Thread: How to calculate explosive pressure in confined spaces?

I am not sure if this question belongs more in the chemistry section but here goes.

Hi, I need to calculate the pressure of the gases generated when a mass of a specific explosive compound is detonated, 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;t to figure out.

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.

I know this may be a weird question but thanks for any help you guys can give me.

http://link.springer.com/article/10....0881122#page-1

You are using the gas to propel a "bullet"? You know the bullet's physical characteristics, i.e., it's diameter and mass? You are using the pressure produced to accelerate a given mass to a known velocity, 900 m/sec., over a known distance? (a "gun barrel").

If one assumes a couple of things, which ordinarily is necessary in such problems to avoid excessively "muddy" waters, i.e., the "bullet" moves through it's acceleration period with no loss of gas, it seems the average pressure exerted by the gas may be calculated while dispensing with all the other gobbledy-gook, no?

Force = Mass X Acceleration, the mass is KNOWN, the acceleration may be calulated using Uniformly Accelerated Motion formula (you find it!), the PRESSURE is easily found by

using the newly-acquired value for the FORCE in: PRESSURE = FORCE per UNIT AREA (>> the bullet's cross-sectional area).

What do you think of that? jocular

Er, that will give the pressure required, not the pressure actually developed, neh?
I.e. if you're starting from scratch - designing a cartridge - then you don't know the pressure yet.

however---- using straight high explosive is not effective since the decomposition is almost instantaneous- you want it to be a propellant ie generating the gas as the bullet accelerates

That's a good point.
You really don't want to be using high explosives as a bullet propellant.

Thanks for the quick reply guys, I am still in sixth form so forgive me if I miss some details.

This is a semi-auto pistol around the same size but slightly longer than a 6 inch desert eagle.

Jocular is kind of right in the sense that i am working it backwards, in other words I know the velocity I want to achieve which equates to the force of thrust on the bullet 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 energy, detonation velocity etc.

However I also understand that acceleration will not be constant due to the increasing volume between the back of the bullet and the back of the propellant case as the bullet travels down the barrel. Perhaps you guys could clarify how to approach calculating the summation of the effect of a these constantly decreasing thrust forces on the bullet (but for that I assume we must first know the maximum generated chamber pressure generated by a particular mass of this explosive.

Just so you guys know...

The bullet weights 45.359237 grams, I have already done the volumetric and mass calculations, I am not quite ready to go into aerodynamics just yet, I have got a general aerodynamic shape, right now I am just kind of rough drafting things.
.50 calibre = 0.5 inches at the bullets maximum diameter which = 12.7 millimeters thus the radius = 6.35 millimeters.

[Tried to upload a picture of what I had done so far but it wouldn't let me, I'll work on that]

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

Thanks for responding so quickly guys.

Originally Posted by ChaosD.Ace

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

Oh, it almost certainly can.
BUT (I luuurve that word) use of high explosive means that (as you noted) you get a ridiculous maximum pressure, which in turn means a ridiculous chamber strength is required, which bumps up the weapon's mass.

I'll see if I have any "Dummy's Guides"-type equations left hanging around on my hard drive, but don't hold your breath, it's a few years since I did this sort of stuff for fun, so even if I have any, finding them (X no. of computers and Y no. of hard drives later) is going to be no picnic.

Not necessarily, what is stronger and lighter than diamond and more flexible than cotton fibres?

Also since I am accelerating this bullet to such a high velocity in a relatively short barrel, the bullet will be in there for less than half the time a normal bullet remains in a normal gun, doesn't that mean that whether the propellant detonates or deflagrates doesn't have too much of an impact on the maximum acceleration.

Because I am sure as heck not going to get a 45 gram bullet to 900 m/s in a desert eagle-likr gun from gunpowder which is why I though to resort to very high explosive instead.

Originally Posted by ChaosD.Ace

Not necessarily, what is stronger and lighter than diamond and more flexible than cotton fibres?

I'm not sure flexible chamber is good idea.

Originally Posted by ChaosD.Ace

Not necessarily, what is stronger and lighter than diamond and more flexible than cotton fibres?

For the first part... carbon nano tubes. However, just because something is better than something else doesn't mean it's economically practical. I.E, you don't see any 40 ton titanium tanks.

