Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: ChaosD.Ace on 02/09/2013 20:58:37
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Hi, I need to calculate the pressure of the gases generated when a mass of a specific explosive compound is detonated inside a confined space, for which I know:
The density of the explosive (1980 kilograms per cubic metre)
Detonation Velocity of explosive (10,100 metres per second)
Volume of said confined space ( a small cylinder, with volume 1.900153047e^-6 metres cubed)
Relative Effectiveness Factor (2.38)
The explosive is octanitrocubane. 20% to 30% more powerful than RDX. One molecule of Octanitrocubane decomposes into 8 molecules of CO2 and 4 molecules of N2 (nitrogen), so it has a pretty large volume of expanding gases. I do not know the enthalpy change of this reaction and thus I don't know the heat released but it shouldn't be too difficult to figure out.
In other words I am looking
These expanding gases will be pushing a bullet which I require to reach 900 metres per second so I do know the total thrust that must be exerted on this bullet by the pressures generated by the explosive.
You may assume that all the materials will be able to withstand the pressures and temperatures generated. The thrust I calculated is 103.3210964 Kilo Newtons.
Using F=ma (and a bullet that weights 0.045359237 kilograms) acceleration is equal to 2277840.264 m/s.
However acceleration isn’t constant due to the increasing volume between the back of the bullet and the back of the propellant casing as the bullet travels down the barrel. Perhaps you guys could clarify how to approach calculating the summation of the effect of these constantly decreasing thrust forces on the bullet (but for that I assume we need the maximum chamber pressure generated by the propellant.
I know this may be a weird question but thanks for any help you guys can give me.
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Very weird question indeed. It is not a good idea to use a high explosive as a propellant - it is more likely to shatter the bullet than to accelerate it. You need a controlled evolution of gas, and match the barrel length to
the duration of the burn.
Assuming your figures are correct (though you have got the dimensions of acceleration wrong), how did you calculate the thrust without knowing the gas pressure?
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This is a semi-auto pistol around the same size but slightly longer than a 6 inch desert eagle. 7 inch barrel.
As I said assume the bullet, chamber and barrel will withstand the pressures and temperatures generated.
The bullet begins at an initial velocity of 0, i need to accelerate it to 900 m/s over a distance of 0.1778 metres. This gives a linear acceleration of 2277840.269966254 metres pre second squared. Thus the bullet took 3.9511111111111111 e-4 seconds to travel that distance.
Using the bullets mass, change in momentum over time = impulse (force). So 0.045359237 kilograms times 900 m/s = 40.8233133 kgm/s divided by 3.9511111111111111 e-4 = 103321.09665338339 newtons. You could also use F=ma, newtons second law. So I'm kinda working it backwards in the sense that I know the velocity I want to achieve which equates to the force of thrust on the bullet which will be required which then equates to a pressure which eventually equates to the mass of the propellant that I will need, so as to figure out the case dimensions and other important stuff. I have just been having trouble finding an equation that relates pressure to an explosive's density, mass, potential chemical energy, detonation velocity etc. Now I am not sure whether the force I calculated is the maximum force after moment of detonation or the overall total additive force down the length of the entire barrel, perhaps you could clarify that.
Unfortunately this constant acceleration is just an estimate since acceleration is not constant because the gases are constantly pushing the bullet with a decreasing amount of thrust force over the time it takes for the bullet to travel the barrel's length due to an increase in the volume the bullet leaves behind it (thus pressure also decreases). How do I account for this? How do I calculate the true acceleration?, the true maximum pressure (chamber pressure) that I am going to require to achieve 900 m/s by taking into account these constant decreases in pressure, thrust etc. Is integration required here? if so how do I begin to approach it.
Also I understand the purpose of deflagration and that is why smokeless powder is used so widely for so long, but do you really mean the most powerful chemical explosive, with a perfect gas ratio and ridiculous maximum pressures can't compensate for the fact it detonates rather than burns, and accelerate a bullet through a 7 inch rifled tube more than gunpowder can, if so then consider me permanently jawdropped!!!!!!!! and i'd best start searching through the chemistry forums to find something better Also, 7 inch barrel, 900 m/s muzzle velocity, far faster than any other semi-auto pistol and I'm still gonna gun for a higher velocity. Surely the bullet doesn't spend enough time in the barrel for deflagration's effect to outweight the fact I am using the highest chemical explosive there is :(, cuz I am sure as heck not gonna get a 45 gram bullet to 900 m/s in 7 inches and a handgun with smokeless powder while keeping cartridge size down, it is the whole point of me trying something far more potent.
