its circumference. By adding some inner support structure to help pull the edge inward, the flywheel can spin faster and store more energy. A disk doesn't have enough support structure, while something like a sphere or a bicone has too much support structure.
The solution turns out to be to use a disk with an exponentially decreasing thickness—like a bicone, but curved slightly inward. Such an arrangement makes the stress at every point of the flywheel equal, maximizing the use of the material. An exponential falloff means the disk is infinitely large, so in-practice of course flywheels cannot be shaped like this. Instead, they are shaped in approximations, or engineers don't bother: other design considerations (like safety^{[4]} or manufacturing ease) usually take precedence to maximum energy density.
Nevertheless, this optimal, inward-curving shape gives rise to the idea of a "mass-efficiency" factor, commonly denoted "\(k\)". A value \(k=1\) corresponds to an ideal flywheel, with real flywheels taking on strictly-smaller values.
Computing the \(k\)-factor for a given geometry is complicated. Resources are scarce and sometimes disagree with each other. The best reference I found is "Flywheels for Low-Speed Kinetic Energy Storages Systems" (sic). Instead, the factor is usually precomputed and referenced (peculiarly, always to three digits) later. As best as I can tell, here are some such factors, along with their associated geometry and expression, if it isn't too complex:
\(k\)-Factor | Expression | Geometry |
---|---|---|
\(1\) | \(k=1\) | Ideal flywheel, where thickness exponentially decreases with radius (exact formula is given by equation 10 in reference). Optimal, but infinitely large—so, impossible. |
\(\approx 0.7\) to \(\approx 0.97\) | (See equation 15 in reference.) | Ideal flywheel, but the end truncated and replaced by "ballast", possibly of some greater thickness. Exact efficiency depends on design parameters. Common in high-performance engineering work. |
\(\approx 0.95\) | (Numerically calculated.) | Bicone. Common in high-performance engineering work. |
\(\approx 0.6\) | \(k=\frac{2}{3+\nu}\), where \(\nu\) is Poisson's ratio. | Flat disk. Note that using \(\nu=0.3\), typical of metals, reproduces the widely-quoted figure \(k \approx 0.606\). Very common in the real-world, due to its simplicity! |
\(0.5\) | \(k = 0.5\) | Idealized flywheel with all of its mass concentrated in an infinitely-thin ring around the edge. Approximations of this type are common in the real world because of their simplicity. Also, because this arrangement has the highest moment of inertia, I suppose it has the least variation in angular speed when adding or removing energy. |
I am aware of a few other values, but they come with little context, and I suspect are unimportant or erroneous. For example, I often see \(k=0.931\) listed as a "constant-stress" disk, while I believe that this is simply a particular example of one of the many types of approximations to constant-stress disks, now being quoted without context.
The practical upshot here is that \(0 < k < 1\) is a constant that determines how close your flywheel system is to optimal. It's easy to get \(0.6\) or a bit higher, and with enough work you can get around \(0.95\) or better, with intermediate designs falling somewhere in-between.
^{[4]} For example, one arrangement is to use a ring-shaped design, with a thin disk supporting it, such that the region of highest stress is on the disk just inside the outer edge. This way, if the flywheel is spun too fast, the disk fails instead of the outer ring. This is much safer than if the outer ring itself had ruptured, since the outer ring contains most of the kinetic energy.