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Need Help With Project. Similar to a sprinkler.

  1. Mar 24, 2014 #1
    What I'm making is kind of like a sprinkler or Aeolipile. I have a 2 blade propeller that I'd like to boost with an air tank(s). It has to go from 100rpm to 300rpm via a nozzle at the tips of blades. I'm trying to determine the minimum pressure and volume requirements for a tank of air that can do this.

    I calculated that I needed a total torque of 1629.06N*m to go from 100 to 300rpm in 2 seconds (the amount of time isn't set in stone) but I've been stuck on the fluids aspect of this problem for a couple days now. I'm sure I'm making this harder than it needs to be by thinking so long on it. I'm just not confident in my assumptions.

    All I want is an equation I can use to calculate the tangential force acting on the blade due to the flow of air from a tank at some pressure and volume through some pipe and out a 90 degree nozzle.

    I'd very much appreciate your help and advice.
  2. jcsd
  3. Mar 25, 2014 #2
    If the absolute peripheral velocity of the tip of the blade is ##\mathbf{u}## and the absolute velocity of the exiting flow from the nozzle is ##\mathbf{v}##, then the force acting on the blade is the thrust force

    where ##\dot{m}## is the mass flow through the nozzle.
    ##\mathbf{v}_{relative}## is the relative velocity of the flow with respect to the moving blade. Large thrust forces are achieved when ##\dot{m}## is as large as possible and when ##\mathbf{v}_{relative}## is as large as possible.

    Further reading: systems with variable mass (rocket motion), e.g. Physics, Resnick and Halliday, vol. I, p. 200.
    All the best! :smile:
  4. Mar 25, 2014 #3
    Thank you for the input. I'm not sure that applies here though. I understand that a jet engine would rely on the relative velocity, because it has to accelerate the air around it. But my rig will have compressed air shooting out from within the system. Wouldn't that only concern the actual velocity? Please correct me if I'm wrong.
  5. Mar 25, 2014 #4
    Let's think of a rocket in outer space, where there is no surrounding medium. By ejecting mass, it moves forward as a consequence of momentum conservation. If you can find a copy of Resnick/Halliday's text, they have a number of examples to illustrate this.
  6. Mar 25, 2014 #5
    Oh, my bad. I was referring to a jet engine like that of an airplane. That type of propulsion would be dependant on its surroundings. But would mine? It functions like a rocket insofar as the work is done by expelling a fuel.
  7. Mar 25, 2014 #6
    Yes, then the force is the one in post #2.
  8. Mar 25, 2014 #7
    Very well then. Thank you. ^_^
  9. Mar 25, 2014 #8
    Where is the compressed gas tank located? Is it moving, or stationary?

    If the tank is stationary, then the gas in it has internal energy (pressure), but its kinetic energy is zero. When this gas reaches the periphery, the device itself must have accelerated the compressed air to peripheral speed, which will slow the device down. Then, upon ejection, the pressure drop in the ejected gas will be delivered as propulsion to the blade, which will accelerate the device rotation.

    If the tank is at periphery and set into motion by an external force, then the delivered thrust will come from both kinetic and pressure energy of the ejected gas.
    Last edited: Mar 25, 2014
  10. Mar 25, 2014 #9
    The tank will be stationary relative to the prop. It would function like a rotating sprinkler. Except with a tank of air rather than an infinite source.
  11. Mar 26, 2014 #10
    Probably best is to consider the energy of unit air mass leaving the tank; how much energy is lost in accelerating this mass to tip speeds; and how much energy this mass will deliver when nozzled out. This analysis will give an idea of device efficiency. All the best! :smile:
  12. Mar 26, 2014 #11
    Last edited: Mar 26, 2014
  13. Mar 26, 2014 #12
    I read it and it seems to be a very helpful thread. Also, they have a link to an engineering calculator for ##\dot{m}## exiting a pressure air tank. The NASA link you have gives you the exit nozzle velocity. Looks pretty good to me :smile:

    I would first estimate the efficiency of this device, if efficiency and work output is what you are after :smile:
  14. Apr 1, 2014 #13
    Actually I want to ask about the work. By my last post's calculations I would need to expel 230gal of air at 130psi to achieve that acceleration. But the work required produce that force over the resulting distance is easily achieved by very small tanks that would not expel 230gal of air. So now I'm confused.
  15. Apr 2, 2014 #14
    One easy way to analyze this system is by using the tip velocity ##v=\omega R##. To put an air mass at the tip, need to spend energy of ##mv^2/2##; To eject this mass via the nozzle, need to perform nozzle analysis (http://www.pipeflowcalculations.com/airflow/) and see what is the needed pressure at the nozzle inlet. This gives an estimate of the needed energy investment. If the absolute velocity at nozzle exit is zero, this gives max. thrust. The energy delivered as propulsion is the kinetic + loss in pressure energy of exiting gas.
  16. Apr 3, 2014 #15
    Do you think that's necessary after my original calculations?
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