Sorry If i made my post unclear, I'm still in sixth form.
So do you think you have an equation that relates the parameters of an explosive compound to the pressure it will generate? or perhaps help me calculate the true acceleration/ maximum pressure required/ rate of change of pressure etc. whatever else you think you can help me with.
Thanks for the reply.
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Detonation velocity is irrelevant to the propellant value - it is the speed of the flame front through the explosive, and if it is higher than the speed of sound (which it is in this case) you will have a "high explosive" that produces a shock wave rather than the smooth release of gas you need to accelerate a bullet without distorting it.
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Dude, IT WILL NOT DISTORT. So do you have an equation?
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Hello ChaosD.Ace, guess it depends on how detailed you want to get, this is a very good book (imo) I used trying to get kids interested in a physical science class, they spent a lot of time talking about guns: Interior Ballistics: How a Gun Converts Chemical Energy Into Projectile Motion Paperback –by E. D. Lowry (Author)http://www.amazon.com/Interior-Ballistics-Converts-Chemical-Projectile/dp/B000NHXH8C (http://www.amazon.com/Interior-Ballistics-Converts-Chemical-Projectile/dp/B000NHXH8C). This book primarily talks about black power and nitrocellulose but has the basic physics of the process, pressure curves, temperatures and such. As a first attempt at your stated problem I would use the density and volume of the cylinder to calculate the amount of octanitrocubane and use the relative effectiveness factor to obtain an equivalent amount of TNT then use this to estimate an energy released, I believe the conversion factor is (corrected this number, was 4.18) 4180 joules/gm TNT but you need to check this. From the energy and neglecting friction, etc. use 1/2mv^2 to estimate v=velocity. I just don't have time today to plug and chug the numbers. Again, this is a very crude analysis but may let you see if the velocity is in the ballpark of your design needs.
Apparently software is available to help with the design process, allows the use of different burn models and equations of state: "software package designed specifically for small arms interior ballistics research" http://www.dsbscience.com/software/chamber/index.php#burn_models (http://www.dsbscience.com/software/chamber/index.php#burn_models). Pretty sophisticated stuff by my standards!
After researching this subject a bit online, including various explosives, I am probably on the NSA watch list. For the record I'm a pacifist.
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"You may assume that all the materials will be able to withstand the pressures and temperatures generated. The thrust I calculated is 103.3210964 Kilo Newtons."
OK, the Calibre doesn't seem to be stated, but lets guess it's of the order of 1cm diameter.
So, to get that force onabout a ten thousandth of a square metre means that the pressure is about 10000 times 100 Kn/square metre
1000000000 N/m^2
That's pretty close to ten thousand atmospheres or a gigapascal.
Or, more pertinently, about half as much again as the yield strength of high tensile steel.
http://en.wikipedia.org/wiki/A514_steel
OK, stronger steels are available, but they tend to be brittle- which is suicidally stupid if you are making a gun barrel and also, that strength is measured at room temperature.
So re "You may assume that all the materials will be able to withstand the pressures and temperatures generated"
You may assume that, but you would be wrong.
"Dude, IT WILL NOT DISTORT"
Dude, it will.
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That is, after all , the intention purpose and function of high explosives. The ultimate HE would produce a tiny amount of energy in a very narrow shockwave, so it would split the barrel or dent the bullet without moving either. Propellants are quite different. At the other end of the scale, a pound of black powder could blow a safe across the room without denting it, whereas a gram of cutting charge would shatter the lock without disturbing the contents.
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Sorry I didn't mention the calibre, it is a .50 cal.
Is steel really the best you can come up with, steel is so last century.
Try a double-walled carbon nanotube, diamond hybrid shell chemically bonded to an osmium-iridium alloy core (and that's just the bullet)
1 gigapascal, is that it? Super-hard phase carbon nanotubes can withstand up to 546 GPa (diamonds can withstand 420 Gpa) http://en.wikipedia.org/wiki/Carbon_nanotube and their lighter diamond.
Do you really think I hadn't thought about this already.
