calbiterol Posted March 11, 2007 Share Posted March 11, 2007 I need a way to relate pressure and volume without temperature. I've tried deriving an equation myself but I can't get one that (seems to) work. Long story short, I need to determine the initial volume of a gas (air) needed to bring a projectile to a specific velocity, without knowing or measuring temperature change. Diameter/radius and length of barrel are both constant in each case - but a general format is needed that keeps barrel length and diameters as variables. In other words, for the purposes of differentiation/integration, diameter and length are constants, but for final equation form, must be variables. I'm perfectly willing to derive the final equations myself, I just need a place to start. Thanks much! Link to comment Share on other sites More sharing options...

Bignose Posted March 12, 2007 Share Posted March 12, 2007 Here is the problem, you need to look at Gibbs' Phase Rule: F = C - P + 2 F= degrees of freedom in the system C = number of chemical species P = number of phases in the system Let's take a single component gas, like pure Argon. C=1, only Argon, P =1, it is assumed to be only a gas. F = 1-1+2 = 2 So, there are two degrees of freedom. That means, in order to describe the system, you have to know at least two variables, like temperature, pressure, or density (which is directly related to volume). So, if you only know pressure, for example, you cannot know exactly what the temperature or density is. You know how they are related, so the temperature and density can be graphed as a line, but you have no knowledge whatsoever of where on that line the state of the system is at. If you know exactly what the pressure and density is, there is only one temperature a pure gas system can be at. Link to comment Share on other sites More sharing options...

calbiterol Posted March 12, 2007 Author Share Posted March 12, 2007 But gas density is dependent on temperature and pressure. If I had an equation for what the density was at each point, I could figure it out. I suppose a better question would be, when volume varies directly with distance x, how do pressure and temperature vary with distance x? Link to comment Share on other sites More sharing options...

calbiterol Posted March 12, 2007 Author Share Posted March 12, 2007 Could one use energy transfer to relate the two? In other words, could one take the potential energy of the gas under pressure, [math]U_{compression} = PV [/math], and the kinetic energy transferred to the projectile, [math]E_k = \frac{mv^2}{2} [/math], and set the two equations equal? Only, from that point, I'm not sure how to find an equation for change in pressure with respect to x. Gah. Link to comment Share on other sites More sharing options...

Bignose Posted March 12, 2007 Share Posted March 12, 2007 If you are just looking for the relationship between temperture, pressure and density, then the Ideal Gas Law is the place to start. P=(n/V)RT P=pressure (n/V)=moles per volume (which is a density) R=gas law constant (very dependent upon which units you use for all the other variables) T=temperature While only a very few gases really follow this rule exactly, and only over a limited temperture/pressure range, it is also a pretty reasonable approximation for a large range of gases over a large range of T & P. To improve the accuracy, you can use different gas laws, like the van der Waals gas law, or Reidlich-Kwong, but the ideal gas law is a very good starting place. It is possible to derive that going all the way back through kinetic theory of gases, but it is unlikely to be a useful exercise. It does have several hundred years of observation behind it. I really doubt that you would get accurate estimates using what you described in post #4. The energy is probably going to be very far from equal, since there are so many energy loses in the system. Heat and noise and the fact that some gas will escape around the projectile are all significant loses. Link to comment Share on other sites More sharing options...

John Cuthber Posted March 12, 2007 Share Posted March 12, 2007 I think PV = 1/2 MV^2 is the best start but it will not really answer the question. You won't drive a projectile to faster than the speed of sound no matter what pressure and volume you use (unless you do some really complicated things). The speed of sound is roughly the same as the speed as the molecules; if the molecules of the gas are only doing 300 m/s they cannot push a projectile that is moving faster than that because they cannot catch up with it. Link to comment Share on other sites More sharing options...

calbiterol Posted March 12, 2007 Author Share Posted March 12, 2007 Is there any law for rates of gas expansion? I got to an equation that would be solved by a gas expansion law. It's come a LONG way from my last post. Link to comment Share on other sites More sharing options...

Klaynos Posted March 12, 2007 Share Posted March 12, 2007 Ah the joys of statistical mechanics: http://theory.phy.umist.ac.uk/~judith/stat_therm/node88.html Link to comment Share on other sites More sharing options...

calbiterol Posted March 13, 2007 Author Share Posted March 13, 2007 Edit: And If I don't know entropy? Link to comment Share on other sites More sharing options...

