- Kinetic Energy = 1/2*I*w*w

Inertial constants for different shapes:

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:

M*w*w*R

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

Example Calculation:

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*w
^{2}. For flywheels I =1/2MR^{2}. - If we measure w in revolutions per second then the
stored energy of a flywheel is approximately 6MR
^{2}x w^{2}(RPS) - For M=140 kg and R=50cm this yields a required w of 500 RPS or
30,000 RPM
- 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.

Compressed Air:

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 salt deposit
- 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 expansion stage

For gases, Pressure is directly related to Temperature (Ideal Gas Law)

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 e-mail: nuts@moo.uoregon.edu