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On a standard combustion gun it is often tossed around that they are only capable of 100 psi at the most. While I have no evidence to the contrary I often scratch my head and think "where did this number come from?" I have done numerous searches that have turned up nothing of value. As a matter of fact from what I can tell there is no evidence what so ever that this number was ever anything factual at all.
A few things concern me with this number, was it attained by someone completely sealing a chamber, connecting a PSI gauge and igniting the mixture inside? If so how is it possible for something as inaccurate as a psi gauge to correctly capture a PSI spike that might last a millisecond or two? Psi gauges are notoriously bad at capturing spikes in psi. For that reason alone I would dispute this "100 psi" claim, however that is just the begininng of my dispute.
What size chamber is this 100 psi max theory based on? Would this necessarily matter? Perhaps not, but perhaps the natural properties of the chamber may vary based on volume, material, and yes even shape after all we are dealing with a millisecond event here. What is the ignition source? Are we talking about an even metered/chamber fanned mixture or just some right gaurd spray and pray?
It just seems to me that this number was randomly plucked from thin air (or somewhere else. ) If I'm wrong please point me to where this number is derived from and the data that was collected.
Thoughts? Anything I missed? Direct evidence?
well you can't get above 100psi on HGDT for a non-hybrid combustion.
what type of a meter would you use to find this, I am thinking that you would have to have an expensive electronic pressure sensor hooked up to a computer to monitor the whole event. Do these sensors even exist?
If I am way off I'm sorry but let me know so I can learn from my mistake
Some hard data, courtesy of jimmy
And while it is a great tool the author of HGDT David Hall said it best when he wrote in his FAQ
This is what logic would dictate. You would need an extraordinarily accurate and sensitive psi meter. The other route to getting a respectable understanding of such an event would be through mathmatical understanding of the reaction of propane and atmospheric air. I don't think that there would be an accurate representation for how all combustion based guns with either method, but at least that number would be based off of some sort of real world understanding and not this amorphous 100 psi assumption.
Wow Jimmy, great article! Very very well written. Excellent job on the data and researched.
I plucked a few lines from this to illustrate my point
this is the crux of the issue. Combustion is a very complex event and the ability to measure accurately when it counts most is elusive at best.
This 80 to 100 blur is concerning in that A. this is a very simple gauge that is designed to measure constant psi not accurately or sensitively capture a spike in pressure. B. In no way is video taping an gauge going to give you an accurate assessment of what's going on in the chamber.
This again illustrates my point, this is not as simple as looking at a gauge of a sealed chamber. There is a very complex reaction going on here.
You can look at it another way. For a given volume of air/propane at 95.98 : 4.02% (optimum mixture) you'll have a certain amount of energy available. This can be calculated.
If you find the specific heat of each combustion product and apply the energy from the reaction to them, you can find the net temperature increase from the reaction.
Figure out how much the gases expand at the new temperature and you'll get results that you can put into BAR/psi.
Or I think that's right at least .
Energy density of propane is approximately 2.45 MJ/M^3 (according to online source). I think this figure would be useful in calculations...not too inclined to do it myself right now.
By that number, an advanced combustion would only have about 250 joules total energy, figuring 6 cubic inches of propane for a 150 inch^3 chamber.
By conservation of energy, that would mean that a golfball would shoot out at about 340 ft/s, if all of the energy was imparted to it...does this number seem low? I don't know about the energy density number, it just came up in a search and had proper units.
That number is very low.
You're looking at the wrong figure. You want the energy change of combustion, which is 2,219,200 J/mol. (And yes, I know that off the top of my head, because I'm a total geek)
In real world terms, this means that per litre of propane/air mix at 0 degrees C (32 F), there are approximately 4000 Joules of energy released on combustion.
At 20 degrees C (68 F), this falls to 3700 Joules.
Interestingly, if you convert these energies into pressures, you get numbers like 550 psi.
I don't think I'm QUITE so tired that my maths has gone that bad (and if it has, I'm screwed, because the reason I'm still up is to do my maths assignment) - so I have to hope there's some factor I'm not seeing.
Does that thing kinda look like a big cat to you?
Some where we should have a huge pdf with ALL the formulas so that any moron like me can go and put the numbers into their calculator and tada have everything in black and white, or maybe some one could even make a program with ALL the formulas we will ever need...just a suggestion...
The easiest way to look at the "max pressure" question is to just do the thermodynamic calculation for propane + air in a closed chamber.
GasEq will do it. Or, as others have posted, you can use the thermodynamic heat of combustion and calculate it yourself.
Bottom line is that in a closed chamber a perfectly measured and perfectly mixed mixture of propane + air, ignoring any possible heat loss (which means the estimate is a bit high) ... ~137 PSIA, ~122 PSIG.
Some might say "but that is a 'theoretical' result" and theory is often different than the real world. That is true, but in thermodynamics the real world always works less well than what the model predicts. In our case, it is unlikely that a real gun will reach the ~122 PSIG predicted value. Heat loss, less than perfect mixing or ratio, leakage etc all drop the actual peak pressure. There aren't really any mechanism that can increase the static pressure above what is predicted by theory.
(The only possible way to get an instantaneous pressure greater than predicted would be because of shock affects. If you get to DDT the peak pressure at certain points in the chamber can be much higher than thermo predicts. But, a shock wave last for a tiny length of time compared to the combustion process. The time domain is so short that the mass and momentum of the chamber will significantly increase the chambers ability to withstand that short duration pressure spike.)
So, worst case scenario would be a round jammed in the barrel. The chamber pressure would spike to about 120 PSIG. Within one second after ignition the gases would have cooled off enough so that the pressure in the closed chamber was back in the range of a few PSIG.
In a functioning gun the peak pressure is significantly less since the round will start to move before peak pressure is obtained. The moving round enlarges the chamber and the peak theoretical pressure drops accordingly. The combustion process is so slow that all rounds I've ever seen are moving well before the peak pressure is obtained. (A perfectly tuned burst disk gun might be able to get pretty close to the ~122 PSIG peak, but it won't ever actually reach the theoretical peak since there will always be some heat loss.)
So, the "100 PSIG peak pressure" actually should be more like 120 PSIG. That is a pretty firm number and is reliable. In a normally functioning spud gun the peak will be significantly less. In my experience, the peak pressure in typical spudguns is about 60 PSIG.
Hmm thanks for the info Jimmy. obviously burst disk and perhaps even a tighter fitting round will increase the likelihood that one may reach a release pressure potentially closer to the theoritical 120 PSIG, but because of the nature of most combustion guns you are looking at a more likely range of 60 or so PSIG.
Most of the time when the 100 psi number comes up is in regards to safety, but my question is more what the practical yield is and your post has answered that to my satisfaction! Thank you.
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