Klaynos Posted March 13, 2007 Share Posted March 13, 2007 Edit: And If I don't know entropy? The grand partition function; http://en.wikipedia.org/wiki/Partition_function_(statistical_mechanics)#Grand_canonical_partition_function But I assume you don't know the energy levels of the system either? So let me get this right, you have a single system that is open? You don't know pressure but you do know/want to know pressure and volume. I shall have a think as to whether I can come up with anything :s Link to comment Share on other sites More sharing options...

calbiterol Posted March 13, 2007 Author Share Posted March 13, 2007 I think I might have figured it out. When I know for sure, I'll throw up the end equation, the derivation, and then start to test it. Cheers. Edit: I suppose I might as well update you on my progress. I have an equation that defines acceleration in terms of x. Before stating that, though, I'll describe the system more fully. Re-Edit: Okay, here goes. First, for readability, I'll define the constant [math]B_k[/math] as [math]\frac{m_{air} C_p}{nR}+1[/math], or reconfigured in terms of density and simplified, [math]\frac{C_pT_i^2R_{air}}{P_i^2}+1[/math], where [math]R_{air}[/math] is the specific gas constant of air and approximately equal to 287.05. [math]C_p[/math] is specific heat, in terms of mass, and [math]T_i[/math] and [math]P_i[/math] are initial temperature and pressure, respectively. Now, derivation ignored because it's long and I want to wait until I finish the equation, I've gotten to this equation: [math]a = -\frac{B_k A x + V_i}{A (B_k + 1)}[/math], where a = acceleration, A = area of barrel and x = displacement of projectile (see above). This equation started as: [math]P_iV_i = m_{air}C_p(T-T_i) +\frac{m_{projectile}v^2}{2}+PV[/math]. (Not the best available symbols, but v is velocity and V is volume.) Link to comment Share on other sites More sharing options...

CPL.Luke Posted March 13, 2007 Share Posted March 13, 2007 pv= E is a good place to start. I assume your using a chemical reaction to place energy behind the projectile, so however much energy is in the gas immediatly after detonation will be equal to pv, from this we can see that EA/V=ma or E/x=F=ma thus E/x=md^2x/dt^2 I need to go take a test now, but I should be able to solve that differential equation later. Link to comment Share on other sites More sharing options...

CPL.Luke Posted March 13, 2007 Share Posted March 13, 2007 I just realized that you could integrate the expression for force and get 1/2mv^2=E ln(x) (from a to b) just use a as the distance from the back of your air cannon to the front, and keep in mind that this formula only works as is for a barrel that contains the pressurized air and the projectile, and has a constant barrel width, it would change slightly if there was a gas resovoir in the back. Link to comment Share on other sites More sharing options...

calbiterol Posted March 14, 2007 Author Share Posted March 14, 2007 CPL.Luke, that ignores energy lost or gained due to temperature change. There's something else I'm forgetting along those lines... Also, somewhere along the line of deriving the equation I have for acceleration in terms of x I had equations that were very friendly to adding in terms to accommodate friction and drag. So long story short, I want to stick with my equation, I just need to turn acceleration in terms of x to velocity in terms of x - which is more complicated than just integrating with respect to x, isn't it? Link to comment Share on other sites More sharing options...

calbiterol Posted March 14, 2007 Author Share Posted March 14, 2007 Scratch that, I took an integral instead of a derivative. I'll fix that later, as of right now, that equation is false. Link to comment Share on other sites More sharing options...

calbiterol Posted March 14, 2007 Author Share Posted March 14, 2007 Okay, fixed the problem, and I'm stuck here: [math]B\frac{dP}{dx}(Ax+V_i)+BAP=-m_{proj}a[/math]. I suppose I should integrate with respect to x, but I'm frankly too tired to figure out the integral of the first term. Link to comment Share on other sites More sharing options...

CPL.Luke Posted March 14, 2007 Share Posted March 14, 2007 asumming no heat is allowed into the system change in temperature is unimportant, it is merely a diferent measure of the energy contained in pv as can be seen in the equation pv=nrt thus the first term in you equation, mc delta t is only necessary if you are alowing heat to enter or leave your system, and considering the rate at which the projectile is going to fly down the barrel I don't think that much heat is going to be allowed out of the system, and even if it were, the quantity would be immeasurable. Link to comment Share on other sites More sharing options...

John Cuthber Posted March 14, 2007 Share Posted March 14, 2007 I don't think the temperature change will be big. While I initially thought that the stored energy was a good way to look at it (and there's a lot to be said for starting any problem with "can I apply the conservation of energy to this question") I'm no so sure any more. Practically speaking, PV is a constant The force on the projectile is its area A (the area of the barrel) times the pressure P. From that you can get an acceleration of the projectile. From that, in turn, you can get a distance traveled (x) in some small time interval The initial volume is V1. The volume after that short time interval is V2 =V1 + Ax With that new volume you can calculate the pressure P2 and start again OK, if you like calculus you can take the limit for a time interval tending to zero and get an exact formula, but I'd just do this by itteration. Link to comment Share on other sites More sharing options...

calbiterol Posted March 14, 2007 Author Share Posted March 14, 2007 I don't think the temperature change will be big. mc delta t is only necessary if you are alowing heat to enter or leave your system, and considering the rate at which the projectile is going to fly down the barrel I don't think that much heat is going to be allowed out of the system, and even if it were, the quantity would be immeasurable. No, heat transfer out of the system isn't an issue. I'm ignoring that. However, I do need to know the rate the projectile is flying down the barrel, for a couple reasons. First, I need to know what the velocity of the projectile is at the moment it leaves the barrel. Second, I need to know the rate at which the gas is expanding, because I am NOT making the assumption that all of the gas has been depressurized by the time the projectile leaves the barrel. Third, instantaneous velocity is needed to include the requisite drag and friction terms, which I will incorporate sometime soon. But here's the biggie, which is also why I quoted John. Temperature change is big. I'm talking pressures ranging from 300 to 800 psi, down a 6-inch to 3-foot barrel. I've done similar experiments without any maths, and dropping only 120 psi down a 2.5-foot barrel produces enough of a temperature drop to create ice fog at the end of the barrel. Oh, and for what it's worth, I'm perfectly comfortable with calculus, as long as I take time to remember it. I certainly appreciate the help and the input - and I definitely think there is a certain kind of beauty to simple solutions - but I also went about this problem the way I did for a reason. I should have clarified that earlier; sorry. Cheers, Calbit Link to comment Share on other sites More sharing options...

