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Idea for an Oxidiser

Post questions and info about combustion (flammable vapor) powered cannons here. This includes discussion about fuels, ratios, ignition systems, safety, and anything else relevant.
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Idea for an Oxidiser

Unread postAuthor: TwitchTheAussie » Sat Feb 24, 2007 9:36 pm

I've been reading about nitrous being an oxidiser, as such I've been Nitrous has to be used with another gas so will these small tanks
http://cgi.ebay.com.au/POCKET-BIKE-NITR ... dZViewItem (sorry crap with links) be good for a mix with butane/propane in a cannon?
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Unread postAuthor: SpudBlaster15 » Sat Feb 24, 2007 10:11 pm

Nitrous Oxide greatly enhances the burn rate of the fuel, and would thus be unacceptable for use in a standard Sch 40 PVC launcher. If one were to build a launcher using copper, steel, Sch 80 PVC, or any such material with a high pressure rating, a stoichiometric mixture with propane or butane would produce much higher performance levels than a stoichiometric mixture of propane/butane and atmospheric air.
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Unread postAuthor: BC Pneumatics » Sun Feb 25, 2007 10:21 am

Well said SpudBlaster.
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Unread postAuthor: LGM » Sat Mar 03, 2007 9:00 pm

DO IT DO IT DO IT!!
That seems like a really good idea, replacing the nitrous cylinders would get annoying though, see if you can rig one up with the tanks used on cars.
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Unread postAuthor: pyrogeek » Sat Mar 03, 2007 10:15 pm

N2O a liquid under pressure and becomes a gas when the pressure drops. So how would you meter it out precisely? So, you definitely can't just spray it in for a set amount of time. And due to the temperture and pressure fluxuation, you can't really use a regular metering pipe like with C3H8.

Maybe a meter pipe with a pipe prior to the regulator would work though. The N2O could be sprayed in, let it warm and stabilize, then meter it out. But, everything will have to be rated so some insane pressure to be safe.
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Unread postAuthor: frankrede » Sun Mar 04, 2007 12:22 am

Its been done, the antimaterial cannon uses onboard nitrous
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Unread postAuthor: willarddaniels » Sun Mar 04, 2007 1:55 am

It would be simple to use a metered system, just like metered propane.
You can use oxygen as an oxidizer as well, it is probably cheaper. It is surely easier to get.
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Unread postAuthor: BC Pneumatics » Sun Mar 04, 2007 4:46 pm

pyrogeek wrote:N2O a liquid under pressure and becomes a gas when the pressure drops. So how would you meter it out precisely? So, you definitely can't just spray it in for a set amount of time. And due to the temperture and pressure fluxuation, you can't really use a regular metering pipe like with C3H8.

Maybe a meter pipe with a pipe prior to the regulator would work though. The N2O could be sprayed in, let it warm and stabilize, then meter it out. But, everything will have to be rated so some insane pressure to be safe.


So does propane, and at lower pressure than nitrous.
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Unread postAuthor: pyromanic13 » Sun Mar 04, 2007 4:54 pm

I don't see why nitrous would have an advantage over pure oxygen :?
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Unread postAuthor: BC Pneumatics » Sun Mar 04, 2007 4:58 pm

Because it won't work as well. the advantage being that you don't fragment your chamber.
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Unread postAuthor: boilingleadbath » Sun Mar 04, 2007 10:31 pm

Here's an interesting analysis:

10N2O + C3H8 -> 3CO2 + 4H2O + 10N2
...or 11 mol -> 17 mol

5O2 + C3H8 -> 3CO2 + 4H2O
or 6 mol -> 7 mol

So, unless the nitros-propane burns alot cooler than the oxygen-propane (like more than 30% cooler), the nitros-propane will acctualy produce more pressure.

Have you guys seen any flame temperature stats for this or similar reactions?
(the nitros-propane, that is - O2-propane stats are everywhere)

...Thanks to SpudBlaster and GasEq, we have the relevent data.
3200*C for N2O-propane
3400*C for O2-propane

which means that the N2O provides a significantly higher pressure than the O2.
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Unread postAuthor: BC Pneumatics » Mon Mar 05, 2007 7:54 am

10N2O + C3H8 -> 3CO2 + 4H2O + 10N2
-3039KJ/mol

5O2 + C3H8 -> 3CO2 + 4H2O
-2219KJ/mol

This means that the lower reaction is only about 73% as energetic as the first. You will have to excuse my prior screw up, I was thinking that formation of N20 had a negative enthalpy value.

Does anyone have the flamefront acceleration for these reactions?
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Unread postAuthor: joannaardway » Mon Mar 05, 2007 8:27 am

That's not the problem. - you can fit less moles of the Nitrous reaction into a chamber. You need 11 units for a full reaction, whereas the C3H8/O2 only needs 6 units.

Adjusting for that factor, the energies involved are actually 33% higher for the Propane/O2 than the Propane/N2O.

But the true advantage is that nitrous is a lot easier to store than oxygen. Nitrous is stored as a liquid, but oxygen is stored as a gas at about 3000 psi.
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Unread postAuthor: BC Pneumatics » Mon Mar 05, 2007 8:32 am

joannaardway, back to high school chemistry for you.
That is just the balanced equation, and only shows the proportions you need. The truth is, gases at equal pressure and equal volume, contain the same number of moles. Thank you Avogadro.
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Unread postAuthor: joannaardway » Mon Mar 05, 2007 8:56 am

No - I'm in "high school chemistry" at the moment (well, not right now)

Now, to turn your own argument back against you, here's the numbers and figures. Let's say a chamber is 24 litres (1 mole of gas at RTP).

For the Nitrous/Propane Reaction, 10/11ths (21.8L) of that is Nitrous, and 1/11th (2.2L) of that is Propane.
For the Oxygen/Propane reaction, 5/6ths (20L) is Oxygen, and 1/6th (4L)is propane.

So for Nitrous/Propane:
We have 1/11th of a mole of propane - (Molar enthalpy change of formation -104.5 kJ a mole) - so that's 9.5 kJ of energy needed to break the bonds.
Also the 10/11ths of N2O (Molar enthalpy change of formation +82 kJ a mole) - We gain 74.5 kJ here.

The energy gained by forming the 3/11ths of CO2 (Molar enthalpy change of formation -393.5 kJ a mole) is 107.3 kJ, and the formation of the 4/11ths of water (Molar enthalpy change of formation -142.9 kJ a mole) produces 51.9 kJ.
And of course, Nitrogen is an element so it has no Molar enthalpy change of formation.

Summing that the total energy change gives ( (-107.3-51.9) - (9.5 + 74.5)) = -224.2 kJ of energy from a 24L chamber. (For those not in the know the little minus sign means it's exothermic - not negative energy)

Doing the same for Oxygen:
Break up of 1/6th of a mole of propane: 17.4 kJ loss
Production of 3/6th (1/2) of a mole of CO2: 196.8 kJ gain
Production of 4/6th (2/3) of a mole of H2O: 95.3 kJ gain.

Summing that, we get -274.7 kJ of energy from a 24 L chamber.

Putting those figures alongside each other:
-224.2 kJ from Nitrous-Propane
-274.7 kJ from Propane/O2 (significantly better)

So, for a chamber of a set size, Oxygen/Propane owns Nitrous-Propane.

Don't assume that I'm thick. I'm slap bang in the middle of an A-level chemistry course, and this is top-billing stuff.
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