FLYWHEELS and ENERGY STORAGE
a wheel winds up through some system of gears and then delivers
rotational energy until friction dissipates it
stored energy = sum of kinetic energy of individual mass elements that comprise the
Kinetic Energy = 1/2*I*w*w
I = moment of inertia --> ability of an object to resist changes in its
w = rotational velocity (rpm)
I = k *M*R*R (M=mass; R=Radius); k = intertial constant (depends
Inertial constants for different shapes:
Wheel loaded at rim (bicycle tire): k =1
solid disk of uniform thickness; k = 1/2
solid sphere; k = 2/5
spherical shell; k = 2/3
thin rectangular rod; k = 1/2
To optimize the energy-to-mass ratio the flywheel needs to spin at
the maximum possible speed. This is because kinetic energy
only increases linerarly with Mass but goes as the square of
the rotation speed.
Rapidly rotating objects are subject to centrifugal forces that can
rip them apart. Centrifugal force for a rotating object goes as:
Thus, while dense material can store more energy it is also subject
to higher centrifugal force and thus fails at lower rotation speeds
than low density material.
Tensile Strength is More important than density of material.
Long rundown times are also required --> frictionless bearings and
a vacuum to minimize air resistance can result in rundown times of
6 months --> steady supply of energy
Flywheels are about 80% efficient (like hydro)
Flywheels do take up much less land than pumped hydro systems
Some Network Resources Related to Flywheels
Consider a solid disc flywheel of radius 50 cm and mass 140 kg. How
fast would it have to spin to have a store the equivalent amount
of energy that is stored in just 10 kg of gasoline when burned in
an internal combustion engine:
- 10 kg of gasoline = 140 KWH
- Engine has 15% efficiency --> 21 KWH of useable energy
- Flywheel has a conversion efficiency of 80%
- Flywheel must therefore store 21/.8 = 26.25 KWH
- Kinetic Energy goes as 1/2*I*w2. For flywheels
- If we measure w in revolutions per second then the
stored energy of a flywheel is approximately 6MR2 x w2 (RPS)
- For M=140 kg and R=50cm this yields a required w of 500 RPS or
- The required energy storage is 26 KWH/140 Kg = .18 KWH/kg
which excees the energy storage density of steel - hence such
a flywheel requires construction out of carbon fiber.
Has high energy storage capacity compared to the alternatives.
About 10 times higher per cubic meter than water.
One example (in Germany) to date:
- Storage reservoir is underground cavity in a natural
- The storage volume is 300,000 cubic meters
- Sheer weight of the salt deposit is able to pressure
confine the air reservoir
- Air is compressed to 70 atm (1000 lbs per square inch)
- Compression is done by electrically driven air compressors
- System delivers 300 Megawatts for 2 hours by using the
compressed air to drive a turbine
- Difficult to measure the efficiency of this system. Two
major contribution to the inefficiency:
- Energy required to cool the air as it is being put
into storage --> this is a critical requirement (see below)
- Energy required (usually from fuel) to expand the cool
air taken from storage as it entires the turbine.
- Desireable design feature would be recycle the waste heat
from the compression stage and use it to reheat the air during
For gases, Pressure is directly related to Temperature (Ideal
If the temperature of the air at 1 atm is 20 C, how much will
the temperature raise if we increase the pressure to 100 atm.
For air, increase in T goes approximately as increase in P to
the 1/4 power when T is measured in Kelvins
100**1/4 is about 3.5, so the temperature of the air increases
by a factor (20 +273) = 293 * 3.5. This is about 1100 K or
830 C --> which would melt the salt reservoir!
The Electronic Universe Project