Sorry I should have mentioned it before, shlunka you are correct, this is by no stretch of the imagination an economical weapon, I did not mean it to be an economical weapon when I started this project, just a superior weapon (lighter, stronger, more reliable, greatest pistol muzzle velocity and energy etc.) Also silicon chips weren't cheap or economical when the technology was first developed, nowadays you can get them very easily, nanotech is still mainly in its early research steps and also you gotta realise, it's carbon were talking about not rare metals, also carbon nanotubes aren't difficult to make it's just that it takes a while to get a decent amount of them, and like I said we are not in the mass production stage. Like silicon technology, refinements will be made to their production processes and they will become a lot more economical, however for now my weapon is definitely not economical, but I am not interested in economics, why did we put a man on the moon, was it economical, no, we did it because we could, my weapon is kind of the same thing.

Dywyddyr, chamber is definitely not going to be flexible, carbon nanotubes alone are not flexible it's their fibers that are flexible and I am definitely not using their fibers, the chamber wouldn't blow up per say, because the material could handle it, however you would end up with the most warped weapon ever. It would be cool to see in slow mo though.

By the way where did your username come from Dyeyddyr

Originally Posted by Dywyddyr

Originally Posted by ChaosD.Ace

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

Oh, it almost certainly can.
BUT (I luuurve that word) use of high explosive means that (as you noted) you get a ridiculous maximum pressure, which in turn means a ridiculous chamber strength is required, which bumps up the weapon's mass.

I'll see if I have any "Dummy's Guides"-type equations left hanging around on my hard drive, but don't hold your breath, it's a few years since I did this sort of stuff for fun, so even if I have any, finding them (X no. of computers and Y no. of hard drives later) is going to be no picnic.

Actually, most knowledgeable ammunition reloaders know that commercially-available smokeless powders are manufactured of high explosive propellants. Avoiding use of the term "gunpowder", which has more or less come to be applied only to black powder, smokeless powders for non-large caliber (over small-arms size) ammunition is invariably composed of nitrocellulose and in the case of many pistol powders, which require smaller quantities of faster-burning constituents, nitroglycerine.

The old familiar "Hercules Bullseye" pistol powder consists of a combination of both of the above materials. Rifle powders, on the other hand, usually contain no nitroglycerine. In any event, BOTH are certainly detonable materials; in fact, it has been stated by many that Bullseye may be detonated (due to it's nitro content) by a #4 blasting cap. They are incorrect, given no confinement to speak of (cardboard container), I could not get Bullseye to detonate thusly.

In any event, a curve depicting burn-rate vs. pressure for such materials always displays a reasonably low combustion rate which is fairly linear, occurring before detonation takes place, at which point pressure increases extremely quickly, allowing no room to "get out of the way"! Thus, small-arms ammunition IS most certainly loaded with detonable materials, which therefore requires that caution be taken in reloading, to ensure operation within the acceptable burn-rate portion of the curve for the powder being used.

Wanna know what little more I have to contribute, as far as burn-rates are concerned? jocular

Jocular, Absolutely, I want to know as much as possible, Also is the reason why it doesn't detonate in confined spaces, contact with air, because octanitrocubane does not require external elements or compounds to decompose, this stuff detonates in the vacuum of space, how would that affect my end goal?

Originally Posted by jocular

Actually, most knowledgeable ammunition reloaders know that commercially-available smokeless powders are manufactured of high explosive propellants.

Then, unfortunately, most "knowledgeable" ammunition loaders turn out be not that knowledgeable.

Nitrocellulose, and other propellant powders, are classed as low explosives.
But nice try anyway.

Originally Posted by Dywyddyr

Originally Posted by jocular

Actually, most knowledgeable ammunition reloaders know that commercially-available smokeless powders are manufactured of high explosive propellants.

Then, unfortunately, most "knowledgeable" ammunition loaders turn out be not that knowledgeable.

Nitrocellulose, and other propellant powders, are classed as low explosives.
But nice try anyway.

Better than a nice try, it's pretty factual, but I don't care to take part in a pissing match based on semantics. The classification system for explosives generally uses the terms "Primary" and "Secondary" for High Explosives; however, the exact nomenclature depends on the "defining authority"

At any rate, how can you possibly state that "other propellant powders".......are classed as "low explosives" when that "other" is nitroglycerine?

Do you have any hands-on personal experience with nitrated esters, or are your well-intentioned assertions based solely on the blathering of Internet Experts? You might want to bone up on Tenney L. Davis's "The Chemistry of Powder and Explosives", printed during WW-II. jocular

Edit: Re: Velocity of Detonation, as outlined in your kindly provided Wiki: Low Explosives : Detonation velocity LESS than the speed of sound, ~ 1100 feet per second? Nitroglycerine, 21,000 feet per second. That's what, ~ 20 times faster?

