Oxy hybrid fueling question

[Combustion Monkey]

I am thinking about making a hand held hybrid using oxygen for portability. The cannon wont see past 4x because I have no desire to frag myself. The question is, what is the ratio of oxygen to propane, assuming that the chamber has 1x of air in it at the start?

[SpudFarm]

never calculated or read that but when you know that there is around 21% O in the air the rest should be simple.

[SpudUke5]

Well its a little more complicated than that.

When propane, C3H8, combusts it requires 5 moles of oxygen to 2 moles of propane. So for a ratio of how much you need, its a 5:2 ratio of oxygen:propane. So you need 80% oxygen and 20% propane. But you need to account for the 21% of oxygen in the atmosphere so im not sure about that. I think you can purge the chamber with oxygen and displace the air inside the chamber and then measure the amount of propane.

Dont take my exact word on it because im not sure. But its not just 21%.

[starman]

Assuming you start out with a real air load, the first X, for every X after that, and assuming a stoichiometric O2 meter, you would need 5 times the per X volume of propane, and about 4.8 times the per X volume of MAPP.

Does that make any sense?

Edit:
I believe the balance equation is: C3H8 + 5O2 => 3CO2 + 4H2O

So 1 mole of propane to 5 moles of O2.

[Combustion Monkey]

I knew I should have payed closer attention in math class

[john bunsenburner]

this is chemistry mate. Propane will burn pretty much no matter what, if exess O2 is provided it burns like starman said if too little O2 it burns to form C, CO, CO2 and H2O:

C3H8 + 3.5 O2 → CO2 + CO + C + 4 H2O + heat.

Really i suggest you have a slight exess of O2, to make sure as i doubt your regultor will be an exact stoichiometric O2 meter. And yes in this case the ratio of C2H8 to O2 is 1:5.

[Combustion Monkey]

Really i suggest you have a slight exess of O2, to make sure as i doubt your regultor will be an exact stoichiometric O2 meter. And yes in this case the ratio of C2H8 to O2 is 1:5.

Does that 1:5 ratio take into account the 1x of air that will be in the chamber to start with?

[john bunsenburner]

well thats is why i suggested to put an exess of O2 in, theoretically you could put propane, which is heavyer than air in first, let the air coem out(so slowly filling, indoors) and then saddign 5 times you chamber volume in O2. that would be 6x though.

[Combustion Monkey]

Hmm not sure I want to be firing this at 6x. But I guess the numbers could always be tinkered with to bring the mix back down to 4x

[john bunsenburner]

well you i guess you could go for 6x, and yes of coarse the numbers can be played with but i assumed you knew you chamber vol and wanted the easy way.

[Combustion Monkey]

Thats true, I do want the easy way and the gun will be able to take 6x. I haven't actually built the chamber yet. I wanted to get the fueling figured out before I built any farther.

[john bunsenburner]

Well i think your on the right track, i am also pretty sure the gun will with stand 6x, only thing to worry about might be the recoil. Otherwise you have things figured out, cant wait to see the gun!

[starman]

Really i suggest you have a slight exess of O2, to make sure as i doubt your regultor will be an exact stoichiometric O2 meter. And yes in this case the ratio of C2H8 to O2 is 1:5.

Does that 1:5 ratio take into account the 1x of air that will be in the chamber to start with?

Yeah, when you load up a 1x mix, the "magic" 4.2% propane volume number comes from the fact the there's 21% (or so) O2 in the chamber. From the ideal formula, 5 (moles of O2) x 4.2% = 21% (percent of O2 in air)

So, first assume a chamber of fresh air, then add in the fuel for total number of X shot you're taking....then add enough O2 to account for X-1, because your first X of O2 is already there in the air.

I've seen hybrids that didn't fool with fresh air at all and just pumped in enough O2 for the whole X shot....the burned and left over gases from the previous shot becoming the "buffer gas" that nitrogen usually acts as.

[Combustion Monkey]

I've seen hybrids that didn't fool with fresh air at all and just pumped in enough O2 for the whole X shot....the burned and left over gases from the previous shot becoming the "buffer gas" that nitrogen usually acts as.

Thats exactly what I was going for, but was unsure about the spent gasses in the chamber.

Thanks for the good advice everyone and keep it coming! If all goes well this should be a cool little gun. I finished the piston assembly yesterday, and once I get the metering nailed down the hard part is over.

[SpudUke5]

Assuming you start out with a real air load, the first X, for every X after that, and assuming a stoichiometric O2 meter, you would need 5 times the per X volume of propane, and about 4.8 times the per X volume of MAPP.

Does that make any sense?

Edit:
I believe the balance equation is: C3H8 + 5O2 => 3CO2 + 4H2O

So 1 mole of propane to 5 moles of O2.

Ah thats right, sorry i was going off memory, which was obviously wrong.