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Pressure at bottom of oil drum

  1. Oct 2, 2016 #1


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    1. The problem statement, all variables and given/known data

    A vertical cylindrical container contains 5590 gallons of gasoline and is 1.38 m in radius. Due to evaporation within the tank, the pressure on the top of the fluid is 2.5 times normal atmospheric pressure. The density of gasoline is 737 kg/m3

    2. Relevant equations

    3. The attempt at a solution

    converted gallons to Liters: 5590 gallons x 3.785L/1 gallon = 21160 L

    Found the height of cylinder: Vcylinder = πr2h

    21160/(π×1.382) = 3536m

    Solve for Pressure using P = Po + ρgh

    = (101300 Pa × 2.5) + (737×9.8×3536)

    Plugged it all in and it was....wrong. :-/

    Only thing I suspect may be wrong is the units I'm suppose to be using ( cm instead of m for example ). Or the possibility of the cylinder being closed when I assumed it was open. But everything else to me looks good. Please help.
  2. jcsd
  3. Oct 2, 2016 #2


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    Welcome to the concept of "gauge pressure."
  4. Oct 2, 2016 #3


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    Staff: Mentor

    Check the units that you're using for the fluid volume. You want to find a cylinder height in meters and the radius is in meters so the cylinder bottom area is in square meters, so the volume should be given in....
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