Oh and I am making it a shaped charge. Just so you guys know
[/quote] alancalverd That is, after all , the intention purpose and function of high explosives. The ultimate HE would produce a tiny amount of energy in a very narrow shockwave, so it would split the barrel or dent the bullet without moving either. Propellants are quite different. At the other end of the scale, a pound of black powder could blow a safe across the room without denting it, whereas a gram of cutting charge would shatter the lock without disturbing the contents. [/quote]
That is an open air scenarion. In the barrel of MY gun, where do the gases go, they magically disappear, or perhaps 1 gigapascal (probably alot more) as mentioned by the bored chemist simply hangs there doing nothing.
Oh and guys please don't bring up economics, I am not interested in how uneconomical this weapon is.
And thanks a lot to distimpson for assisting me, I would like to get extremely detailed, which is why I am here, also does this book have any nitty gritty equations? but again thanks a lot.
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ChaosD.Ace wrote
I would like to get extremely detailed, which is why I am here, also does this book have any nitty gritty equations?
I don't have a copy of this book at the moment, just my old notes, my rendition of a pressure versus time diagram, I believe it was a shotgun chamber, figure attached below.
Also, in my previous note above, the correct value for energy is 4180 joules/gram TNT (I was only off by a factor of 1000). So, a back of the envelop calculation: volume=1.9x10-6m3 times the density 1980 kg/m3 gives 0.00376kg, using RE=2.38 this is equivalent to 0.00895kg of TNT or an energy of 37400 joules. If all energy was converted to m=0.045kg projectile velocity 0.5mv2= 37400 joules, velocity would equal a blistering 1290 m/s, in real life there are always losses.
For the p-sci class, we were discussing good old PV=nRT, and used the ideal gas equation to take adiabatic expansion of the process into account, so didn't really pursue this any further. This may help get started but the devil is in the details and an accurate simulation under such extreme conditions will take a lot of work, the ideal gas equation is just a teaching tool. Hope this helps. As always, check my numbers and calculations if they are important to you.
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Here are some more notes, an energy budget for the process. The maximum pressures were on the order of 10K-20K psi for nitrocellulose gun powers, velocities hundreds of m/s.
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1 gigapascal, is that it?
No, that's an average- the peak would obviously be rather higher.
Also the figure of 540 you quote is a bulk modulus, rather than a strrength.
Not to mention that the measurements are made cold and the gun barrel will be hot.
So, once again, and whether you like it or not, Dude, it will.
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Nitrocellulose as mentioned by distimpson, has a sensible flame rate which can be optimised with additives to suit the projectile and barrel dimensions
http://www.google.com/patents/EP0608488A1?cl=en
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Sorry for the massive post.
Hi Distimpson thanks a lot, looks awesome, I must ask though how accurate is that energy conversion factor for Octanitrocubane, The researcher who pioneered the molecule is a prof at the uni of chicago, I have his email and number, and will try to contact him to see If hopefully he'll give me an energy conversion factor.
I guess we kinda think alike, after a while of research I also remembered PV=nRT However there are problems (more on that a little further down), Since the detonation velocity is so fast and the propellant will be in such a small volume of space in crystal form, the full sublimation into gases should be almost instantaneous. So it detonates and releases all gas before the bullet begins to dislodge or at least displaces a small amount, I can allways account for that later. Now freeze hold this moment in your mind.
Distimpson just as you did, Using the internal volume for the casing (8.7614922 e-7 m3) and the density of the solid explosive (1980) I got the mass that would fit in that volume (1.7347755 e-3 kg). Now using mass over Mr (464.13) I can get the moles (3.7376931 e-6 moles of solid Octanitrocubane). Multiply by avogadro's constant and I have the number of molecules of the explosive in the chamber (2.2508916 e18 molecules of solid ONC).
I know that 1 molecule of the explosive becomes 8 molecules of CO2 and 4 of N2. So multiply the number of molecules of the explosive I calculated earlier by 8 and 4 respectivly. Now I know the number of molecules of CO2 (1.8007133 e19 molecules) and N2 (9.0035664 e18 molecules) that would be generated by the amount of the explosive that fitted in that space. Divide both by avogadro's constant and I have the number of moles of both. Add them up and now I know the number of moles of total gas in the chamber (4.4852317 e-5 moles). However if one were to use the Volume in Dm3 divided by 24 formula, the number of moles is much smaller value (3.6506217 e-5 moles). How do I figure out the compressibility factor in order to use PV=ZnRT.
Also since these are extreme temps and pressures I am going to use the Redlich–Kwong model for corrections. http://en.wikipedia.org/wiki/Real_gas.
Unfortunately using the reddlich-kwong corrections, I am having trouble finding the table of references that display the values for a and b for specific gases (in my case CO2 and N2). I'm not sure if any of you have had to work with this before but I thought to ask anyway.
