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Airgun shoots ball: Pressure = ?

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data

    An airgun shoots a ball made from lead, through air of high pressure that pushes the ballforward like a piston. Because this happens very quickly, Q = 0 and the process is adiabatic. The initial volume of the air is Vi = 12.0 cm3 = 12.0 * 10-6 m3, which behaves as an ideal gas with γ = 1.40. The air decompresses and pushes the ball as a piston through the gun's barrel, which has a length L = 50.0 cm = 0.500 m. The ball has a cross section area A = 0.0300 cm2 = 0.0300 * 10-6 m2and has a mass m = 1.10 g = 1.10 * 10-3 kg. Which is the initial pressure Pi so that the ball can exit the barrel with v = 120 m/s ? Ignore any friction from the barrel or the win'ds resistance.

    2. Relevant equations

    PiViγ = PfVfγ
    ∑F = ma
    P = F/A
    p =m/V
    V=π‧r2‧h (cylinder)
    V=4/3 * π‧r3 (ball)

    Lead's Density: p = 11.34 g/cm3 = 11.34 * 103 kg/m3

    3. The attempt at a solution

    Alright, my first instict was that I'd use this formula: PiViγ = PfVfγ

    Now, to use this, I need to find Vf & Pf, so that coupled with Vi & γ (which I already know), I can find Pi.

    So, I can use L & A to find Vf, like this:

    >p = m/V <=> 11.34 * 103 kg/m3 = 1.10 * 10-3 kg / V <=> V = 9.7 * 10-8 m3 (the ball's volume)

    >Ball: V=4/3 * π‧r3 <=> ... <=> r = 2.85 * 10-3 m

    >Cylinder: V=π‧r2‧L = 1.28 * 10-5 m3

    So Vf = 1.28 * 10-5 m3

    The problem I'm facing is that I can't figure out how to connect Pf with the v (velocity).

    Any help is appreciated!

    PS: Could somebody check the calculations a bit? If I take the V = A*h = A* L formula, I get V = 1.5 * 10-6 m3, which is the volume of the whole barrel, minus the ball. Now, I get Vf = 1.28 * 10-5 m3 which is the volume of barrel with the ball blocking a part of it. So, logically, the volume of the ball should be Vb = 22.0 * 10-6 m3, but as we can see above, using the density and mass I get Vb = V = 9.7 * 10-8 m3.

    Can somebody spot the mistake before I move on and do a mess at the folowing calculations?
  2. jcsd
  3. Aug 6, 2017 #2
    The ball is going to be accelerating, so you are going to need a force balance on the ball. Please write down your force balance on the ball.
  4. Aug 6, 2017 #3
    Well, the general formula is: ΣF = m*a

    The mass is 1.10 * 10-3 kg.

    The force, F, can be modeled as P*A, where we know A and P is the pressure.
  5. Aug 6, 2017 #4
    You are going to have to solve this equation of motion for the ball location as a function of time, given that the pressure on the ball is varying with time. So you are going to have to flesh out this equation some more.
  6. Aug 6, 2017 #5
    Actually, once you have the equation fleshed out, you can solve for the velocity as a function of position along the barrel. Let x be the position and Po be the initial pressure. In terms of x, what are the gas volume and pressure?
  7. Aug 6, 2017 #6
    I searched around a bit, and came across this formula: vf2 = vi2 + 2a(xf - xi). It connects location, velocity and acceleration. It's a mix of the main ones (one for velocity and one for location, as functions of time and acceleration), but I don't think i should prove it to use it, right? So, we have:

    a = (vf2 -vi2)/(2(xf - xi))

    Plus, we also have:

    ΣF = m*a

    P*A = m*[(vf2 -vi2)/(2(xf - xi))]

    Now, we have:

    xf = L
    xi = 0
    vi = 0
    vf = 120 m/s
    A = 0.0300 * 10-6 m2
    m = 1.10 * 10-3 kg

    So, using those, we can find the pressure Pf, right when the ball exits the barrel of the airgun. Pf = 528 * 106 Pa

    PiViγ = PfVfγ <=> ... <=> Pi = 578 * 106 Pa. The book's answer is 5.74 * 106 Pa, so obviously I did wrong something in the computations. Problem is, I can't spot it.
  8. Aug 6, 2017 #7
    The equation you used for the acceleration assumes constant acceleration, and, in this problem, the acceleration is decreasing as the ball moves down the barrel.
  9. Aug 6, 2017 #8
    Oh, yeah, I missed that. So, do I have to "fix" it so that I can integrate the two parts of the equation? Use the same formula from my previous post, but integrate for x?
  10. Aug 6, 2017 #9
    Yes. I'd like to help further, but I'm away from my computer and am trying to do this on my iPhone. I'll be back on my computer later. See you then.

