Flywheel Energy Storage Calculator

A flywheel is a battery made of momentum: a motor spins a rotor up, and the energy sits in its rotation until a generator slows it back down. The stored energy is the rotational kinetic energy

$E = \tfrac{1}{2}I\omega^2$

Symbols and SI units:

• $E$ — stored energy, joules (J); divide by $3.6 \times 10^6$ for kWh
• $I$ — moment of inertia about the spin axis, kg·m²
• $\omega$ — angular velocity, radians per second (rad/s)

Convert from rotational speed in rev/min with

$\omega = \frac{2\pi N}{60}$

The inertia depends on how the mass is distributed:

with $M$ in kg and $R$ in m. Mass at the rim counts far more than mass near the hub, which is why real rotors are heavy at the edge.

The assumption people forget: $\omega$ must be in rad/s, not rpm. Substituting $6000$ directly overstates the energy by a factor of about $91$.

Energy density is stored energy per unit rotor mass, in joules per kilogram (J/kg) or watt-hours per kilogram (Wh/kg):

$e = \frac{E}{M} = \frac{I\omega^2}{2M}$

For a solid disc, $I = \tfrac{1}{2}MR^2$ gives $e = \tfrac{1}{4}R^2\omega^2$ — the mass cancels. Capacity therefore comes from speed and radius, not from bulk, and the real ceiling is the tensile strength of the rim material, since hoop stress also grows with $R^2\omega^2$.

Usable energy: a flywheel cannot be run down to zero, so only the band between the maximum and minimum working speeds is available:

$E_{\text{usable}} = \tfrac{1}{2}I\left(\omega_{\max}^2 - \omega_{\min}^2\right)$

Because $E$ goes as $\omega^2$, halving the speed leaves a quarter of the energy, so a $2{:}1$ speed range delivers $75\%$ of the stored total.

The assumption people forget: this is the stored energy. Real systems lose a few percent per hour to bearing and windage drag, so flywheels suit short-duration, high-power duty rather than long-term storage.

• Feeding rpm into $\tfrac{1}{2}I\omega^2$ — convert with $\omega = 2\pi N/60$ first.
• Using $\tfrac{1}{2}mv^2$ with the rim speed — that ignores the slower inner material; use $I$ instead.
• Taking $I = MR^2$ for a solid disc — that is the thin-rim value and doubles the answer.
• Using the diameter as $R$ — it is the radius, and the error is a factor of four.
• Assuming doubling the mass doubles the energy density — for a disc the mass cancels; only $R$ and $\omega$ raise it.
• Quoting stored energy as deliverable — subtract the residual at minimum speed, plus standby and conversion losses.
• Mixing units in kWh conversions — $1$ kWh $= 3.6 \times 10^6$ J.