# Flywheel Kinetic Energy

## Kinetic energy and moment of inertia of a flywheel

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A flywheel is used to smooth the energy fluctuations in combustion engines and make the energy flow uniform. The flywheels store energy mechanically as kinetic energy.

### Kinetic Energy

The kinetic energy of a flywheel can be expressed as

E_{f}= 1/2 I ω^{2}(1)

where

E_{f}= flywheel kinetic energy (Nm (Joule), ft lb)

I = moment of inertia (kg m^{2}, lb ft^{2})

ω = angular velocity (rad/s)

*Angular velocity converting units*

*1 rad = 360*^{o}/ 2 π =~ 57.29578^{o}*1 rad/s = 9.55 r/min (rpm) = 0.159 r/s (rps)*

### Moment of Inertia

Moment of inertia quantifies the rotational inertia of a rigid body and can be expressed as

I = k m r^{2}(2)

where

k = inertial constant - depends on the shape of the flywheel

m = mass of flywheel (kg, lb_{m})

r = radius (m, ft)

Inertial constants of some common types of flywheels

- wheel loaded at rim like a bicycle tire -
*k =1* - flat solid disk of uniform thickness -
*k = 0.606* - flat disk with center hole -
*k = ~0.3* - solid sphere -
*k = 2/5* - thin rim
*- k = 0.5* - radial rod
*- k = 1/3* - circular brush -
*k = 1/3* - thin-walled hollow sphere -
*k = 2/3* - thin rectangular rod -
*k = 1/2*

*Moment of Inertia Converting Units*

*1 kg m*^{2}= 10000 kg cm^{2}= 54675 ounce in^{2}= 3417.2 lb in^{2 }= 23.73 lb ft^{2}

### Flywheel Rotor Materials

Material | Density(kg/m)^{3} | Design Stress ( MN/m)^{2} | Specific Energy ( kWh/kg) |

Aluminum alloy | 2700 | ||

Birch plywood | 700 | 30 | |

Composite carbon fiber - 40% epoxy | 1550 | 750 | 0.052 |

E-glass fiber - 40% epoxy | 1900 | 250 | 0.014 |

Kevlar fiber - 40% epoxy | 1400 | 1000 | 0.076 |

Maraging steel | 8000 | 900 | 0.024 |

Titanium Alloy | 4500 | 650 | 0.031 |

"Super paper" | 1100 | ||

S-glass fiber/epoxy | 1900 | 350 | 0.020 |

- Maraging steels are carbon free iron-nickel alloys with additions of cobalt, molybdenum, titanium and aluminum. The term maraging is derived from the strengthening mechanism, which is transforming the alloy to martensite with subsequent age hardening.

### Example - Energy in a Rotating Bicycle Wheel

The typical *26-inch* bicycle wheel rim has a diameter of *559 mm (22.0")* and an outside tire diameter of about *26.2" (665 mm)*. For our calculation we approximate the radius - *r* - of the wheel to

*r = ((665 mm) + (559 mm) / 2) / 2 *

* = 306 mm*

* = 0.306 m*

The weight of the wheel with tire is *2.3 kg* and the inertial constant is *k = 1*.

The Moment of Inertia for the wheel can be calculated

*I = (1) (2.3 kg) (0.306 m) ^{2} *

* = 0.22 kg m^{2}*

The speed of the bicycle is *25 km/h* (*6.94 m/s)*. The wheel circular velocity (rps, *revolutions*/s) - *n _{rps} *- can be calculated as

*n _{rps} = (6.94 m/s) / (2 π (0.665 m) / 2)*

* = 3.32 revolutions/s*

The angular velocity of the wheels can be calculated as

*ω = (3.32 revolutions/s) (2 π rad/revolution)*

* = 20.9 rad/s*

The kinetic energy of the rotating bicycle wheel can then be calculated to

*E _{f} = 0.5 (0.22 kg m^{2}) (20.9 rad/s)^{2} *

* = 47.9 J*

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