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I Rigid column fluid mechanics

  1. Jan 2, 2017 #1
    Hello, please see attached text below - this concerns ''rigid column theory''. When a valve is closed, the pressure rises at the valve; the pressure rises above the reservoir pressure by ΔP with a height increase ΔH above H. When the valve is opened, this right hand pressure will disappear and the fluid will start to flow through the open valve.

    Why is ΔH = H - Hf?



    upload_2017-1-2_9-52-28.png

    upload_2017-1-2_9-52-43.png

    upload_2017-1-2_9-52-59.png


    upload_2017-1-2_9-53-11.png
     
    Last edited: Jan 2, 2017
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  3. Jan 2, 2017 #2

    BvU

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    So the driving force / ##\left ( {\pi\over 4} d^2 \rho g \right ) \ \ ## ΔH is the head H at the entrance of the pipe minus the head loss in the pipe.
     
  4. Jan 2, 2017 #3
    I've drawn so many diagrams and simply cannot show that delta H = H - Hf!!
     
  5. Jan 2, 2017 #4

    BvU

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    That a question ?
     
  6. Jan 5, 2017 #5
    The explanations given in the attached document (the first post in this thread) are poor in my opinion.
    Here is how I interpret it: when the valve closes, it seems to imply that a piezometer at the valve with show height H + ΔH
    upload_2017-1-5_13-45-15.png


    Thus, if we define positive direction to the right,

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

    (ρgH + ρ0).A - (ρg(H+ΔH) + ρ0).A = 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 when delta H is positive, then there will be a deceleration of flow.


    Similarly, upon valve opening, we will see:
    upload_2017-1-5_13-45-44.png
    If positive direction still to the right,
    ∑forces on the fluid in horizontal pipe = m.a

    (ρgH + ρ0)A - (ρg(H-ΔH) + ρ0)A = ρAL. dv/dt

    thus ΔH = L/g dv/dt which implies that when delta H is as shown on the diagram, then there will be a deceleration of flow.


    Now, I always though that head loss was as shown in the following diagram:
    upload_2017-1-5_13-54-5.png

    I've drawn the energy grade line (EGL) and hydraulic grade line (HGL) as the piezometer. ΔH, as defined in diagram 2 on this post is shown in white. Delta H is not equal to H - Hf (Hf also called HL).

    upload_2017-1-5_13-49-58.png

    So, where have I gone wrong?
     

    Attached Files:

  7. Jan 5, 2017 #6

    BvU

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    To be honest, I so far only reacted to the part of your question I could recognize (##H_f##). For the pressure surge I have to read up too.
    Text says they close the valve gradually, so the stuff keeps flowing. Instead of assuming a linear deceleration as in the link, the book stops with (2.2a), basically ## F = ma ##. So I agree with you, up to and including
    Note that in fact so far we haven't said anything about the magnitude of ##\Delta H##.
    Rememer that ##\Delta H## is a difference.

    Upon valve opening, the inital head at he valve is H and the valve has to counter it with a force ##\rho g H A##. Any form of opening the valve lowers the force the valve can exercise on the liquid (the head at the valve), so there is a negative ##\Delta H## available to accelerate the cylinder of liquid.

    So imho the baseline for ##\Delta H## in your second picture is the top line and ##\Delta H < 0##.

    As such, I see no wrong in the treatment in the text: there are no excessive claims, just manipulation of a few equations.

    Your third diagram depicts a steady state: ##{du\over dt}=0## so there is no ##\Delta H##.
     
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