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Friction factor & velocity

  1. Jan 1, 2009 #1
    There's a couple of recent threads on friction - rather than hijack those here's a new thread with a few questions I have.

    What I'm trying to understand is the relationship between the friction force (on an object moving through a fluid) and the velocity of the object. A few places I see "friction force is linear with velocity for laminar flow and goes as velocity squared for turbulent flow." I'm trying to undestand this on a physical level.

    I'm more familiar with fluid flow through pipes. In this case, the problems are usually solved by using a friction factor (f) multipied by velocity squared: f*(v^2)/2g. The factor f is found from the Moody chart, which shows f as a function of reynolds number. At low Re, f is proportional to 1/Re; at fully turbulent flow (high Re), f is constant. Since Re is proportional to v, this gives us the described behavior: when Re is low the losses go as v^2/v (ie, linear in v) and when Re is high the losses go as v^2.

    OK, so now my questions: 1) why is the loss =f*v^2? and (2) why is the shape of f vs Re as shown on the Moody chart? I'm looking for the physics behind this. Anyone able to give me a clue?

    Thanks
     
  2. jcsd
  3. Jan 1, 2009 #2

    A.T.

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    The velocity squared dependency can be explained as follows in simplified terms: If you double your velocity, you double the number of molecules that hit you per time unit. But the average collision velocity doubles as well. So the momentum transferred to you per time unit(force) is four times bigger.
     
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