  11. Aug 6, 2017 #10
    No rush! It's nearing midnight here, so I was going to check in tomorrow anyway (try my hand at it again, with a fresh mind).

    Until tomorrow, thanks a ton for the help thus far!
  12. Aug 6, 2017 #11
    The easiest way to do this problem is to recognize, via the work-energy theorem, that the exit kinetic energy of the ball is equal to the work done by the gas in expanding from its initial volume to its final volume minus the work to push back the atmosphere ahead of the ball. In terms of the initial pressure Po, what is the the work done by the gas in expanding adiabaticly and reversibly to its final volume Vo+AL?
  13. Aug 7, 2017 #12
    Okay, so we have: Wext = ΔK = Kf - Ki = Kf

    Now, we have for the Work: W = -P∫VfVidv, one for the air in front of the ball, and one for the air behind the ball.

    Behind the ball: Vi = 12.0 * 10-6 m3 & Vf = 1.28 * 10-5 m3 (see OP)

    In front of the ball: Vi = A*L - 12.0 * 10-6 m3 & Vf = A*L - 1.28 * 10-5 m3 (see OP)

    Now, I'll take the P from here: P*A = m*[(vf2 - vi2)/(2(xf - xi))] (see post #6)

    I'm a tad stuck here though. Yesterday you said I should take the P*A = m*a equation and treat a as not a constant, since the force that is applied to the ball changes, and thus the acceleration is not constant. So, I should integrate something, right? In the Work's formula, I have to integrate based on volumes. Here I have velocities and locations. I thought about A*x = V (volume), but in my case I have xf & xi, which are "special coordinates" and cannot be used as variables so that I can integrate based on them.
  14. Aug 7, 2017 #13
    This isn't quite right. Let me help. It is easiest to work this problem using cgs units, and then convert the final result to metric.

    The initial volume of gas behind the ball is ##V_i=12\ cm^3##
    The change in volume of gas behind the ball is ##\Delta V=(50)(0.03)=1.5\ cm^3##
    The final volume of gas behind the ball is ##V_f=V_i+\Delta V=13.5\ cm^3##
    So the gas behind the ball expands adiabatically and (assumed) reversibly from ##12.0\ cm^3## to ##13.5\ cm^3##

    During the expansion of the gas, its pressure is $$P=\frac{P_iV_i^{\gamma}}{V^{\gamma}}$$where V is the instantaneous volume somewhere between the initial and final volume. The work done by the expanding gas on the ball is given by $$W_{gas}=\int_{P_i}^{P_f}PdV$$
    I'll let you integrate this out and see what you get.

    The work that the ball does on the atmosphere in pushing the air ahead of it out of the barrel is $$W_{atm}=P_{atm}\Delta V=(1.01325\times 10^6\ \frac{dynes}{cm^2})(3\ cm^3)=3.04\times 10^6\ ergs$$

    So, from a mechanical energy balance $$KE=W_{gas}-W_{atm}$$where $$KE=\frac{1}{2}mv^2$$

    You can use these equations to solve for ##P_i## in dynes/cm^2 and then convert to Pa.
  15. Aug 7, 2017 #14
    Well, I don't recognize the dynes and ergs, so I just went with the SI Units, if that's alright.

    Starting with the integration (I assume you meant Vi & Vf instead of Pi & Pf), we get: W = Pi*(1.38*10-6m3 || Now, the problem I'm facing is that in my book I have W = -P∫VfViDv , where W < 0 means that the gas does work on the ball/piston, and where W > 0 is work done on the gas. Anyway, let's go with the above mode (W = + P...) since that's valid.

