Mechanical Engineering
Flywheel Energy Storage
Calculate kinetic energy stored in a rotating flywheel and the speed drop under a given load torque impulse.
Units:
Flywheels store energy as E = 0.5 x J x omega^2. Doubling the speed quadruples the energy stored. Solid disc: J = 0.5 x m x r^2. Rim flywheel (most material at rim): J approx m x r^2.
E = 0.5 x J x omega^2 | J_disc = 0.5 x m x r^2 | delta_n = T x t / J
Input
Flywheel mass iTotal mass of the flywheel in kg.
kg
Effective radius iRadius of gyration (solid disc: 0.707 x outer radius, rim flywheel: outer radius).
mm
Operating speed iRotational speed in rpm.
rpm
Minimum allowable speed iMinimum speed before system problems occur. Coefficient of fluctuation = (n_max - n_min) / n_mean.
rpm
Load torque impulse iPeak load torque applied over the pulse duration.
N·m
Pulse duration iDuration of the load torque pulse in seconds.
s
Result
Moment of inertia J [kg·m²]
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Angular velocity ω [rad/s]
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Kinetic energy stored [J]
—
Kinetic energy stored [kJ]
—
Speed drop under pulse [rpm]
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Speed after pulse [rpm]
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Assessment
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