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I Nozzle Reaction Forces

  1. Oct 13, 2016 #1
    I have been reading many examples of control volume calculations regarding nozzles

    They all end up with a reaction force on the nozzle in the direction of the flow.

    upload_2016-10-13_18-1-30.png


    I cant understand this from a Conservation of momentum perspective.
    I would expect that the reaction force, by Newtons third law, would be opposite the mass that was accelerated in. This is how many people explain it but it doesn't equate with the control volume result.
    Can anyone review my attempt at a solution and critique

    Attempt at a solution
    It appears the resultant force has nothing to do with the flow acceleration or pressure drop
    Much like the flow around a bend, it is due to the change in flow direction.
    More specifically, it is the x direction components of the resultant force due to the angle of the nozzle changing the flows direction.

    img00235.gif

    upload_2016-10-13_18-38-44.png

    (Y direction components being equal and opposite directions cancel out.)

    Why doesn't the flow "acceleration" require a resultant force?

    Even though the velocity of the fluid has increased at the nozzle, the flow has experienced no 'real' acceleration so no resultant force is required.
    While the velocity at P2 has increased, the volumetric flow rate Q has not. It has only changed in area and by continuity- velocity has increased.

    The x-direction flow rate Q into and out of the control volume is the same - so there is no acceleration.

    The pressure gradient force?

    The flow rate Q is fully accounted for by the pressure difference P1-P2 over the original area

    Why do firefighters brace a hose forward ?


    I don't think they do I think they actually brace it down.
    upload_2016-10-13_18-33-19.png

    This is because a hose initially runs along the ground then up the firefighter.
    So again there is resultant forces from changing a fluids direction initially up (the resultant force is provided by gravity and against the ground) once it is travelling up to make it horizontal again the firefighter must provide a downward force above the hose to change its direction downward again.
     
    Last edited: Oct 13, 2016
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  3. Oct 13, 2016 #2

    BvU

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    How so ?
    Check this one

    Nobody would try to use a nozzle that isn't solidly connected to the hose, right ?

    The firefighter has nothing to do with it. Unless he has to use a hose where the nozzle isn't fixed -- and that's hopeless.
     
  4. Oct 13, 2016 #3

    sophiecentaur

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    If the nozzle is of smaller diameter than the hose (usual setup?) then the water will be accelerating as it goes through the taper and emerge at a higher speed than it had in the hose, which involves a forward force acting on the water as it flows through. That will require a constant reaction force on the hose to stop it moving backwards. This is the same principle as a rocket or jet engine.
    It is not necessary to consider the details of the passage of the water and the details of the pipe / nozzle to get a good, near enough answer. All that's necessary is to know the entry and exit diameters and the flow rate (and density of course).
    Edit: the direction of the nozzle will affect what actual force is acting on the nozzle. If it is pointed at right angles to the hose then forces on the hose will cancel and all the reaction force will be on the nozzle. If the hose and nozzle are totally aligned, the reaction force will be upstream, on the compressor only. So you could say that the fireman is needed in order to control any lateral force.
    That link above explains it well.
     
    Last edited: Oct 13, 2016
  5. Oct 13, 2016 #4
    Hi @BvU yes that link gives the same answer- it is to the right. You will have the same volumetric flow rate no matter what the size of your nozzle is or whether you have any shape in the control volume at all - in the x -direction the only thing that can change the control volume flow rate Q is the original diameter in and P1-P2

    @sophiecentaur yes that is what I thought as well but the reaction force is to the right -forwards not backwards.
     
  6. Oct 13, 2016 #5

    sophiecentaur

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    The answer is 'to the right' and that is 'backwards' i.e. in the opposite direction to the direction of the water jet. Let's face it, whatever the details, Newton's Third Law has to apply. What I was meaning is that there will be a backwards 'reaction force' because the water is moving forwards BUT the fireman has to resist this force with a forwards Reaction force of his own, to stope the nozzle going backwards due the the 'first' reaction force. :smile:
     
  7. Oct 13, 2016 #6
    I am so confused by this- to the right is in the direction of the flow. I had a look at @BvU example and that says the same what am I missing that this somehow becomes to the left?
     
  8. Oct 13, 2016 #7

    sophiecentaur

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    I look at the picture of the fireman and he is facing left??
     
  9. Oct 13, 2016 #8
    @sophiecentaur the nozzle examples in my attempt at solution are all facing right- The fireman picture in the next section is to explain what I think the fireman is actually bracing against

    Have a look at the link @BvU provided if you want to be as confused as I am

     
  10. Oct 13, 2016 #9

    BvU

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    How can you say that ? ##\dot m## is the same and ##v## increases considerably. That's what we call acceleration.
     
  11. Oct 13, 2016 #10
    The upstream pressure is higher than the downstream pressure (Bernoulli). If fluid pressure is acting on the wall of the nozzle (this is the only force acting on the wall), it acts perpendicular to the wall. This means that there is a resultant pressure force on the nozzle wall in the same direction that the fluid is flowing.
     
  12. Oct 13, 2016 #11

    Jonathan Scott

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    Just imagine what would happen if you cut the pipe just behind the nozzle. I think it's obvious that the nozzle would be propelled strongly in the direction of the water by the forward water pressure inside the walls of the restricted part. However, if you take the whole system of supply, pipe and firmly-attached nozzle you would expect it to be propelled the other way. I don't see any contradiction in that.
     
  13. Oct 13, 2016 #12
    In the latter case, if the hose were perfectly straight, the nozzle would not be propelled at all.
     