    Watm = 3.04 * 106 ergs = 0.304 J

    KE = 7.92 kg(m/s)2 = 7.92 J

    So: 7.92 J = Pi*(1.38*10-6m3 - 0.304 J <=> Pi = 5.96 * 106 Pa

    Well, my book gives an answer of 5.74 * 106 Pa, so I'm guessing it's just an issue of Significant Digits which I'll sort out when I write down properly (can you confirm the book's answer if you've run the numbers?).
  16. Aug 7, 2017 #15
    Some books use the sign convention that ##\Delta U=Q-W##, where W is the work done by the system on the surroundings and ##W=\int{PdV}##, while others use the sign convention that ##\Delta U=Q+W##, where W is the work done by the surroundings on the system and ##W=-\int{PdV}##. In either case, the P is inside the integral sign (unlike the way you wrote it).
    No. Your book left out the work required to push back the atmosphere, by assuming that there is vacuum ahead of the ball. If you do the same, you will get their result.

    I should point out that your book expected you assume adiabatic reversible expansion of the gas in solving this problem. The real-world expansion would actually be highly irreversible, and, as a result, these calculated results would be expected to be highly inaccurate.
  17. Aug 7, 2017 #16
    Oh, that explains it. My bad on the P, my keyboard's a tad busted, and sometimes words get stuck and whatnot. I've got a question though: Why is there a (-) before the Watm in your equation? In my book the only thing I can find about Work/Kinetic Energy relations is the all-encompasing ΔΕmechanical = -fkd + ΣWother forces where fkd is the friction.

    I'm asking because, based on my theory, Wgas should be negative (gas expands, it is work done by the gas on the ball) and Watm should be positive (gas is being compressed, it's work done by the ball on the gas). Plus, there's ΔΚ = Kf -Ki = Kf = 1/2mv2. If I put all that in the afformentioned formula, I get:

    1/2mv2 = Watm + Wgas, where Kf = 7.92 J, Watm = +0.152 J & Wgas = -1.381*10-6Pi m3. After all's said and done, I get Pi = - 5.62 Pa.

    Should I treat the Watm kinda like "friction", because it's work done on another gas? I've not come across any such exercises and it's not in the theory.

    Ah, yeah. I tried it without Watm and I got their result. It's in the OP as well (must've missed due to getting carried away).

    Yeah, most exercises tend to be set in "specific circumstances".
  18. Aug 7, 2017 #17
    The gas is doing work on the ball, not the other way around. So, Wgas is positive. How do we know? Is the displacement of the ball in the same direction as the force exerted by the gas on the ball, or in the opposite direction. Watm is negative. Is the force that the atmosphere exerts on the ball in the same direction as the displacement of the ball, or in the opposite direction?
  19. Aug 8, 2017 #18
    So, the gas expands, and does work on the ball, that is Wgas. The displacement of the gas is in the same direction as the force exterted by the gas on the ball. Therefore, according to your equations, Wgas is positive. As for Watm, the displacement of the ball has a different direction than the force exterted by the atmosphere on the ball. So, before it we put a (-) in the KE = ΣW formula. Correct?

    I'm wondering about this though:

    In my book, I have W = -∫PdV. So, for the displacement of the ball by the gas, I'll have Wgas = -Pi*1.38*10-6m3, which is negative. Kf = 7.92 J. So, disregarding Watm like the book does, I get: 7.92 J = -Pi*1.38*10-6m3, <=> Pi = -5.74 Pa. How do I explain the (-)?

    Like I said, I'm tackling this one based on my theory and the exercise's parameters (disregard the atmosphere in front of the ball, or any other friction), so I'm wondering what's the explaination for the (-).
  20. Aug 8, 2017 #19
    The equation from your book with the minus sign is for the work done by the ball (the surroundings) on the gas (not the work done by the gas on the ball). Look at the wording that your book uses very carefully. The work done by the ball on the gas is (-) the work done by the gas on the ball. The force that the ball is exerting on the gas is equal in magnitude and opposite in direction to the force that the gas is exerting on the ball.
  21. Aug 8, 2017 #20
    Ah, now I get it. I'll check over the book again to make sure.

    Thanks a ton for the patience and help, I appreciate it!
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