1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pressure in a bent pipe

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data
    the water flow in the pipe in upward direction . the pipe with internal diameter 60mm carries water with velocity = 3m/s under head of 30cm . Determine the resultant force on the bend of 70 degree and its direction.

    2. Relevant equations


    3. The attempt at a solution
    my question is what does the head refer to ? from where to where ? red pen refer to from datum to the inlet , blue pen means from inlet to outlet. which is correct , blue or red?

    second is is P1 = P2 ????? ( assuming no friction loss)
     

    Attached Files:

  2. jcsd
  3. Apr 6, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, for the head you can take the center of the flow lines, so where you wrote the 'circumcircled' F1 and F2

    And the pressures are not identical: ##\Delta p = \rho g \Delta h## (Archimedes, Bernoulli)

    [edit] Oops, I suppose 'head of 30 cm' refers to the pressure at F1. So at F1 the pressure is ##p = \rho g \times 0.3 {\rm \ m} ##
     
  4. Apr 6, 2016 #3
    if inlet diameter = outlet diameter , the pressure at both inlet and outlet also not the same ? assuming no friction loss
     
  5. Apr 6, 2016 #4
    do you mean the red line is correct?
     
  6. Apr 6, 2016 #5
    The head H is defined as ##H=\frac{p}{\rho g}+z##, where z is the elevation above a fixed datum. In this case, the datum for z = 0 is to be taken as the elevation of the pipe inlet (red location). So, at this location, H = 0.3 m. At the outlet, the pressure may be different and, of course, the elevation is different. This datum applies to the entire pipe and bend.

    The head H can be interpreted physically as the height to which fluid in a static column would rise above the datum elevation (if the column were attached to the pipe) at any given location along the pipe.
     
  7. Apr 6, 2016 #6
    so , the 0.3 refers to the distance of inlet and outlet ? ( blue line ) ?
     
  8. Apr 6, 2016 #7
    No.
    The 0.3m refers to how high water would rise vertically within a tube that is attached to the pipe at its inlet cross section. There is no tube actually attached there, but, if there were, this is how high the fluid would rise.
     
  9. Apr 6, 2016 #8
    ok, so the Fx = rho(g)(v2-v1) - P1A1 +P2A2cos(theta) ?
    Fx= 1000(0.848)(3cos70 -3) - (0.3)(9.81 x1000)(A1) , but how to find P2 ?
     
  10. Apr 6, 2016 #9
    I'm not checking your equation but, if it is a section of pipe in which there is just a bend in the pipe, the assumption is that the exit cross section area is equal to the inlet cross section area. What does the Bernoulli equation tell you about the relationship between P1 and P2? (The equation you are using to find the force is not the Bernoulli equation; it is the macroscopic momentum balance equation). What does the Bernoulli equation tell you about H1 and H2?
     
  11. Apr 6, 2016 #10
    so P 1= P 2?

    since bernoullis equation =
    P1 / (rho)(g) + z1 + (V1)^2 / 2g = P2 / (rho)(g) + z2 + (V2)^2 / 2g

    since Q=Av = constant , cancelling off z and v , P 1= P2 ?
     
  12. Apr 6, 2016 #11
    No. In this problem, z1 and z2 are not equal. The inlet and outlet are at different elevations. If you rewrite Bernoulli's equation in terms of head H, then
    $$H_1+\frac{v_1^2}{2g}=H_2+\frac{v_2^2}{2g}$$
    What does this tell you about H1 and H2?
     
  13. Apr 6, 2016 #12
    Like this? Sorry, I am not in front of computer now
     

    Attached Files:

  14. Apr 6, 2016 #13
    Huh?? What are you asking?
     
  15. Apr 6, 2016 #14
    H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
    2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
    P1- P2 = (V1 ^2 ) - (V2 ^2) ??
    like this ?
     
  16. Apr 6, 2016 #15
    i asked my senior , he told me that this is a closed pipe , so the Q =constant , as area for inlet and outlet same velocity = constant (v1=v2) , this act like the hydarulic jack , so the pressure transferred is constant throughout the pipe .... i doubt is the statement correct ? can someone explain ?
     
  17. Apr 7, 2016 #16
    If that's what he said, it's not correct. Suppose that Q = 0. Certainly your analysis should be able to handle this specific case. If Q = 0, the system is in hydrostatic equilibirum.

    Multiple Choice Question:
    Which of the following answers are correct for a system in hydrostatic equilibrium (like a hydraulic jack)?

    (a) the pressure is uniform throughout
    (b) the pressure is constant horizontally
    (c) the pressure varies vertically
    (d) the head is uniform throughout

    Chet
     
  18. Apr 7, 2016 #17
    for MCQ , my ans is (a) the pressure is uniform throughout
    so , the conclusion is P1 are not the same with P2 ?
    relationship between P1 and P2 must b e
    H1 - H2 =[ ( (V1 ^2 ) - (V2 ^2) ) / 2g ]
    2g(H1 - H2) = (V1 ^2 ) - (V2 ^2)
    P1- P2 = (V1 ^2 ) - (V2 ^2) ??
     
  19. Apr 7, 2016 #18
    (b), (c), and (d) are all correct. (a) is incorrect.
     
  20. Apr 7, 2016 #19
    in hydraulic jack , why the pressure isnt uniform throughout ? the hydarulic jack has the same pressure at both ends , enabling the small force applied to carry big force , right ?
     
  21. Apr 7, 2016 #20
    In a system at hydrostatic equilibrium, does the pressure increase with depth or doesn't it? Why should the pressure within a hydraulic jack be uniform?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Pressure in a bent pipe
  1. Pipe Pressure (Replies: 4)

  2. Pressure in a pipe. (Replies: 18)

  3. Pressure in bent pipe (Replies: 17)

  4. Bent pipe (Replies: 2)

  5. Force in bent pipe (Replies: 7)

Loading...