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Impulse thrust calculations from pressure bottle

  1. Nov 24, 2004 #1
    Hi,

    Am wondering how to calculate the thrust and impulse thrust for gas escaping a pressure bottle - ie a scuba tank or the like. My maths is a lil shady so go easy on integration and such.

    Like say if I wanted 1MJ of energy output - what size bottle at what pressure with what nozzle diameter would be required.

    Thanks in advance
    Ben
     
  2. jcsd
  3. Nov 24, 2004 #2
    P = F/A;

    Change in momentum = Impulse = Force * time

    Change in energy = Work = (Force)(distance)(cos (theta))....since this a rocket..the force will most likely be in the same direction as the movement...so...cos(0) = 1....and you have...

    Work = Force(distance).

    Or, since you know the momentum, you know the velocity; you can find energy, since KE = (1/2) mv^2

    For future reference, in case your force is not constant or something like that...you may need to use calculus...

    ...then... Fdt = dP....and F *(dot product) dr = dW
    (note that r = distance)

    Essentially, this means that to find Impulse, you have to integrate Fdt, and to find work, you would have to integrate the dot product of F and dr.
     
    Last edited: Nov 24, 2004
  4. Nov 24, 2004 #3

    Clausius2

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    Very easy. You have to use the 2nd law of motion. But there is an special reformulation of that by means of the integral momentum equation of fluid dynamics:

    [tex] \int_S \rho v v dS=-\int_S P dS [/tex]

    which means in first quasy-steady aproximation that the force produces with the flow of gas outwards is balanced with pressure forces:

    If the hole of exhaust has an area of A, and the flow is exhausted at velocity U, and the external pressure is P_a and the internal one P_o:

    the exhaust velocity can be estimated as:

    [tex] U\approx \sqrt{\frac{P_o-P_a}{\rho_o}} [/tex]

    With that, you can make some figures. A more detail analysis will employ the unsteady process of exhausting, but the rocket equation can be employed by you to have a decent solution. I have made only a rude estimation.
     
  5. Nov 24, 2004 #4

    dextercioby

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    With a small adaggio.It should be 2 around here
    [tex] \int_S \frac{\rho v^2}{2} dS=-\int_S P dS [/tex]

    :wink:
     
  6. Nov 25, 2004 #5

    Clausius2

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    Of course That's false. The complete momentum integral equation is:

    [tex] \frac{d}{dt}\int_V \rho \overline {v} dV +\int_S \rho \overline {v} \overline {v}\cdot \overline{dS}=-\int_S P\overline{dS} + \int_S \tau \overline{dS} + \int_V \rho f_m dV [/tex]

    It is impossible I'm wrong here boy. I have used this equation a million of times. :wink: :!!)

    Take a look at some Fluid Mechanics book. What you posted have no sense.
     
  7. Nov 25, 2004 #6

    dextercioby

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    Sorry,my mistake,the integral i posted appears in the enregy law.I mixed up equations.Damn,i was stupid... :mad: :cry:
     
  8. Nov 25, 2004 #7

    Clausius2

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    It doesn't matter. If someone gives me one euro for each stupidity I have said here, I would be the richest guy on the world. :rofl:
     
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