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Mean velocity in a circular pipe

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Frictionless, inviscous flow in a circular pipe.

    Velocity profile, [itex]v(r) = 6 - 6r^{1.828}[/itex]
    Volume flowrate, [itex]Q = 9 \frac{\text{m}^3}{\text{s}}[/itex]
    Pipe radius, [itex]R = 1[/itex] m
    Given velocities, [itex]v(0) = 6[/itex] m/s, [itex]v(R) = 0[/itex] m/s.

    Find mean velocity [itex]v_{av}[/itex]

    2. The attempt at a solution
    If I just divide the flowrate by area, I get the correct answer - 9/[itex]\pi[/itex] (correct according to the tutorial solutions anyway). It also seems to make sense.

    However, if I integrate the velocity along [itex]r[/itex] from 0 to [itex]R[/itex] and divide everything by [itex]R[/itex], I get a different value.

    [tex]
    v_{av} = \frac{1}{R} \int_0^R (6 - 6r^{1.828}) \mathrm{d}r = 3.858
    [/tex]

    Can anyone explain the discrepancy? To me both approaches make sense but I can't work out why the results are different.
     
  2. jcsd
  3. May 13, 2012 #2

    tiny-tim

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    Science Advisor
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    hi kbaumen! :smile:

    for an average, shouldn't the integral be (1/πR2) ∫ 2πrdr etc ?
     
  4. May 14, 2012 #3
    Cheers, that actually makes sense. Integrate over area and divide by area.
     
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