Internal energy of compressed gas

[Bhope69199]

Hi,

I've been reading about compressed air energy storage and keep coming across that in 300 bar containers the achievable energy is 0.1MJ/L. Is this 0.1MJ/L of the volume of the air it is compressed to or of the total L of air that was initially used? (E.g If 1500L is compressed to 300 bar into a 5L cylinder will the total energy be 0.5MJ or 150MJ)

I have been trying to work this out but cannot seem to come up with the the 0.1MJ/L value.

Using the internal energy calculation U = 5/2PV I get 0.375MJ which is 0.00025MJ/L . (5L volume at 300 bar = 1500L total air. If I compress 1500L of air at 300 bar it can compress to 5L with total internal energy of 0.375MJ)

Could someone explain how they worked out the 0.1MJ/L value and where I am going wrong?

Thanks.

 

[scottdave]

Would you be able to give a link to the article that you read?

 

[256bits]

Look at the Wiki of compressed energy storage.
https://en.wikipedia.org/wiki/Compressed_air_energy_storage
Isothermal storage or adiabatic can theoretically give a 100% return, but most likely there will be losses for the shaft work.

Yes, a link to what you are reading would be wonderful.

 

[Bhope69199]

So plugging in the numbers in the Work equation in the wiki article I get 0.7MJ of work for a 5L cylinder. This seems far from the 0.1MJ/L quoted.

There isn't much detail in any of the articles of how they came to the 0.1MJ/L figure but here are two refs;

http://www.ijrmet.com/vol5issue1/2/11-SwadhinPatnaik.pdf

and

https://www.google.co.uk/url?sa=t&r.../696/0&usg=AFQjCNGwDqqq66y-6ChdBl3YMfuE4wCVHw

 

[scottdave]

I read the IJRMET article. They mention about 0.1 MJ/L, so 0.075 MJ/L I guess is in the ballpark (25% off). It also tosses out 0.1 MJ/kg, which at 300 bar, air has a density of 0.3 kg/L, so that figure is more achievable. It was really showing as a comparison to other ways of storing energy such as batteries or fuel, so being 25% off, doesn't seem too bad for those comparisons.

 

[Bhope69199]

OK thanks but where did you get the 0.075 MJ/L figure from?

I calculated the total being 0.7MJ so works out as either 0.14MJ/L if only considering the 5L (which is similar) or 0.00047 MJ/L if considering the total gas used of 1500L.

If it is the 0.14MJ/L then I am assuming it is per L of compressed gas.

 

[scottdave]

OK thanks but where did you get the 0.075 MJ/L figure from?

I calculated the total being 0.7MJ so works out as either 0.14MJ/L if only considering the 5L (which is similar) or 0.00047 MJ/L if considering the total gas used of 1500L.

If it is the 0.14MJ/L then I am assuming it is per L of compressed gas.

You had a figure of 0.375MJ for 5 liters, which I divided. From the context of the article, it looks like it is the compressed volume they are talking about. I remember reading an article in Popular Science, probably 30 years ago, where they had tested something to drive a small vehicle around a warehouse or something, then there would be stations at different locations to swap out the air tank (which could refill). The "engine" was all plastic parts, if I remember. I'm going from memory.

 

[Bhope69199]

OK so they only consider the 5L final volume rather than the initial volume of air used at the start (1500L).

Thanks for your help.

 

[scottdave]

OK so they only consider the 5L final volume rather than the initial volume of air used at the start (1500L).

Thanks for your help.

Yes. Suppose we want to compare compressed air and liquid propane (LP) and maybe compressed natural gas as possible energy sources for running a golf cart. You would look at these, as they are going to be installed at the equipment, not in the uncompressed state.