Originally Posted by jocular

Better than a nice try, it's pretty factual, but I don't care to take part in a pissing match based on semantics. The classification system for explosives generally uses the terms "Primary" and "Secondary" for High Explosives; however, the exact nomenclature depends on the "defining authority"

Yet there IS, as you note later, a difference.

At any rate, how can you possibly state that "other propellant powders".......are classed as "low explosives" when that "other" is nitroglycerine?

Huh?
Where's the nitroglycerine? Where is it mentioned in "others" or as a "low explosive".

Do you have any hands-on personal experience with nitrated esters, or are your well-intentioned assertions based solely on the blathering of Internet Experts? You might want to bone up on Tenney L. Davis's "The Chemistry of Powder and Explosives", printed during WW-II. jocular

Ah right. "I've touched the stuff so I'm an expert". As opposed to my "I've spent years doing the basic theory and calculations so I'm an expert".
(I.e. don't assume I get my information from "blathering internet experts" - something you should have noticed if you'd bothered reading, for example, post #8 - my resume includes a year or so working for a process engineering firm that ran a large explosives plant in the UK).

Edit: Re: Velocity of Detonation, as outlined in your kindly provided Wiki: Low Explosives : Detonation velocity LESS than the speed of sound, ~ 1100 feet per second? Nitroglycerine, 21,000 feet per second. That's what, ~ 20 times faster?

Your point?

Originally Posted by ChaosD.Ace

Dywyddyr, chamber is definitely not going to be flexible, carbon nanotubes alone are not flexible it's their fibers that are flexible and I am definitely not using their fibers

You're going to have to explain the difference between the nanotubes and their fibres for me.

The strength and flexibility of carbon nanotubes makes them of potential use in ... everyday items like clothes and sports gear to combat jackets

(Wiki)
Carbon Nanotube Muscles Strong as Diamond, Flexible as Rubber

however you would end up with the most warped weapon ever. It would be cool to see in slow mo though.

Huh?
Splitting cartridge case?

By the way where did your username come from Dyeyddyr

I made it up.

Here is a really good video that shows you single nanotubes and how they are formed into fibres

Smart Material: Carbon Nanotubes - YouTube

Here is an article on the hardness toughness, youngs modulus, etc. of single carbon nanotubes (just one)

Carbon nanotubes and the pursuit of the ultimate body armor

Also they conduct electricity due to each carbon atom being bonded to three others thus leaving each carbon atom with one delocalised electron, remember this video demonstrates carbon nanotube FIBRES, single carbon nanotubes have the same exact conductive properties as demonstrated in this video.

Spinning nanotube fibers at Rice University - YouTube

Now imagine taking each of these delocalised electrons and bonding them to carbon atoms of a slightly wider carbon nanotube, in essence having on carbon nanotube inside another one, both bonded together.

http://nanotechweb.org/cws/article/tech/21379

Not quite as uneconomical as shlunka originally thought. Nanotech is definitely making fast progress.

Hope these four posts help.

It's late now so I'll see you guys tomorrow.

Thanks again for the great replies.

So, basically, nothing that says CNT are rigid?

Also here is that work I did I was talking about earlier so you have a better idea.

http://s1344.photobucket.com/user/ChaosDAce/media/scan0003_zps2e6fa6b4.jpg.html?filters[user]=137245797&filters[recent]=1&sort=1&o=0

Remarkably short bullet.
A typical 50 cal projectile is 2 - 3 inches long.
You're going to have stabilisation problems.
Oh, and IF, as you claim, CNT doesn't flex you're also going to have problems getting it to spin - how will engage the rifling?

First link doesn't work, is this what you meant?

By the way, you've got approximately $3,200 worth of osmium in that one bullet. Just sayin'.
Who do you expect to buy it, and for what role?

Originally Posted by Dywyddyr

Your point?

You're deliberately trying to weasel out of saying, in other misleading wording, that nitroglycerine as used in small arms propellants, is NOT a high explosive. You and I both know, as well as does the listening walls, that nitro is not a low explosive, wherever the hell one finds or uses it. jocular

Originally Posted by jocular

Originally Posted by Dywyddyr

Your point?

