Registered users: Google [Bot], Yahoo [Bot]
 
User Information


Site Menu


Sponsored


Who is online
In total there are 42 users online :: 2 registered, 0 hidden and 40 guests Most users ever online was 155 on Mon Aug 15, 2016 1:40 am Registered users: Google [Bot], Yahoo [Bot] based on users active over the past 5 minutes 

The Team
Administrators
Global Moderators


Sponsored


Calculating combustion pressureI would like to calculate the pressure created by the combustion inside a combustion cannon, therefore the pressure that starts pushing the projectile through the barrel. I thought that:
W = p * dv (W for work, p for pressure and v for volume) And if we consider that: W = Ec (kinetic energy of the projectile) then: ((p0 * v0) / vf) * (vf  v0) = 0,5 * m * vp^(2) (p0 = initial pressure, v0 = chamber volume, vf = chamber + barrel volume, m = mass of the projectile, vp = muzzle velocity of the projectile) If we take the point where the projectile is at the end of the barrel as our point of study, I get an initial pressure of around 40000 Pa, which is obviously under atmospheric pressure. What am I doing wrong? Or is there another way to calculate initial pressure?
i know it might not be possible for you (considering you may have to take apart your launcher) but in the past, i've seen people replace their barrel for a screw on endcap with a pressure gauge and then "fire" the gun. its a little risky because the chamber could just explode if the psi goes over the chamber's limit.
sorry if this doesn't help you, just trying to give you some ideas
http://www.thehallsinbfe.com/HGDT/index.html
http://www.inpharmix.com/jps/Closed_Cha ... udies.html
You appear to be assuming the process relation p=constant.
The work done on the projectile by the propellant gas is is the integral of the force on the projectile's base over the distance it travels down the barrel's long axis. If you ignore the mass of the gas and the fact that the combustion is a relatively slow process, and assume an adiabatic process relation (PV<sup>ɣ</sup>=P<sub>0</sub>V<sub>0</sub><sup>ɣ</sup>) in the propellant gas, you can easily obtain a second order ODE describing the projectile's motion inside the gun. This ODE is not, so far as I know, analytically solvable, but is easy to solve by a numerical method. This solution is obtained very quickly to any practical level of accuracy, to the point that it is quite easy to use a naive method like "guess and check" to find the initial chamber pressure if a value of the adiabatic index is assumed (this is taken as constant in the simplest version of the calculator, but it's easy enough to add a correction for real data, if you have the numbers handy). This method will estimate pressure lower than actual, due to the fact that the gas has mass, and a particle speed which is on the same order of magnitude as the projectile speed (although this is alleviated somewhat by the aforementioned noninstantaneous combustion process). Also, for the record, asking questions requiring knowledge beyond an eigth grade level on here will get you lots of nonsequitur responses like the two above mine. I'm not entirely sure why people bother adding useless sludge to topics they haven't the faintest knowledge of, but there you have it...
Last edited by DYI on Mon Jul 16, 2012 7:44 am, edited 1 time in total.
Spudfiles' resident expert on all things that sail through the air at improbable speeds, trailing an incandescent wake of ionized air, dissociated polymers and metal oxides.
FWIW: The combustion pressure model in HGDT is *very* simple. HGDT neglects all chemistry changes and does nothing more than ask this question: "How hot is a stoichiometric burn for this fuel/air?" After that, it's nothing more than pv = RT. *BASIC* stuff. But it works surprisingly well.
Not sure DYI's "nonsequitur" comment is valid. Certainly jack's response is responsive to the vague question.
To answer the question it first has to be posted as a valid question. What exactly are you trying to calculate and what exactly is the available data? Brain dead simple answer is that the final pressure (ammo just about to leave the muzzle) is roughly v0/vf*pMax. For propane in air pMax is about 150 PSIG or so. This ignores heat loss and various other things but it is close enough for many uses. You can estimate the effective pressure based on the muzzle velocity and the mass of the ammo. F=m*a=pressure*areafriction. If you use the average pressure, pAvg ((p0pf)/2) and assume that the average acceleration and average velocity is adequate, and you have an estimate of the dynamic friction, then you can estimate the pressure that gives rise to a particular muzzle velocity (calculated via the acceleration). The average velocity is just the muzzle velocity over 2 since v0 is zero. The average acceleration is the muzzle velocity divided by the time it takes for the ammo to move the length of the barrel at the average velocity. IIRC, for most combustion guns at 1X the max pressure is 60 or 70 PSIG, which is a bit less than half the theoretical peak pressure of propane in air in a closed chamber. There is an old post here (or perhaps on the old SpudTech site) that used this approach to estimate the working pressure in a BBMG. IIRC, BBMGs are typically operating at an effective (chamber) pressure of only 20 or 30 PSIG. (EDIT: when operated with a 120 PSIG pressure source)
I just found this: http://s87572736.onlinehome.us/Writeup.html and thought that it might be interesting to consider the force of atmospheric pressure so that:
Wc  Wa = Ec (Wc = work done by combustion force, Wa = work done by the force of atmospheric pressure) Since W = F*x and F = p*S (p0*v0)/(v0+S*x) * S * x  pa * S * x = 0.5 * m * v^2 (p0 = initial pressure, v0 = chamber volume, S = crosssectional area of the barrel, x = distance the projectile has travelled inside the barrel, pa = atmospheric pressure, m = mass of the projectile, v = velocity of the projectile) If I consider x as 0.75m (the length of the barrel, therefore when the projectile is at the end of it) and since I know all other values except p0, I can calculate the initial pressure, and I get around 200000 Pa. Does that sound reasonable for a weak cannon (muzzle velocity of around 35 m/s)?
The link you posted assumes an isothermal process, which is incorrect  the process is too fast for that.
Spudfiles' resident expert on all things that sail through the air at improbable speeds, trailing an incandescent wake of ionized air, dissociated polymers and metal oxides.
Yeah, I hadn't realised that and it's obvious there's a temperature change during combustion. I was reading your other reply and I wonder if you could use an example, since I'm not sure if I understand what you mean. Thanks.
I calculated pressures from muzzle velocity data way back...
Scroll down to the bottom of the page (search for "Draculon") and perhaps there is something in that last section that'll help calculate whatever it is you need. http://www.inpharmix.com/jps/diaphram_valve_BBMG.html
I think I have found a way to find the pressure after all. I found it before I read your post (but I read it anyway), thanks anyway for the link.
 
Who is onlineRegistered users: Google [Bot], Yahoo [Bot] 
