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Pressure in bent pipe

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

    We have the original equation P1A1 –P2A2(cos theta) +Fx = ρ Q( V2 –V1) ,

    Why in the solution there , the author ignore P1A1 and P2A2cos(theta) ?


    2. Relevant equations


    3. The attempt at a solution
    is the book wrong ? but the P1 and P2 are not given also....
     

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  2. jcsd
  3. Apr 4, 2016 #2
    The jet is assumed to be a free jet (unconfined by a tube) so the (gauge) pressures at the very inlet and the very outlet are assumed to be zero.
     
  4. Apr 6, 2016 #3
    Unconfined tube? What is that? i could see the pipe there
     
  5. Apr 6, 2016 #4
    I don't think they meant to show it as a pipe. I think they meant to show it as an unconfined jet. This is my interpretation anyway.
     
  6. Apr 6, 2016 #5
    unconfined jet ? what is that ? what's the difference between unconfined jet and pipe?
    ?
     
  7. Apr 6, 2016 #6
    When you shoot water out of a hose, that's an unconfined jet.
     
  8. Apr 6, 2016 #7
    why in unconfined tube (jet) the pressure at both end are assumed to be 0 ?
     
  9. Apr 6, 2016 #8
    With a jet from a garden hose, the air pressure at the free surface of the jet is constant at 1 atmosphere (gauge). This pressure is present throughout the jet cross section. So the pressure throughout any straight section of a jet (more then a few diameters away from the hose exit) is 0 gauge. In your problem, there is a straight section of jet coming toward the blade, and straight sections at A and B leaving the blade. The gauge pressure at these locations is 0. However, in the region where the jet is changing direction in contact with the blade (non-straight section), the fluid pressure varies rapidly from a high value at the blade surface to 0 at the free surface. This pressure variation is what enables the jet to change direction. However, the details of this flow and pressure variation in the region of curved flow are circumvented when you do the overall momentum balance. This enables you to determine the net force of the blade on the jet without ever knowing the details of the flow in the curved region.

    Chet
     
  10. Apr 6, 2016 #9
    form your explaination above , do you mean the jet coming from a hose which enter the stationary blade?
    why the fluid pressure will become 0 to enable the jet to change direction ? IMO, when the jet change direction , it will slow down , pressure will increase , right ? why you said it become 0 ?
     
  11. Apr 6, 2016 #10
    You have the right idea. What I'm saying is that it will increase at the blade surface, but, at the free surface, the jet is still in contact with the air, so the pressure at the free surface will still be zero. In the region where the jet is in contact with the blade (and changing direction), the fluid pressure will vary across the diameter of the jet, from a high value at the blade surface to zero at its free surface.
     
  12. Apr 6, 2016 #11
    do you mean both my idea aforementioned correct ?
    Chet form your explaination above , do you mean the jet coming from a hose which enter the stationary blade?
    why the fluid pressure will become 0 to enable the jet to change direction ? IMO, when the jet change direction , it will slow down , pressure will increase , right ? why you said it become 0 ?


    do you mean water pressure ?
     
  13. Apr 6, 2016 #12

    Yes, if you want to think of it that way.
    I don't know how to say it any better. In the region where the liquid is changing direction when in contact with the blade, the pressure within the liquid varies across the cross section of the jet from a high value at the blade surface (as you said, pressure will increase ") to a value of zero at the free surface of the jet, all within the same cross section of the jet.
     
  14. Apr 6, 2016 #13
    do yuou mean the water pressure increases from the inlet to the bent part of the pipe , then it decreases to 0 at the outlet of the pipe ? since at the outlet of the pipe , the pressure acting is no longer water pressure , but atmospheric pressure?
     
  15. Apr 6, 2016 #14
    No, no, no. I said it increases across the diameter of the jet. A pressure gradient across the diameter of the jet is required to change the direction of all parcels of fluid comprising the jet (both near the wall and near the free surface), because each parcel is being accelerated (i.e.,changing direction).
    You need to understand 2 things:
    1. Pressure is continuous at a free surface, and the pressure of the water must match the pressure of the air at a free surface.
    2. Pressure does not only vary along streamlines. It can also vary in the direction perpendicular to streamlines.
     
  16. Apr 6, 2016 #15
    I thought when the water is flowing and friction is ignored , the velocity of water across the straight pipe would be constant ? thus , the pressure gradient would be constant ? ( consequence of beroulli's principle ) ?
     
  17. Apr 6, 2016 #16
    That's only if the streamlines are straight. If the streamlines are curved, the pressure varies between streamlines. How else could the parcels of fluid be accelerated normal to a curved path (centripetal acceleration)? After the streamlines straighten out again, the pressure becomes uniform across the cross section again.
     
  18. Apr 6, 2016 #17
    at the free surface , the pressure acting is the atmospheric pressure , right ? why you said it's 0 ? is it because of gauge pressure? why cant we use absolute pressure in the calculation?
     
  19. Apr 6, 2016 #18
    Yes.
    We can, but it's easier to use gauge pressure.
     
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