You're deliberately trying to weasel out of saying, in other misleading wording, that nitroglycerine as used in small arms propellants, is NOT a high explosive. You and I both know, as well as does the listening walls, that nitro is not a low explosive, wherever the hell one finds or uses it. jocular

Not at all.
Because I'm fully aware that
A) the nitroglycerine is used in small quantities as supplement, rather than a propellant per se, and
B) it's actually used a solvent i.e. when it IS used in a double base propellant the fact that it forms a solution denatures it to some extent. (You know, sort of the same way that the nitro in nitrocellulose isn't a high explosive any more).

I am definately not an expert but the reason you don't use the highest possible explosive to propel a bullet is the reason you don't push a beer down the bar by hitting it with a hammer. You want the projectile to arrive intact not as a cloud of fragments. Now if the cloud of fragments is an acceptable out come and you don't mind having to carry your "gun" in a motor vehicle, then you can use the fastest explosive you can get.

Dude this is a desert eagle-like pistol, I am not firing sniper rounds out of a hand gun, this is in the realm of .50 AE not .50 BMG, In fact the name of mine is .50 NS (.50 NanoShell (get it))

.50 AE cartridge, The whole cartridge itself isn't even 2 inches long, if anything my bullet alone is way too long, I'll have to shorten it.

http://en.wikipedia.org/wiki/.50_Action_Express

The rifling thing I have already thought about. I am sectioning the bullet. Lead leaves to much residue, which is why they jacketed the bullets in copper, soft enough to conform to the barrel and form a proper seal but with a melting point high enough to not leave as much residue. My point is, only the part of the bullet that needs to engage the rifling needs to be soft. So the cylinder would be made out of a dense weave of double walled carbon nanotube fibers so they would conform against the polygonal rifling. the tip of the bullet on the other hand would be very rigid and not structurally deform.

This brings up another thing, polygonal rifling vs traditional rifling, for some reason I think polygonal rifling may be more suited for my nanotube weave, and it is claimed that it is superior, however it's never been verified. Thing is Glock, Desert Eagles, H&K and other big names all use it. But this is probably more for a gun forum.

Hey at the end of the day we could always use diamond for the tip.

Yep that is the right link.

Actually I'm thinking more of an iridium osmium alloy, I'm not sure if that is possible, or how soft I need the core to be I just needed to add extra weight.

Sealeaf you are correct, we are definitely not using traditional bullets here, not unless you want a molten metal shard gun.

$3,200, is that it? I expected a lot more. what about iridium?

Something to ponder-Synthesis[edit source | editbeta]

Although octanitrocubane is predicted to be one of the most effective explosives, the difficulty of its synthesis inhibits practical use. Philip Eaton’s synthesis was difficult and lengthy, and required cubane (rare enough to begin with) as a starting point. As a result, octanitrocubane is more valuable, gram for gram, than gold.^[5] A proposed path to synthesis is the cyclotetramerization of the as yet undiscovered and presumably unstable dinitroacetylene

Yeah I know, kind of a bummer that they can't get the predicted density (yet, hopefully), If I remember correctly, something like 40 steps are required right? and each one is quite lengthy by themselves, but you must admit it is an awesome molecule and the gas product ratio alone makes it worth it. Still the synthesis difficulty is quite annoying, Eaton is currently working on a solution for this, and you know the american military (DARPA in specific) their like bloodhounds for technology of this type, they won't let go of something like this so it will most likely be worked out, if not I've already got some substitute candidates.

Dywyddyr, I am also using a match barrel, which will reduce rifling issues.

Dywyddyr, been doing some research, turns out radial (shear modulus) they are very soft and cut easily, Longitudinally (youngs modulud, bulk modulus etc.) they are harder than diamonds, so if I wanted to use them for a hardened tip I would need to allign them so they are protruding outwards towards the direction of motion of the bullet, but then again I supose I can also look for a substitute substance.

Hardness
Standard single-walled carbon nanotubes can withstand a pressure up to 25 GPa without deformation. They then undergo a transformation to superhard phase nanotubes. Maximum pressures measured using current experimental techniques are around 55 GPa. However, these new superhard phase nanotubes collapse at an even higher, albeit unknown, pressure. The bulk modulus of superhard phase nanotubes is 462 to 546 GPa, even higher than that of diamond (420 GPa for single diamond crystal)

So no-one here can help me?

Originally Posted by ChaosD.Ace

So no-one here can help me?

Well, this is a science forum. Guns are not that much of a priority - at any rate for those of us who don't live in the rural USA. I understand your propellant is more costly, on a weight basis, than gold. Is this practical, or just a theoretical exercise?