John Cuthber Posted March 15, 2007 Share Posted March 15, 2007 If the temperature change is big then you certainly can't use the ideal gas laws. You might find tables of this sort of data somewhere on the net, but I wouldn't really know what to search for. Link to comment Share on other sites More sharing options...

calbiterol Posted March 15, 2007 Author Share Posted March 15, 2007 Yes, you can you ideal gas laws, you just have to treat temperature as a variable. If I had to guess, I'd say the temperature drop is around 60 degrees F. Link to comment Share on other sites More sharing options...

calbiterol Posted March 20, 2007 Author Share Posted March 20, 2007 It seems I might have overanalyzed the problem. Here, how does this sound? [math]R_{air}[/math] is the specific gas constant of air and approximately equal to 287.05. [math]C_p[/math] is specific heat, in terms of mass, and [math]T_i[/math] and [math]P_i[/math] are initial temperature and pressure, respectively. V is volume, v is velocity. Adjustment of [math]B_k[/math]: Manipulation of Ideal Gas Law: [math]PV=nRT[/math] [math]\frac{PV}{T}=nR[/math] Density equations: [math]\rho_i=\frac{m}{V}[/math] [math]\rho_{i_{air}}=\frac{P}{R_{air}T}[/math] [math]B_k=\frac{m_{air} C_p}{nR}+1[/math] [math]B_k=\frac{m_{air} C_p T_i}{P_i V_i}+1[/math] Substitution [math]B_k=\frac{\rho C_p T}{P}+1[/math] Substitution [math]B_k=\frac{C_p}{R_{air}}+1[/math] Definition of density, simplification [math]B_k~\frac{1.0035}{287.05}+1[/math] Derivation: [math]P_i V_i = m_{air}C_p(T-T_i) +\frac{m_{projectile}v^2}{2}+PV[/math] [math]P_i V_i = m_{air}C_p(\frac{PV}{nR}-T_i) +\frac{m_{projectile}v^2}{2}+PV[/math] Substitution [math]P_i V_i + m_{air}C_p T_i = PV (\frac{m_{air} C_p}{nR}+1)+\frac{m_{projectile}v^2}{2}[/math] Distribute, rearrange [math]P_i V_i + m_{air}C_p T_i = P (A x) (B_k)+\frac{m_{projectile}v^2}{2}[/math] Substitution [math]P_i V_i + m_{air}C_p T_i = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Substitution (from above) [math]P_i V_i + \rho_i V_i C_p T_i = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Definition of density [math]P_i V_i + (\frac{P_i}{R_{air} T_i}) V_i C_p T_i = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Substitution [math]P_i V_i + \frac{P_i V_i C_p}{R_{air}} = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Substitution [math]P_i V_i (\frac{C_p}{R_{air}} + 1) = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Rearrange [math]P_i V_i = P (A x) + \frac{m_{projectile}v^2}{2(\frac{C_p}{R_{air}} + 1)}[/math] Division I suppose the question is, is that valid? Link to comment Share on other sites More sharing options...

calbiterol Posted March 27, 2007 Author Share Posted March 27, 2007 ... Is that valid? It seems like it might be a little... counterintuitive... if it is. If it's valid, though, I'll throw in a term for drag and a term for friction and call it a day. Link to comment Share on other sites More sharing options...

CPL.Luke Posted March 28, 2007 Share Posted March 28, 2007 calbiterol you are adding in an extra term, temperature does cahnge by the term cm(delta t) is unneccesary as this would be heat, which we are asuming is zero matheatially we use the first law of thermodynamics delta E=Q+W delta E=W= nr delta T=delta PV=w (on bullet) now to solve for W and thus delta E we have to find the force on the bullet E_0=PV E_0/x=F then we can integrate to get the work done between the starting point, and the end of the barrel (E_0ln(b)-E_0ln(a))=W since we can chose a system of units where a=e we can see that that is the proper expression under the first law of thermodynamics for the work done on the bullet Link to comment Share on other sites More sharing options...

CPL.Luke Posted March 28, 2007 Share Posted March 28, 2007 note that while T does change from a to b we don't need to consider it because there is no heat transfered out of the gas/bullet in practice there will be heat loss due to friction, and the barrel will heat up absorbing some of the energy. This would be accounted for by saying that Q does not equal 0 and takes on that cm (delta T) form that you were putting in before, however dthis expression would be in refference to the temperature change of the bullet/barrel, not the gas. Link to comment Share on other sites More sharing options...

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