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Name: Aaron
Status: educator
Grade: 9-12
Location: CA
Country: USA
Date: Summer 2013

I have heard of some small groups creating compressed air flywheels, or a device that uses compressed air as stored energy to run a turbine when energy is needed around January of this year. I am curious not only of how efficient this would be but also how efficient a vacuum tube flywheel would be; a device that pushes molecules out of a container until energy is needed which is given by opening the container slightly and use the air rushing in to move a turbine.

Hi Aaron,

Do not confuse "efficiency" with the amount of stored energy. Such a system as you describe (a tank evacuated to a perfect vacuum, with the in-rushing air driving a turbine) could no doubt be designed to be reasonably efficient, but it will not store very much energy.

Look at it this way..... A tank containing a perfect vacuum is at a pressure of only 1 atmosphere (14.7 psi) LOWER than the outside air. It would only store exactly the same amount of energy as the same tank pressurized to a mere 14.7 psi ABOVE atmospheric pressure (only about half the pressure that a car tire has!). The result, either way, would be not enough energy stored, to do any useful work.

Regards, Bob Wilson

Hi, Aaron

Thanks for your question. I am an electrical engineer. I do know much of an answer to your question and will provide that.

When you compress a gas, it becomes hot. If you rapidly compress air in an insulated tank, the temperature of the air will increase, so that it will be hotter than it was before it was compressed. Regarding the gas in the container, there is a governing rule involved; PV=nRT. Pressure x Volume = (n = number of moles of gas) x (R which is a constant) x (Absolute Temperture in degrees Kelvin).

see regarding moles and the number n.

Due to the formula and the heating of the gas, as you increase the number of molecules and the temperature in the container the pressure will increase by an amount that is more than proportional to the amount of air. If you quickly release the pressure before the gas cools, you should be able to recover an amount of energy almost equal to what you put in. However the more likely scenario is that the gas will cool as it sits for a while, thus decreasing the pressure. This means that the pressure available for energy recovery will be less than the pressure required to fill the container. In fact regardless of the gas temperature when you start to release the pressure, the temperature will drop (thus decreasing the pressure further) as you release the pressure. If the compressed gas was at room temperature before release, it will chill and actually become cold during release. This further reduces the pressure available for energy recovery according to the formula. The work you get out will be significantly less than that which you put in.

If you use negative pressure (vacuum) instead of positive pressure, you will reverse the sequence but be stuck with the same problem. In addition, with positive pressure a strong container can hold many "atmospheres" of pressure. But with vacuum you can only pump out one "atmosphere", so the useful capacity of a container size will be greatly reduced.

The result is that using gas compression or vacuum for energy storage is relatively inefficient.

On the subject of the turbine; I think that turbines require a relatively rapid flow rate in order to transfer energy efficiently. I think that a piston type motor will more efficiently transfer energy at slow and variable flow rates.

Best regards, Bob Zwicker

Hi Aaron,

Thanks for the questions. Compressing air is a method of storing energy. If one calculates the energy of a typical compressed gas cylinder, it has about as much energy as a stick of dynamite. (There are a couple of ways of doing this calculation.) For that reason, compressed gases are to be handled with due caution. The compression process is not highly efficient since a motor is needed to run a compressor. Much heat is produced when a gas is compressed--you may get burned if you touch the output high pressure line.

I hope this helps. Please let me know if you have more questions. Thanks Jeff Grell

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