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Force on pressure pipe

  1. Apr 4, 2016 #1
    1. The problem statement, all variables and given/known data

    The formula of the force exerted on the fluid by the pipe is P1A1 –Fx – P2A2 , I don’t understand why the P1A1 is positive ?


    2. Relevant equations


    3. The attempt at a solution

    Assuming to the right is positive , shouldn’t the force acting on the water = -P1A1 ?

    Because the water is flowing from left to right , the force acting on the water by the pipe should be in oppostite direction( which is –P1A1 ) , am I right ?
     
  2. jcsd
  3. Apr 4, 2016 #2
    Picture below
     

    Attached Files:

  4. Apr 4, 2016 #3
    as water is flowing to the right force exerted by pipe on water should be left to right=its perhaps helping the water to maintain the flow
    Things don't just start moving by themselves. There must be a force acting on the water in the pipe for it to move, and the obvious one is its own weight. Water is quite heavy stuff - a litre of it weighs a kilogram. The water in the tank pushes down on the water in the pipe.
    what is your Fx ?
    it appears your pipe diameter is changing!
     
  5. Apr 4, 2016 #4
    if so , why F2 is in opposite direction ? according to you , F2 should help the water to flow , so it should in the same direction as the water flow , right ? which is directed to the right ?
     
  6. Apr 4, 2016 #5
    in previous post i did'nt see the diagram of 'reducer'.... i think the contraption is pushing water through a nozzle of reduced diameter and its not a 'streamline flow' arrangement - the change in momentum is being presented as a force Fx-
    so the water at the nozzle is being pushed as a 'reaction' of the volume of water being pushed to the right- its a macro-model made by the set up;
    its something like pushing burnt gas out of nozzle of rocket and a back thrust .
     
  7. Apr 4, 2016 #6
    i still dont understand why the F1 is acting as the same direction of the low of water , can you explain further?
     
  8. Apr 4, 2016 #7
    or do you mean since the diameter is decreasing along the pipe , so the F1 is to help the water flow (since there is not much obstruction ) , F2 is the force which try to prevent the water flow (since the water is difficult to move thru smaller diameter of pipe), so it is in opposite direction?
     
  9. Apr 4, 2016 #8
  10. Apr 4, 2016 #9
  11. Apr 4, 2016 #10
  12. Apr 4, 2016 #11
    The expression you gave is not the force exerted by the pipe on the fluid. Only -Fx is the force exerted by the pipe on the fluid. P1A1 is the force exerted by the fluid behind the control volume on the fluid in the control volume, and -P2A2 is the force exerted by the fluid ahead of the control volume on the fluid in the control volume. Since the cross sectional area has changed, there also should be a rate of change of momentum term on the other side of the equation.
     
  13. Apr 4, 2016 #12
    Why F1 is behind the control volume? F2 is ahead of control volume?
     
  14. Apr 4, 2016 #13
    If there is pressure within the fluid, the fluid behind exerts a forward force on the fluid ahead of it, and the fluid ahead exerts a backward force on the fluid behind it.
     
  15. Apr 4, 2016 #14
    Why, can you explain??
     
  16. Apr 4, 2016 #15
    Are you aware of Pascal's Law that, at a given location in a fluid, pressure acts equally (isotropically) in all directions?
     
  17. Apr 4, 2016 #16
    Yes, how is the condition above link to pascal law
     
  18. Apr 4, 2016 #17
    Is there pressure in the fluid that is situated behind the fluid in the control volume? According to Pascal's Law, does it push forward on the fluid within the control volume? Is there pressure in the fluid that is situated ahead of the fluid in the control volume? According to Pascal's Law, does it push backwards on the fluid within the control volume?
     
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