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Momentum of a rocket with pressurized gas

  1. Jun 16, 2011 #1
    I think that the momentum achieved by a rocket that uses a pressurized ideal gas as it's only propellant, stored in a chamber of fixed volume before being rocketed away. will achieve a final momentum proportional to the Pressure of the gas times the fixed volume of the chamber. momentum ∝ (Pressure Initial) • (Fixed Volume). So if the gas pressure was twice as much or if the volume was twice as much you would get twice the momentum as before.

    Notice it is a proportional and not a = sign, I know the units are not right. Can some one tell why this proportionality is true or false as my Physics Book Haliday Resnick does not make it easy to figure out.
     
  2. jcsd
  3. Jun 16, 2011 #2

    Andrew Mason

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    The change in momentum experienced by the rocket is equal in magnitude and opposite in direction to the change in momentum of the expelled gas. These changes in momentum must all be measured in the same non-accelerating frame of reference. So:

    [tex]\Delta p_{rocket} = \int F_{rocket} dt = - \Delta p_{gas} = - \int PAdt[/tex]

    where p = momentum and P = pressure, A = cross-sectional area of the hole through which the gas is expelled.

    This is not an easy thing to calculate because as the gas escapes from the rocket, the pressure decreases, so the force ejecting the gas from the rocket, hence the speed and rate of change of momentum of that gas, decreases. Also the mass of the rocket decreases as mass is expelled from the engine.

    AM
     
  4. Jun 16, 2011 #3

    cjl

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    I don't completely agree with that, no. The physics are actually fairly complex, however simply showing that it isn't a direct proportionality is relatively straightforward.

    For flow out of a rocket, there are really two cases to consider: choked flow, and non-choked flow. Choked flow is when the flow speed reaches mach 1 at the throat of the nozzle, and is generally the case for a gas with a chamber pressure at least twice the ambient pressure surrounding the nozzle (details vary slightly, but I'll ignore that for now). In the case of choked flow, the flow speed is relatively trivial to compute, since it depends only on the gas in the chamber and the temperature in the chamber. Interestingly enough, it does not depend at all on the pressure, so long as the pressure is sufficient to maintain the choked condition. For a rocket without a diverging nozzle section, the flow speed will be mach 1 for all choked conditions. Thus, the momentum of the air leaving the nozzle will indeed be dependent only on the mass flow through the nozzle. Since the mass in the bottle is directly linearly proportional to the initial pressure, this would seem to support your assumption so long as the flow was primarily choked (which is a safe assumption if Pchamber >> Pexternal). However, this is ignoring the fact that there is also a force from the pressure at the nozzle. Since the pressure at the nozzle will be directly proportional to the pressure inside the bottle, the total momentum change to the bottle will be slightly larger than twice as much for twice the initial pressure. I'll also note that this is making a couple of other assumptions - namely that the bottle mass is much larger than the propellant mass (so I can neglect the change in mass throughout the expulsion of the propellant), and that the chamber is isothermal throughout the expulsion of the propellant, which is clearly wrong (but an easy assumption to make). The end result, however, is that the proportionality is clearly not linear, although you could get it fairly close to linear through the use of a very high expansion ratio nozzle (which lowers exit pressure and increases exit velocity) combined with a very low flow rate (to keep the chamber as close to steady-state isothermal as possible). However, this isn't a very realistic case (as it would require either near-vacuum exit conditions, or an incredibly high chamber pressure).

    In the situation where the flow is not primarily choked (in other words, Pchamber is not much much larger than Pexternal), it gets even farther from a direct proportionality. This is because in this case, the exit speed depends on the pressure in the chamber, with higher pressure causing (obviously) higher exit speed. It isn't a linear proportionality, but we'll ignore the math for now and just look at it conceptually. If you start with some pressure such that the flow isn't choked, you'll get some total momentum change from it. Now, if you double the pressure (and assume the flow is still not choked), you have twice the mass in the same volume of chamber. The first half of the mass will leave the chamber more quickly than the second half, since the pressure is higher. After the first half has left, then you are left with the starting conditions from before (with half the pressure). However, since the exit velocity (and thus the momentum per unit mass) is higher during this first half, the momentum imparted to the rocket will be more than doubled by doubling the chamber pressure.

    (Hopefully this makes sense)


    (Andrew: your equation isn't quite right... you have to consider both pressure thrust and momentum thrust of a nozzle, thus the thrust coming out of a nozzle is equal to dm/dt*vexhaust + Pexhaust*Aexit)
     
  5. Jun 16, 2011 #4
    I have to digest these comments but I think from what I've read the answer is momentum is not proporional to pressure because the momentum imparted to the rocket will be more than doubled by doubling the chamber pressure. It makes sense thanks.
     
  6. Jun 28, 2011 #5
    cjl You wrote that.
    Gettting the pressure up by adding gas to the volume is one way of doing it. How about the case where the mass stays the same but the pressure went up by increasing the gas Temp.
    If the heat capacity is linear, 1.006 kJ/kg•K for dry air, wouldn't that make the velocity of the momentum linear also?
     
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