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Calculating Pipe Roughness

  1. Mar 15, 2010 #1

    danago

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    Gold Member

    In my fluid mechanics class, we did an experiment where we were asked to measure some values and then, using a Moody Diagram, estimate the roughness of the pipe. Briefly, each trial consisted of allowing water to flow down a thin tube (diameter D=9.5mm) and into a discharge cylinder. A manometer was connected to two points along the tube (separated by a distance of 1m) so that we could measure the pressure differential.

    As an example, i will use the data from one of the trials we conducted:

    Flow speed (V): 52.3 cm/s
    This was measured by observing the rate at which the water level rose in the discharge cylinder and then using this, along with the tube geometry, to calculate a flow speed.

    Head Loss (h) over 1m of tube: 4cm
    Calculated by measuring the height differential of the manometer.

    I then calculated the friction factor:

    [tex]
    f=\frac{2hDg}{LV^2}=0.027
    [/tex]

    and the Reynolds Number:

    [tex]
    Re=\frac{VD}{\upsilon}=4950
    [/tex]

    Now when i plot this on a Moody Diagram, it falls below the line for a smooth pipe, which as far as i thought, was impossible? This happened with all of our trials.

    Does anybody know what could possibly have gone wrong? Have i made a mistake in my calculations?

    Any help is greatly appreciated.

    Thanks,
    Dan.
     
  2. jcsd
  3. Mar 16, 2010 #2
    Firstly let us look at you calculations.

    A velocity of 52.3 cm/sc implies a collection rate of 37 cc/s is this correct?

    I agree with your calculation of Reynolds number.

    However you calculation of friction factor is not well defined.
    By this I mean that the the use of 'f' for friction factor has lead to confusion since the American definition is four times (yesx4) the value of the British one.
    Therefore you have to be careful whether you are entering an American or British (Moody worked in Britain) Moody chart.

    However the British have resolved the situation by changing to use lambda ([tex]\lambda [/tex] for the American definition.

    So British Moody charts are now published using

    [tex]\lambda \quad = \quad 4f[/tex]

    But even allowing for the possibilty of an f / 4f confusion your result still falls closer to laminar flow line
    [tex]\begin{array}{l}
    \lambda \quad = \quad \frac{{64}}{{{\mathop{\rm Re}\nolimits} }} \\
    f\quad = \quad \frac{{16}}{{{\mathop{\rm Re}\nolimits} }} \\
    \end{array}[/tex]

    Than the line for smooth pipes.

    Edit: I should have been more clear.

    Your formula for f corresponds to the American definition. I agree with your arithmetic so
    [tex]\lambda [/tex] = 0.027.
     
    Last edited: Mar 16, 2010
  4. Mar 16, 2010 #3

    danago

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    Gold Member

    Yes, 37 cc/s is correct.

    Sorry, i did forget to specify which friction factor i was using. It is in the Darcy–Weisbach factor, which i guess would mean it is the American one? Not sure how i will go about estimating a pipe roughness though :( I spoke to somebody else today, and their results showed the exact same problem.

    Thank you for replying by the way :)
     
  5. Mar 16, 2010 #4
    Did you see the edit to my first post?

    9.5mm is a very small bore tube, but within the range of this experiment:

    http://www.jfccivilengineer.com/pipe_friction_loss.htm

    So I can only conclude your flow rate was still too low.

    FYI
    Moody also gives the following formula instead of the chart - but this does not help in your case as it makes the roughness negative.

    [tex]f = 0.001375\left[ {1 + \left( {\sqrt[3]{{\frac{{20000k}}{d} + \frac{{1000000}}{{{R_e}}}}}} \right)} \right][/tex]

    Where f is calculated in the British way and is your f divided by 4

    I will be interested to learn the explanation.
     
  6. Mar 16, 2010 #5

    danago

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    Gold Member

    Yea i did try using an equation for the calculation -- It wasn't that one, but the Colebrook equation, however it also indicated a negative roughness.

    Anyway thanks for your help, i might just have to make a discussion of my results in my report, and explain the results that i expected to obtain.
     
  7. Mar 16, 2010 #6
    Last edited by a moderator: Apr 24, 2017
  8. Apr 6, 2010 #7
    Solving for friction factor

    Hey I'm needing to solve the Colebrook equation for the friction factor. Given that the equation is 1 / (f^1/2) = -2 log [ 2.51 / (Re (f^1/2)) + ((epsilon) / dh) / 3.72 ] . I need to implicitly solve for f. We have to show all algebraic steps. I was wondering if someone could give some insight on where to start or maybe a walkthrough of how to get to the friction factor as my skills with logarithms isn't too good. Any help would be greatly appreciated.
     
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