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Calculating pipe diameter for half-flow condition/mannings equation

  1. May 29, 2014 #1
    Hello,

    My knowledge on fluid mechanics is on a introductory level. When explaining, it would be great if you could explain in simple English and also like teaching to a student who isn't very good at the subject area!
    I've been reading a lecture slide and have been trying to follow the example to solve my given problem. I don't really understand how the lecturer got certain parts of the solution though (equation hasn't been explained/defined on any of the slides). I don't understand how the lecturer found the diameter or the suitable velocity is for half-flow.

    EDIT: The question doesn't require me to find whether the calculated velocity is suitable, but I would like to know how you do find it.

    I have attached the slide that I'm using as an example below.

    I used A = pi*D^2/8 and P = pi*D/2 as my problem is for half flow condition.

    What is wrong? How do I do this correctly?

    Thank you for reading. Any help would be appreciated, thank you.


    1. The problem statement, all variables and given/known data

    If water flows from Point D at Kinglake to Point E at Sugarloaf in a lined circular channel laid on a slope of 0.001. If Manning’s n = 0.013 and if the design discharge is 0.07 m^3/s, what diameter channel should be designed for a half-flow condition? What will be the mean velocity in the channel for this condition?

    Manning’s n = 0.013
    Slope = 0.001
    Q, Discharge = 0.03m^3/s

    Find diameter of channel for half flow condition.
    Find mean velocity of channel in half flow condition.

    2. Relevant equations

    Mannings equation:


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    gif.gif
    gif.gif
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    Velocity equation:

    gif.gif
    gif.gif
    s.gif
    s.gif
    gif.gif

    3. The attempt at a solution

    2%7D.gif
    2pq3qlc.gif
    abl9oi.gif
    A = pi*D^2/8 and P = pi*D/2
    Diamter, D = 0.53m^(8/3)

    How do I find the diameter required for the channel from this? Is this (above) the diameter? I don't understand on the attachment below, how 0.52m^(8/3) becomes a diameter of 1.21 metres?

    ------

    gif.gif
    V = 0.07m^3/s/((pi*(0.53)^2)/8) = 0.634 m/s

    How do I find whether the velocity I calculated is suitable?
     

    Attached Files:

    Last edited by a moderator: Apr 27, 2017
  2. jcsd
  3. May 29, 2014 #2
    You can't change the formula for the perimeter and/or area of a circle. If you want a half flow condition than divide the flow by 2.
     
  4. May 30, 2014 #3
    Ok thank you.
     
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