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I'm currently working on an idea to use stored compressed liquid oxygen in a scramjet-style engine for suborbital travel. However, I'm not finding information about how much gaseous oxygen is stored in a particular unit of compressed liquid oxygen. I'm also interested in the best way to convert said compressed liquid ox into a gaseous state. Any answers with background/supporting info would be greatly appreciated!

2006-06-26 16:51:26 · 5 answers · asked by N3VJA 3 in Science & Mathematics Chemistry

OK, OK, I understand the mechanics of combustion, and I understand the mechanics of storing the liquid O2. My question deals with quantity, not mechanics.

2006-06-26 17:22:58 · update #1

5 answers

Density of LOX is 1140 kg/m3 at -183 C, 1 atm
Density of GOX is 1.29 kg/m3 at 0 C, 1 atm (same as air, ideal gas)

Therefore 1 m3 LOX = 884 m3 GOX

Dense fluid expander would work for LOX to GOX however this would probably be too heavy for a spaceship.

Note of caution, if to do use pure gaseous oxygen, it is a major fire risk. Do not get hurt screwing around.

2006-06-26 17:37:58 · answer #1 · answered by sara 1 · 0 0

You can't have liquid oxygen unless it is under a great deal of pressure, or it's very, very cold. The answer above applies - open the valve, and the liquid will be come a gas almost instantaneously. You'll probably have to heat the valve and the tank to make sure they don't freeze up from the Joule-Thomson effect.

Be warned that every gram of fuel will cost a gram of payload.

2006-06-26 17:13:59 · answer #2 · answered by rb42redsuns 6 · 0 0

Release the pressure on it in a vacuum and it will convert to gas very quickly. However it would be more energetic to mix it with something and use the combustion for additional energy.

2006-06-26 16:57:59 · answer #3 · answered by Anonymous · 0 0

I am sure they would pass plenty of hospitals, because I assume that is the only place they fill them up.. legally.

2016-03-27 05:30:37 · answer #4 · answered by Anonymous · 0 0

i would use stoiciomitry

2006-07-09 17:01:57 · answer #5 · answered by Anonymous · 0 0

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