  14. Oct 13, 2016 #13

    sophiecentaur

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    That argument sounds convincing until you consider the momentum transfer situation. On the way out of the nozzle, the velocity has increased and the mass flow rate is the same so doesn't that mean momentum has increased on the way through the nozzle? That requires pressure / force and, if the nozzle is pointing off the axis of the hose, there would be a force away from the direction of the jet. The same thing happens with a rocket (doesn't it?) where net reaction force on the inside of the combustion chamber produces a driving force.
    The pressure consideration also needs fluid velocity and/or an area in order to work out the Force
     
  15. Oct 13, 2016 #14

    A.T.

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    The combustion chamber is closed at one end, and the nozzle is expanding in the flow direction, not narrowing like on the hose. That's where the propulsive force acts:

    rocket_physics_15.png


    In the case of the hose those pressure forces act on the upstream fluid, not the nozzle. The net pressure force on the hose-nozzle itself is with the flow.
     
  16. Oct 13, 2016 #15
    The only force that the fluid exerts on the nozzle body is pressure. The momentum increase of the fluid affects the fluid pressure upstream within the nozzle, but, treating the nozzle as a free body, it is only the fluid pressure that determines the axial force on the nozzle (aside from the tension at the hose connection). As long as the fluid pressure within the nozzle is higher than the pressure outside the nozzle, the axial force that the fluid exerts on the nozzle will be in the positive flow direction, irrespective of what is happening with the momentum exchanges of the fluid.

    Certainly the momentum increase affects the pressure of the fluid within the nozzle. But, as far as the nozzle body knows, it is only the fluid pressure that it feels.
     
  17. Oct 13, 2016 #16

    sophiecentaur

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    Yes - there is a difference but isn't that because of the expanding gases in the case of a rocket? There is still acceleration of the water in the nozzle and acceleration requires a force in the direction of that acceleration. That force must have an N3 companion. Where does that act? I suggest backwards, or we have a new method of propulsion.
    This video might be of interest if people think the force on the nozzle is forward. The end of the hose does just what the above link predicts.
     
  18. Oct 13, 2016 #17

    A.T.

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    On the upstream fluid, as already explained.

    This tells you nothing about the pressure force on the nozzle, because there is also hose tension acting on the nozzle. The relevant experiment would be to separate the nozzle off the hose, as already explained.
     
  19. Oct 13, 2016 #18

    sophiecentaur

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    Those two quantities are not the same so they can't be equated like that. The force is, of course, due to the water pressure but not just the static pressure. The fact is, there has to be an N3 pair somewhere, to explain the acceleration of the water exiting the nozzle. You have to agree that it is going faster at the tip than through the hose. If you can't address the N3 issue then you have no valid explanation.
    You are right in your initial assumption about what happens when the end of the hose is blocked off (the static pressure will cause a force that could push the end off the hose or just burst it).

    I have written a number of replies to this but they have all sampled too much. Bottom line is that there is a force acting on the hose in that video (and many more), that makes it snake from side to side. You can do the experiment with a normal garden hose, with a natural curve in it and held a metre or so from the nozzle. The nozzle moves away from the direction of the jet. A force acting only on the inside of the nozzle would simply stretch the hose and make it straight - as in blowing up a sausage balloon. The force which causes this snaking is from side to side and not forward. It has to arise from the Bernoulli effect at the nozzle exit and it is not in the direction of the hose.
    You could achieve the same sort of motion as in a Hero Steam Engine, using water instead of steam. The same N3 explanation would apply.

    The above link with the curved nozzle calculation is pretty straightforward and accounts for all of this, doesn't it?
     
  20. Oct 13, 2016 #19
    Agreed I could have said that better. Ok in your control volume - As others have suggested, take away the nozzle all together (or replace it with a solid wood block)- assuming the entry fixed is and the exit parameters are fixed so keeping the pressure difference-The "acceleration" the parcel experiences is only due to P1 and P2.

    but the control volume experiences the acceleration within it , we should see the Resultant force in the total force calculations but we don't
    There simply isn't a resultant force to the left shooting upstream as assumed.

    I think the problem with these control volume examples is thru continuity P1/u1 isn't fixed it is a direct relationship to P2/u2 and A1/A2
    Some supposed further upstream pressure P initial is fixed however and so is the atmosphere. P final.

    In a perfectly straight pipe two things can happen when you close/attach the nozzle - either the Pinitial works harder against the resultant force downstream to keep u1 ~P1 increases or P1 P initial stays the same and u1 drops

    P1,u1 will be different whether the nozzle is there or not. The resultant force of the fluid acceleration is always accounted for by the pressure difference That is frustratingly never explained in Bernoulli's
     
    Last edited: Oct 13, 2016
  21. Oct 13, 2016 #20
    I stand by what I said. There is a pressure difference between the inlet and outlet of the hose, with the upstream pressure higher than the downstream pressure. Part of this pressure difference is expended in balancing the backward force exerted by the nozzle body on the fluid. The remainder of the pressure difference is used to accelerate the fluid from the low upstream velocity to the higher fluid velocity at the exit jet.

    The action-reaction pair you are asking about is (a) the forward pressure force that the fluid exerts on the nozzle body and (b) the backward force exerted by the nozzle body on the fluid.

    If you would like, I will develop the equations that can be used to accurately quantify the forward pressure force that the fluid exerts on the nozzle body. Please let me know if you are interested in seeing the analysis.

    Chet
     
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