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Water Powered Rockets

  1. Mar 17, 2005 #1
    What kind of energy is stored when u pump air
    into a bottle ie plastic bottle rocket. When it is launched, it has KE,
    but where does the KE comes from? the compressed air? what kind of
    enery then? n how do u calculate the enerygy stored inside?
     
  2. jcsd
  3. Mar 17, 2005 #2

    Pengwuino

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    Pressure is built up when you pump air into a plastic bottle rocket. Kinetic energy doesnt "come from" any place, its simple a state that the bottle is in. The compressed air will create a force that will result in a change in kinetic energy though. Pressure is created by the compressed air, thats pretty much what your looking for. I dont know a very simple way of calculating it though.
     
  4. Mar 17, 2005 #3

    FredGarvin

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    The energy stored is potential energy. When you release the bottle, the potential energy is converted to kinetic energy. The potential energy due to the pressure will be PE=p*V where p = pressure and V = volume.
     
  5. Mar 17, 2005 #4

    russ_watters

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    The fun thing about water rockets is figuring out how much water to put in: since they work via action-reaction, you want a lot of water to be thrown out the nozzle. But since the force is due to air pressure, you want as much air (and as high a pressure) as you can get. I suppose, given the maximum pressure, you could calculate the exact ratio between water and air...
     
  6. Mar 17, 2005 #5
    Actually Fred, no work is done at constant volume (like pushing against a wall does no work). All of the potential energy is stored in the walls of the container, which expand imperceptibly. (relatively high bulk modulus)
     
  7. Mar 17, 2005 #6
    The air that is pumped into the container, builds up pressure on the top of the bottle. The water is sitting by the neck, building up more pressure which is applied by the air. When the pressure is too great, it breaks from its stand and the air pressure pushes the water from the bottom, creating thrust and propelling the bottle skyward.

    If you can work out the volume, I believe you should be able to work out the pressure ( P = I R ) and get a ratio from that. Then all being well you should be able to get the perfect amount for max height.

    :)
     
  8. Mar 17, 2005 #7

    FredGarvin

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    Crosson,
    Good catch. I am slapping myself for that.
     
  9. Mar 17, 2005 #8

    Andrew Mason

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    This is not correct. Fred was right. The air pressure increases and the increased pressure represents energy/unit volume. The energy is [itex]\Delta (PV) = \Delta PV[/itex]. The work is done by the pressure in expanding the volume by pushing on the water, which provides rearward momentum to the water and forward momentum to the bottle.

    The walls may stretch a little and give some additional potential energy (and increase the volume a little, thereby decreasing air pressure) but it will be very small compared to the energy stored as air pressure.

    AM
     
  10. Mar 17, 2005 #9

    russ_watters

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    Slap yourself for that. You were right the first time.

    The walls of the container matter if they can store a lot of energy, but in the case of a semi-rigid plastic water-rocket, they don't. In a balloon (air only), yes...
     
  11. Mar 17, 2005 #10
    Yes sorry, it appears that I need a slap.

    It is true that no work can be done at constant volume, but increasing pressure at constant volume increases the temperature of the gas.
     
  12. Mar 17, 2005 #11

    Andrew Mason

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    I am not sure what you mean by constant volume. The volume of the trapped air is reduced as the water is pushed into the bottle. Work is done in compressing the air and it is converted into potential energy (pressure x volume). That pressure is then converted back into work by the air expanding and pushing the water out of the bottle.

    AM
     
  13. Mar 18, 2005 #12

    FredGarvin

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    Ahh that sweet spectre of self doubt...
     
  14. Mar 18, 2005 #13
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