Butane Combustion Question

First, look up stoichiometry. Then look up LEL and UEL of butane.

Is this homework?

No. This is not homework.

My youngest boy's best friend is a paintball enthusiast. The friend and his father have built what they call a paintball shotgun out heavy gauge PVC pipe. It is quite impressive! It shoots about twenty paintballs per load near what they said was about 200 feet per second at the barrel.

My questions are motivated out of concern for safety. I am not a chemist, but I am good enough with numbers and tables to know when something might be approaching a danger point. Pressures and such, ratings and that kind of thing. I am a practical man, but I am not a scientist or an engineer.

It does not seem to me that what they built is dangerous per se, but I do want to do a little legwork "behind the scenes" to back up with fact any valid concerns I may have that need to be brought to their attention.

My friend's dad has a good head on his shoulders, and I trust his judgement. If I'm honest with myself, I might just be wanting to know the answers to my questions for my own comfort as much for any safety concerns I may have.

I could look up "stoichiometry" and LEL and UEL and hoe a long row the hard way, but I'd rather not if I didn't have to. So I am asking you fellas for a little help with something that concerns me somewhat.

Fair enough?

Fair enough.

We'll need to know the size and schedule of the pipe so we can check the burst pressure. Also the combustion chamber volume etc. I'm too old for paintball but I'm sure there are guys here who know.

As far as LEL/UEL, that would give you the outer limits of combustion so the mid way point in the slope should be the optimum mixture.

Here are some numbers.

I'm looking for the pressure numbers.

More details please.

Sorry,

The table in post #3 came from:www.engineeringtoolbox.com/explosive-concentration-limits-d_423.html

Guess you'll have to cut and paste.

Thanks.

Air consists in about 20% O₂, so given one liter of air couldn't we figure how much butane would be needed to consume exactly that much oxygen, ideally? Or is it more complicated than that? This guy's prolly looking for worst-case temps and pressures.

"so given one liter of air couldn't we figure how much butane would be needed to consume exactly that much oxygen,"

Yes, there is a stoichiometric amount of butane, defined by the mid point of the combustion levels already suggested that would give the most energy per unit volume.

I, obviously am not capable of explaining it correctly. So, I will leave it to others to help.

We still have not resolved the pressure question.

Hello Guest,

To make this fairly straightforward, I am starting with the assumption that the combustion is "perfect". By that I mean that the reaction produces ONLY water and carbon dioxide. Another assumption is that both the air and the butane, as well as the reaction products, are ideal gases.

The stoichiometry is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Because the volume of one mole of an ideal gas at STP is 22.4L and, since both are seeing the same conditions, we can scale according to the stoichiometric equation. Since air is only 21% oxygen, the air required would be 13/0.21 = 61.9 moles. Scaling this to 1L of air, the volume of butane would be 2 moles butane/61.9 moles air = 0.032L = 32mL butane.

OK, I've answered the first part. In the stoichiometric equation, the total number of moles of the reactants is lower than the number of moles of the products. Including the other nonreacting components of the air, moles of reactants is 63.9 and moles of products is 66.9. Because of this there is a slight increase in volume, hence the pressure in a closed chamber. Pressure will increase to a much greater degree because of the "instantaneous" increase in temperature due to the combustion. To calculate the pressure and temperature of this reaction is beyond my knowledge. Maybe someone here can help you with that part.

Mike

Sorry, I may have misled you by assuming that the OP was doing homework.

The real question here is how much pressure is required to make the device fail and what would be the failure mechanism.

Then, the question is can butane and air achieve this pressure.

How about we assume that, as an approximation, the new gas temp is simply the combustion temp of oxygen + butane and from that estimate the corresponding change in pressure? Could we get 'reasonably' close (whatever that means) to worst-case values if we did it this way? A kind of first draft, as it were?

Okaaaayyy...

So, what is the combustion temp of C4/O2

Went here:

http://www.derose.net/steve/resources/engtables/flametemp.html

Got 1,977ºC = 2350K (don't you guys Google? - OK, I'm being facetious; I don't really know how reliable this even is).

With the volume staying the same we have:

n₁RT₁/P₁ = n₂RT₂/P₂

P₂ = n₂T₂P₁/n₁T₁ = (66.9mol)(2350K)(1bar)/(61.9mol)(293.15K) = 8.66 bar

Mike

Actually, I was Googling about the same time you were scolding us ne'er-do-wells.

^{_{I'm a little slow, maybe, but I'm still way ahead of dead. I think. Hey! What's that Light?!}}

I found this bit on combustion temps, but the numbers assume:

1) Reactants enter the combustion process at 25C @ 1 atm

2) Products leave the combustion process at 1 atm,

3) Process is adiabatic,

4) Combustion is stoichiometric ...

5) ... and without 'excess air'

Hi Eu,

I think you know I was only kidding about the Googling thing. With the assumptions you brought up, along with the two I made, make this an extremely complex problem to solve with any accuracy.

_{It's not the "light at the end of the tunnel", but the headlight of a train coming to run you over!}

Mike

"I think you know I was only kidding about the Googling thing..."

_^<hehe> You were?

"With the assumptions you brought up, along with the two I made, make this an extremely complex problem to solve with any accuracy"

Poor Guest. "Ask an engineer the time, and he'll tell you how to build a watch!"

I was hoping certain assumptions would simplify the problem. Where's a dadgum rocket scientist when you need one? Those guys work these kinds of problems* - combustion under pressure - all the time.

^{_{* And this is what happens when they screw up}}

8.66 barx14.7PSI=127PSI