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Fluid resistance via poiseulle method

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Why is the friction in the small tube higher than in the big tube.

    2. Relevant equations

    F = n * A * dv/dx, where n is internal friction factor, A the area and dv/dx the speed gradient

    V= (Pi * p * r^4 * t) / ( 8 l n ), where V is the volume of water, p is the pressure difference, r the radius of the small tube, t elapsed time, l small tube length


    3. The attempt at a solution

    If i took that each "area" of the fluid that is coaxial to each other and summarized them together, then there would be less friction because there are less layers of water there hence resulting that in the small tube there should be lower friction than in the bigger tube.
    Logic however would say that due to the fact that the water is moving at a higher speed in a smaller area should encounter more resistance than when its moving in a big tube slowly.

    The last idea i had on this subject was that due to the fact that the area in the bigger tube is so much larger, then the resistance per Area is smaller eg (A / sum F) in the big tube than in the small tube.

    n = (Pi * p * r^4 * t) / ( V* 8 * l)

    F = (Pi * p * r^4 * t * A * dv/dx) / ( V * 8 *l )

    and that doesn't even look close to what the real formula is (http://hyperphysics.phy-astr.gsu.edu/*hbase/pfric.html#tube [Broken])

    This was part of lab work to determine the validity of the poiseulle law and calculate the internal friction force n. That was the question i was asked to "defend" but i am literally stumped about how the friction works in there.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
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