Pressure at bottom of oil drum

  • Thread starter tensor0910
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  • #1
tensor0910
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Homework Statement


[/B]
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


Homework Equations




The Attempt at a Solution


[/B]
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.
 

Answers and Replies

  • #2
Bystander
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Welcome to the concept of "gauge pressure."
 
  • #3
gneill
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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.
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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