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Fluid Dynamics: Static Thrust on a Conv-Div Nozzle

  1. Oct 14, 2015 #1
    The problem statement, all variables and given/known data
    convdiv.png

    The attempt at a solution

    Hi all, I am tasked with finding the static thrust generated by this convergent-divergent nozzle. The stagnation pressure is known the inlet.

    Now, personally, I would've thought drawn a control volume around the entire nozzle. And then said:

    Net_Thurst = Pressure_inlet *A_inlet -Pressure_outlet*A_outlet.

    Now, this does not yield the correct answer. I am not asking for the answer, rather, what is wrong with my understanding of the problem?

    Thanks for the help!
     
  2. jcsd
  3. Oct 15, 2015 #2
    p0c > pc & Ain > Aout → that would mean the net thrust would push the rocket backwards. You have to take into account the forces in the jacket and the acceleration of the fluid.

    Using the control volume like suggested in the hint makes it easier to solve the problem (think about the pressure distribution on the surface of the control volume).
     
  4. Oct 15, 2015 #3
    Okay, that makes sense.

    I have, to this far, thought of the pressure as driving the fluid flow.

    Thus, I would've thought that the conservation of momentum equation applies:

    p1A + p1A*V1^2 = p2A + p2A*V2^2

    Though, it seems that we need to introduce a 'thrust' component. I don't really understand the physics of this. Where does the thrust component evolve from? And doesn't it violate the conservation of momentum equation.
     
  5. Oct 15, 2015 #4
    Sorry, a misunderstanding - maybe a language problem - I thought with thrust you ment the force pushing the rocket forwards.

    If you make a FBD of the fluid in the nozzle you also have borders at the non-horizontal walls of the nozzle and the "diffusor", which you need for the sum of the forces in horizontal direction - so your equation is not complete.

    The thrust will be generated by the acceleration of the fluid, i.e. a force affected it. The same force, but in the opposite direction accelerates the rocket (Newton's 3rd law). So I think that is what you ment with the equation in your first post, which I obviously misread.
     
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