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Fluid Mechanics of oil

  1. Dec 14, 2006 #1
    1. The problem statement, all variables and given/known data
    Medium lubricating oil, of specific gravity 0.860, is pumped through 300 m of horizontal 50-mm-diameter pipe at the rate of 0.00114 m3/s. If the drop in the pressure is 200 kPa, the absolute viscosity of the oil in N×s/m2 is...____________?

    specific gravity (SG) = 0.86
    length (l) = 300 m
    diameter (d) = .05 m
    flow rate (Q) = 0.00114 m^3/s
    Pressure drop (P) = 200E3 Pa

    2. Relevant equations
    I know that:
    shear stress (t) = viscosity (mu) * velocity gradient

    velocity gradient = (change in velocity) / (change in height)

    3. The attempt at a solution

    I'm not quite sure how to begin with this. I have a solution from my prof. (this is a question on a sample/practice exam for our final)... but I have NO clue where he's coming up with this equation:

    P = 32 * (mu) & (l) * [(v)/(.05m)^2]

    Where on Earth does the 32 come from? What significance does the (.05m)^2 have... without pi/4 anyway? Can anyone make any sense of this? The solution is supposed to be... approximately 0.09 (N s) / (m)^2. Let me know if anybody has any ideas.

  2. jcsd
  3. Dec 14, 2006 #2


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    Science Advisor

    You would have to check Reynolds number, but your prof references an equation for pressure drop in the laminar flow regime. You need to back through the proof for the equation, but the 32 comes from:

    [tex]Q = \frac{\pi R^2 V_c}{2}[/tex]

    [tex]V = \frac{\pi R^2 V_c}{2 \pi R^2}[/tex]

    [tex]V = \frac{V_c}{2}[/tex]


    [tex]V = \frac{\Delta p D^2}{32 \mu L}[/tex]

    In stead of going through the entire derivation, go here and look under the section "Volumetric Flow Rate."

    Last edited: Dec 14, 2006
  4. Dec 14, 2006 #3
    Thank you sooo much. This does clear up the issue with the "32" that I had a problem with.

  5. Dec 20, 2006 #4
    Glad to be of assistance archeryguru2000.
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