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Aerospace Friction factor and Turbulent flow

  1. Aug 10, 2008 #1


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    Do the head losses on turbulent flow are always greater then on a laminar flow? Why does it seem from the Moody chart that the friction factor becomes smaller when the flow becomes more turbulent?
  2. jcsd
  3. Aug 10, 2008 #2
    Look at the definition of the friction factor and where the velocity term is located in that equation.

    It is a little bit confusing but the drop in friction factor does not need less of a head loss since you will have to multiply it by V^2/2 and L/d first in order to obtain the later.
  4. Aug 11, 2008 #3


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    I still don't understand why if there are more friction losses, the friction factor decreases. What is the reason to define the friction factor at this way?
  5. Aug 11, 2008 #4
    The friction factor a I understand relates the length of a pipe, its diameter and the dynamic pressure with head loss. It is very typical to relate phenomena such as these in relation to the dynamic pressure (think of drag coefficient). Since the Pd = 1/2 V^2 rho,the velocity is squared in in it.
  6. Aug 11, 2008 #5


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    The friction factor is a function of Re number and the relative roughness of the pipe (e/D).
  7. Aug 11, 2008 #6
    Ok your correct about the e/D part I just looked it up. Anyway, the friction factor is defined as follows


    Note that it is a function of velocity squared were as you plot it against Re, you plot it against a first order velocity dependency. Hence, there is a velocity term left which causes the friction factor to go down when velocity (turbulence) goes up. the fully turbulent part on a Moody diagram seems non linear and I am not sure what causes that do.
  8. Aug 12, 2008 #7


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    The friction factor gets smaller until the Reynolds number gets really large. Then it's just a function of the relative pipe roughness and the friction factor remains relatively constant. This is indicated by the almost horizontal line on the far right side of a Moody chart.

    The Moody chart indicates this pretty clearly and all of the numerical calcs I've done agree with that (for Newtonian Fluids).

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