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Fluid mechanics: finding friction factor, should be simple

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Given: water at 20°C flows through a 3cmID smooth brass tube at 0.002 m^3/s.

    2. Relevant equations
    V=Q/A
    f=64/Re
    Re=VD/(nu)

    3. The attempt at a solution
    V=Q/A=(0.002 m^3/s)/((∏/4)*(0.03 m)^2) = 2.829 m/s

    Re=VD/(nu)
    = (2.829 m/s)*(0.03 m)/(1x10^-6 m^2/s)=84882.64

    f=64/Re=0.0075


    I know this is wrong for two reasons:
    1. The numeric answer is in the back of the book: 0.0185
    2. Looking at the Moody Diagram: there is no f value on the chart below 0.008.

    Any help in finding my mistake would be appreciated.
    Thank you.
     
  2. jcsd
  3. Apr 11, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Out of interest I've had a look at this, but I've not touched fluid flow since I was a student.

    Here is a very similar example: http://answers.yahoo.com/question/index?qid=20071223185604AAiTXv9

    How does your working compare? I see you haven't explicitly used density. (BTW, you've lost one of the 0's from your ƒ.)

    Here's a Moody diagram http://www.engineeringtoolbox.com/moody-diagram-d_618.html

    https://www.physicsforums.com/images/icons/icon2.gif [Broken] It seems to indicate that for Reynold's numbers up to about 2,500 the flow is laminar, the graph is a straight line so the Darcy–Weisbach friction factor formula ƒ=Re/64 applies. But for Re exceeding about 2,500 the flow is turbulent and you must consult the Moody diagram. If you do, I think you'll get your 0.018 figure. :cool:

    How have I done? :smile:
     
    Last edited by a moderator: May 5, 2017
  4. Apr 11, 2012 #3
    We had another problem that explicitly told us to use the Moody Diagram, so I guess I didn't realize it would be necessary, but you are right. I didn't use density because it is included in the kinematic viscosity, nu(didn't bother to find a symbol for it). I did get the right answer from the Moody diagram.
    Thank you.
     
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