Originally Posted by ChaosD.Ace

So no-one here can help me?

So post #2 was no good?

Have fun Don Quxiote

using the newly-acquired value for the FORCE in: PRESSURE = FORCE per UNIT AREA (>> the bullet's cross-sectional area).

Work Done is force times distance, So 103321.0964 Newtons times 0.1778 metres (length of barrel) = 18370.49094 joules.

So if F=ma then F = kilogram metres per second squared. And if Pressure = F divided by Area, the Pressure = kilogram per metre, seconds squared. Is this correct?

Thrust force to achieve 900 m/s for this 0.045359237 kg bullet in a 7 inch (0.1778 metres) barrel = 103321.0964 Newtons (kgm/s).

Radius = 0.00635 m, so pressure on the back of the bullet at moment after detonation must be 815627166.7 kg/ms^2

However this is assuming acceleration is constant, which it isn't 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 volume (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 while taking into account these constant decreases in pressure, thrust etc. Is integration required here? if so how do I begin, and guys, please don't tell me this is an empirical process, there must be something I can do.

Also I don't live in the USA. Britain is my country, and you really think someone at a gun forum will be able to answer my question.

Originally Posted by ChaosD.Ace

using the newly-acquired value for the FORCE in: PRESSURE = FORCE per UNIT AREA (>> the bullet's cross-sectional area).

Work Done is force times distance, So 103321.0964 Newtons times 0.1778 metres (length of barrel) = 18370.49094 joules.

So if F=ma then F = kilogram metres per second squared. And if Pressure = F divided by Area, the Pressure = kilogram per metre, seconds squared. Is this correct?

Thrust force to achieve 900 m/s for this 0.045359237 kg bullet in a 7 inch (0.1778 metres) barrel = 103321.0964 Newtons (kgm/s).

Radius = 0.00635 m, so pressure on the back of the bullet at moment after detonation must be 815627166.7 kg/ms^2

However this is assuming acceleration is constant, which it isn't 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 volume (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 while taking into account these constant decreases in pressure, thrust etc. Is integration required here? if so how do I begin, and guys, please don't tell me this is an empirical process, there must be something I can do.


Also I don't live in the USA. Britain is my country, and you really think someone at a gun forum will be able to answer my question.

Not necessarily so. In a typical non-theoretical application (such as yours), the burning rate of the propellant is factored in such a way that pressure is still increasing at the instant the projectile leaves the confinement of the barrel, despite presence of the other variables you mention (increasing volume, etc.). I think. jocular

But it still doesn't output a greater overall pressure than my high explosive does it. Either way way acceleration isn't constant because the pressures and forces also aren't constant. Do you know how I can account for this.

the only way you are going to get 900m/sec out of a pistol barrel is perhaps the velocity of the fragments when I blows up from your ridicilious explosive propellant.

Dude haven't you been listening the bullet, barrel and chamber will all easily withstand the pressures and temperatures generated.

Ok guys, so I think I may have something. PV=nRT. 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. Using the volume for the casing and the density of the solid explosive I got the mass that would fit in that volume. Now using mass over Mr I can get the moles. Multiply by avogadro's constant and I have the number of molecules of the explosive in the chamber.

Now I know that 1 molecule of the explosive becomes 8 molecules of CO2 and 4 of N2. So multyply the number of molecules of the explosive I calculated earlier by 8 and 4 respectivly. Now I know the number of molecules of CO2 and N2 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. However if one were to use the Volume in Dm3 divided by 24 formula, the number of moles is much smaller. How do I figure out the compressibility factor.

Also since these are extreme temps and pressures I am going to use the Redlich–Kwong model for corrections. Real gas - Wikipedia, the free encyclopedia

This whole molar volume thing is a bit confusing to me.

So iam 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 i am currently trying to get in contact with the researcher who pioneered the explosive molecule, he is a professor at the uni of chicago, to get an energy conversion factor for octanitrocubane (how many joules are released per kilogram of the substance detonated) so I can figure out the maximum temperature in that chamber (get a value for T). I have tried searching for it for a while but haven't found anything (only sublimation temperatures, heat of formation etc.)

Also 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. I'm not sure if any of you have had to work with this before so I thought you guys might know where I can find a table.

Also if you guys would like all the values that i got (volume, no. of molecules, no. of moles etc.) I can put them up in a seperate post so not to make this one too massive.

Thanks for all the replies guys.