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Fluid mechanics: Rigid column theory

  1. Dec 31, 2016 #1

    Please see attached page from a textbook. Can someone explain why H = 4f.le.vo^2/2d.g and why delta H is given by the expression in the book? Note that the figure it mentions is on the top of the page. I have tried for days here.


    Attached Files:

  2. jcsd
  3. Jan 1, 2017 #2


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    I am not familiar with rigid column theory and you didn't exactly provide a lot of information to try to decipher it, but this equation looks like it is rather empirical, so it probably isn't derived from first principles. More likely is that it is dimensionally consistent and includes all the factors upon which the answer should depend, so it works provided you know the value of ##\ell_e##.
  4. Jan 2, 2017 #3


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    You have ##\Delta p = \rho g \Delta H## and Darcy - Weisbach ## {\Delta p\over l} = f_D {\rho\over 2} {v2\over d} ##
    Fuirthermore ## f_d = 4 f##, the Fanning friction factor which seems to be the one your book uses.
    They combine to $$
    \Delta H = 4f {l v^2\over 2dg}
    $$for the head loss in the pipe
  5. Jan 4, 2017 #4
    I hope you are able to see the book page ok on the attachment.

    When the valve closes, it seems to imply that a piezometer at the valve with show height H + ΔH

    Thus, if we define positive direction to the right,

    ∑forces on the fluid in horizontal pipe = m.a

    (ρgH + ρ0) - (ρg(H+ΔH) + ρ0) = ρAL. dv/dt where: ρAL is the mass of fluid in the horizontal pipe and ρ0 is atmospheric pressure (A is pipe cross section)

    This gives ΔH = - L/g dv/dt which implies that then delta H is positive, then there will be a deceleration of flow. Why has there not been any friction mentioned here?
  6. Jan 5, 2017 #5
    Just read the passage again, it says the fluid is frictionless
  7. Jan 5, 2017 #6


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    In that case I wonder what they mean with ##f##. But then again, that's in a subsequent paragraph ...
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