MEC-08

Flywheel fluctuation energy

ΔE = I ω² Cs with Cs = (ωmax−ωmin)/ω.

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Energy storageShigley

Governing equation

\Delta E = I\omega^2 C_s,\quad C_s=\dfrac{\omega_{\max}-\omega_{\min}}{\omega}

where

I: Inertia (kg·m²)
\omega: Mean speed (rad/s)
C_s: Coefficient of fluctuation (—)
\Delta E: Energy fluctuation (kJ)

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-08 — Flywheel fluctuation energy) is the form associated with Shigley. Working symbols: $I$ , $\omega$ , $C_s$ $\rightarrow$ $\Delta E$ . A flywheel stores ½ I ω². The energy that can be given or taken while speed fluctuates by Cs is I ω² Cs.

Purpose

Purpose: compute

\Delta E

from

I

\omega

C_s

in Mechanical via

\Delta E = I\omega^2 C_s,\quad C_s=\dfrac{\omega_{\max}-\omega_{\min}}{\omega}

ΔE = I ω² Cs with Cs = (ωmax−ωmin)/ω. Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

I = 4.000\,\mathrm{kg·m^{2}}

\omega = 150.000\,\mathrm{rad/s}

C_s = 0.050\,\mathrm{—}

, the governing relation

\Delta E = I\omega^2 C_s,\quad C_s=\dfrac{\omega_{\max}-\omega_{\min}}{\omega}

yields

\Delta E = 4.500\,\mathrm{kJ}

. Mean ω, given Cs. Solid-disk I is an input. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Inertia I4 kg·m²Mean speed \omega150 rad/sCoefficient of fluctuation C_s0.05 —

Outputs

Energy fluctuation \Delta E4.500 kJ

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Narration of this film

Mean ω, given Cs. Solid-disk I is an input.

A flywheel stores ½ I ω². The energy that can be given or taken while speed fluctuates by Cs is I ω² Cs.

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Watch on YouTube