So I am basically rearranging to calculate the pressure at the chamber.
So now I have a value for n (number of moles), V (internal volume of the casing), R is a constant. And like I said, I am trying to track down an energy conversion factor for Octanitrocubane so I can figure out the maximum temperature in that chamber (get a value for T). But for now I'll use yours.
My next goal is to calculate rate of change of pressure with change in volume (as the bullet travels down the barrel.) and rate of change of pressure with change in temperature (as heat is transferred to the surroundings and the gases cool down), and put those two together, but first I need that P.
Also, like you said, there will allways be losses. Unfortunately as much as I like to belive that it's possible to calculate a drag coefficient without conducting experiments, It seems as thought it is the case, so I will just have to find an exhisting coeffiecient for the bullet that is most closely shaped to mine.
Does surface tension of a material affect how much friction it will generate?
Bored Chemist, please define which strengths you are reffering to?
Thanks a lot for all the help, I will have a look at theat book.
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Estimate of the initial pressure from Phys. Fluids 2, 217 (1959); Blast Wave from a Spherical Charge, Harold L. Brode: E=PV/(gamma-1), gamma is the ratio of specific heats, using 1.4 for this exercise (assumed final pressure >> than initial). V is the volume of the casing, E is the energy released. Using the estimate for the energy as E=RE x 4.18 x 106 joules/kg TNT x density x V gives:
P=0.4 x 2.38 x 4.18 x 106joules/kg x 1980 kg/m3= 7.9 x 109 pascals or about 1 million psi.
This estimate assumes the propellant is packed to the density given and fills the entire initial volume, no free space, extra volume could give a bit of a cushion. A more accurate model would need to incorporate a burn rate for the propellant and motion of the projectile.
For the rest, this fellow has done all the work in a lot more detail, be a good place to state:http://e-ballistics.com/ebook/internal%20ballistics%20-%20lumped%20parameter%20model.htm (http://e-ballistics.com/ebook/internal%20ballistics%20-%20lumped%20parameter%20model.htm)
Apparently the human race has spent quite a bit of effort on weapons research.
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Apparently the human race has spent quite a bit of effort on weapons research.
....but never used a high explosive as a propellant. The rest of humanity are blind fools! Simply because there is no point or practical possibility of doing something, they have neglected to investigate it.
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Distimpson.
I LOVE YOU.
LOL.
What is your honest opinion, on having 7.9 gigapascals in a handgun. Recoil and Sound-wise.
Mean-time, I got a lot of reading to do.
Thanks again.
Peace (get it?)
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What is your honest opinion, on having 7.9 gigapascals in a handgun. Recoil and Sound-wise.
kaboom! ;D
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Apparently the human race has spent quite a bit of effort on weapons research.
....but never used a high explosive as a propellant. The rest of humanity are blind fools! Simply because there is no point or practical possibility of doing something, they have neglected to investigate it.
yep, I think this fellow sums it up pretty well, and it is a great web site for this type of interest http://www.frfrogspad.com/intballi.htm (http://www.frfrogspad.com/intballi.htm)
"Remember, Watcha-U-Back
(...or, once again, don't do anything stupid)
Unless you are well versed in internal ballistics leave the development of loading data with new powders to the experts with the proper tools, and don't play around with unknown powders. More than one firearm has been destroyed when the wrong powder was used without proper knowledge. As an example, in the 50s and 60s there were a lot of blow ups caused by people trying to use the powder from GI .30-06 blanks in pistols and rifles. What these poor souls didn't realize was that the "EC" powder in those blanks burned so fast that it was also used as a grenade filler. These folks turned their firearms into "grenades" with quite small charges of this powder. By the same token many powders of vastly different burning rates look similar. If it's not in the original can, don't use it!
Leave new load development to the pros and stick with the data in reliable loading manuals."
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Hi distimpson, I live in the UK by the way.
I've been reading through the e-ballistics website.
Since I am using a high-explosive does that mean burn rate is the detonation velocity. Also Octanitrocubane does not require external elements or compounds to sublime, so it doesn't com bust (I think) which means burn rate doesn't apply does it?
Thanks for Christmas present.
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Well, my use of the term burn rate isn't really appropriate, it's a detonation, some sort of kinetic description of the process would be useful/needed for a model.
Anyhoo, interesting subject, thank all for the discussion, I learned some things here and that's always a good deal.