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Fluid Dynamics–Find pipe diameter,head loss & actual mean velocity.

  1. Jul 17, 2013 #1
    1. The problem statement, all variables and given/known data
    a) Calculate the theoretical diameter of pipe, nominal pipe diameter and mean velocity of nominal pipe diameter.

    Maximum velocity = 1.8m/s
    Q(vol. flow rate) = 0.01M^3-1

    b) Find the Reynolds number for the process and then the head loss due to friction, state any assumptions made.

    Please see attached diagram***

    2. Relevant equations

    D=sqrrt 4*q/Pi*U
    q=volumetric flow rate
    D=theoretical diameter

    Q=V*A
    V=Q/A
    V=mean velocity
    A=cros sect area
    Q=vol. flow rate

    Re=um*d*P/U
    Where um=mean velocity
    d = diameter of pipe
    p = density
    U = dynamic viscosity of fluid
    Re= Reynolds number

    Head loss due to friction Hf=32*U*L*um/P*g*d^2
    U = dynamic viscosity of fluid
    L length of pipe from the pump = 27m
    P= density taken from previous questions I have done for oil= 860Kgm-3
    g = gravity 9.81m/s
    d = diameter of theoretical pipe =120mm

    3. The attempt at a solution

    I have worked out the theoretical diameter to be 120mm from the first equation assuming that the mean velocity is roughly half the maximum velocity ie 0.9m/s

    So sqrrt 4*q/Pi*U = sqrrt 4*0.01/3.142*0.9=0.12m
    Then choosing from the table nominal pipe 100mm the actual velocity for the nominal pipe is 1.27m/s using this eq.

    Q=V*A
    V=Q/A
    V=0.01/7.85*10^-3=1.27m/s

    To work out the Reynolds number I have used the density and dyn. viscosity from previous questions (860kgm & 0.032)
    Re=0.9*0.12*860/0.032=2902.5

    To work out the head loss I have used these density and dyn. Viscosity figures again and worked out from the diagram that (from the pump) the length of pipe is 27M.
    So Hf=32*0.032*27*0.9/860*9.81*0.12^2=0.205M

    So head loss due to friction is 0.205M.

    I am not sure if I should use these assumptions i.e density and viscosity that are not given in the question. Am I on the right track or away on a crazy tangent??!!

    Any help would be much appreciated – thank you in advance.

    Cheers
     

    Attached Files:

  2. jcsd
  3. Jul 17, 2013 #2
    You're on the right track, but you used the pressure drop equation for laminar flow, and the flow is turbulent (for the viscosity and density you assumed).
     
  4. Jul 19, 2013 #3
    Thanks for your comment Chestermiller much appreciated.

    So I have now used the following Darcy equation for turbulent flow
    Hf=4LCfUm^2/2gd

    (Cf obtained from Moody diagram)

    So = 4*27*0.004*0.9^2 / 2*9.81*0.12
    Hf=0.143m of fluid.

    Is this the correct method??

    Cheers.
     
  5. Jul 19, 2013 #4
    Yes. This looks right. (But, I haven't checked your arithmetic)
     
  6. Oct 24, 2013 #5
    Hi there,

    I am looking to understand the methods used in questions c) and d)

    Can you provide any insight into how i can answer a question similar to these
     
  7. Jan 23, 2017 #6
    Why is the mean velocity roughly half of the maximum velocity? Does this apply to laminar/turbulent/transitional flows?

    Thanks in advanced
     
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