Work Done on Horizontal Tank pumping oil

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SUMMARY

The discussion focuses on calculating the work required to pump gasoline from a horizontal cylindrical tank, specifically a tank with a radius of 3 ft and a length of 14 ft. The density of gasoline is established at 42 lb/ft³. The user attempts to derive the volume of a slice of gasoline and formulates an integral to compute the work, but acknowledges a potential error in their volume calculation. The correct expression for the volume of the slice is clarified to be 14(√(3-y²))² dy, which is essential for accurate work computation.

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  • Understanding of calculus, specifically integration techniques.
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Students in physics or engineering, particularly those studying fluid mechanics, as well as educators looking for practical examples of work calculations involving fluids in cylindrical tanks.

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Homework Statement



A gas station stores its gasoline in a tank underground. The tank is a cylinder lying horizontally on its side. The radius is 3 ft, the length is 14 ft, and the top of the tank is 10 feet under the ground. The density of gasoline is 42 lb/ft3

a) Consider a slice of gasoline that is Δy ft thick and located y ft above the center of the cylinder. Write an expression giving the work required to pump the slice out. Give your answer using the form below. (Use Deltay for Δy if necessary.)

Density · Volume of slice · Displacement of slice

The Attempt at a Solution



Density = 42 lb/ft3
Volume of Slice:

<br /> x^2 + y^2 = 3\\<br /> x = \sqrt{3-y^2}\\<br /> = 14\sqrt{3-y^2}\\<br /> <br /> \int 42(16-h)*14\sqrt{3-y^2}\,dy\\<br />
With the integral going from 0 to 6

Since I know my volume is wrong the integral can't be right, but for this sake, would the integral be right assuming the volume Is what I put?

------------------------------------------------------
EDIT:

Would the volume for the slice be.
<br /> <br /> = 14\left( \sqrt{3-y^2} \right)^2dy\\<br />This would be so easy if they would let me stand the cylinder upright...
 
Last edited:
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