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Liquid being pushed out of a nozzle - how to calculate it?

  1. Sep 20, 2012 #1
    Hi guys,
    I'm relatively new to physics, so please bear with me. :)

    Let's image we have a cylinder like this, filled with 8760 mm3 of liquid, with a nozzle on one end and a plunger on the other.
    An explosive charge goes off behind the plunger, releasing 45 joules of energy, and pushing the liquid out of the nozzle.

    62a879eccfc5.png

    I want to calculate
    1. how far will the liquid go, before dispersing into mist?
    2. what is the optimal shape and size of the nozzle, to make the liquid go as far as possible?

    Where do you think should I start?

    Thanks! :smile:
     
    Last edited: Sep 20, 2012
  2. jcsd
  3. Sep 20, 2012 #2

    haruspex

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    I don't think there's enough information. You need to know what pressure will be generated. If the explosion is in a relatively large chamber then the pressure will be low but persist, so the jet will be weak but last some seconds. With a smaller chamber you'll get a stronger but briefer spurt.
     
  4. Sep 20, 2012 #3
    As you can see from the drawing, the charge ( that copper-color thing ) is basically next to the plunger, so at least at first there isn't any chamber. The goal is the make the jet as strong and as fast as possible.

    And one more thing :) does the conical shape of the front end of the plunger help or basically the result would be the same as with a simple flat plunger?
     
    Last edited: Sep 20, 2012
  5. Sep 20, 2012 #4
    This sounds complicated.

    To calculate anything, you will need to know/assume:

    -Viscosity of the liquid
    -Does the charge transfer all energy into horizontal kinetic energy for the plunger?
    -Friction? (probably just neglect it)
    -Outside pressure (probably atmospheric?)
    -Gravity (earth's gravity)?
    -Temperatures (probably assume a constant temperature for all parts)
    -Heat of vaporization of the liquid
    -Vapor pressure of the liquid
    -density of the liquid
    -atmospheric heat capacity

    The way I'm thinking of it is, you don't know how much liquid will get shot out, and at what speed. This depends on how much energy is transferred, and how dense/viscous the liquid is.

    Once the liquid is shooting out, you will now have a mass transfer problem where the liquid is exposed to the atmosphere, and you will have to determine how quickly it evaporates. This is mainly dependent on the vapor pressure of the liquid, surface area, and the speed of the liquid.

    Assuming you are far above the ground, the actual trajectory probably doesn't matter, as long as you assume a cylindrical stream. This won't be completely accurate, but I'm not sure how you would predict the stream shape.

    2. The shape of the nozzle, I'm not sure. You probably want a trajectory angle of 45°, and I don't know about shape. You want a narrow stream for maximum initial velocity, but you want a broad stream for minimum surface area to volume ratio. The best solution would depend on values of your variables.
     
  6. Sep 20, 2012 #5
    I don't think the shape matters, because we want to assume complete/instantaneous energy transfer, and no friction, for both cases (to make it easier!). If we didn't make those assumptions, then yeah, it would matter.
     
  7. Sep 20, 2012 #6
    I don't think it does matter either. This would be a different story if the liquid were compressible, which is usually assumed not to be. You can see that no matter the shape of the inside of the plunger, the dV/dx of the mass inside the tube will be the same, because the other side of the plunger is uniform and so the volume space increased on that end is independent of plunger shape, ie its linear with position change (assuming a cyclinder).

    This is very much a thermodynamics problem. You wish to maximize the work done by the explosion to transfer as kinetic energy to the mass leaving the other end. I think the ideal nozzle size would be as small as possible without losing the ability to maintain the pressure change, ie without exploding (to consider it in hyperbole, if you could minimize the nozzle to pass just 1 particle with an energy of 45 joules, that would be the best you could get). In that case it becomes a question of how much mass do you want to push out. If you can push out a little bit of it with a high average kinetic energy summed to 45 joules, or all of it with a lower average kinetic energy summed to 45 joules, which would you choose? That will tell you how big to make your nozzle I believe.
     
    Last edited: Sep 20, 2012
  8. Sep 21, 2012 #7

    haruspex

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    Small as it is, the chamber (the vol occupied by the charge) is part of the calculation. The energy released will initially take the form of hot gases at high pressure occupying that volume. Subsequent developments will follow adiabatic gas expansion laws.
     
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