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Flywheels - Kinetic Energy

The kinetic energy stored in flywheels - the moment of inertia.

A flywheel can be used to smooth energy fluctuations and make the energy flow intermittent operating machine more uniform. Flywheels are used in most combustion piston engines.

Energy is stored mechanically in a flywheel as kinetic energy.

Kinetic Energy

Kinetic energy in a flywheel can be expressed as

Ef = 1/2 I ω2(1)

where

Ef = flywheel kinetic energy (Nm, Joule, ft lb)

I = moment of inertia (kg m2, lb ft2)

ω = angular velocity ( rad /s)

Angular Velocity - Convert Units

  • 1 rad = 360 o / 2 π =~ 57.29578 o
  • 1 rad/s = 9.55 rev/min (rpm) = 0.159 rev/s (rps)

Moment of Inertia

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

I = k m r2(2)

where

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

m = mass of flywheel (kg, lbm )

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 - Convert Units

  • 1 kg m2= 10000 kg cm2= 54675 ounce in2= 3417.2 lb in2= 23.73 lb ft2

Flywheel Rotor Materials

Flywheels - Kinetic Energy
MaterialDensity
(kg/m3 )
Design Stress
( MPa)
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
  • 1 MPa = 106 Pa = 106 N/m2= 145 psi
  • 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

A 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 the 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 m2

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 wheel 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

Ef = 0.5 (0.22 kg m2) ( 20.9 rad/s )2

= 47.9 J

Related Topics

  • Dynamics

    Motion of bodies and the action of forces in producing or changing their motion - velocity and acceleration, forces and torque.
  • Mechanics

    The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.

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