Work Done on Horizontal Tank pumping oil

In summary, the conversation discusses a gas station's underground tank, which is a horizontally lying cylinder with a radius of 3 ft and a length of 14 ft. The top of the tank is 10 ft below ground and the density of gasoline is 42 lb/ft3. The conversation goes on to discuss the work required to pump out a slice of gasoline located at a certain distance above the center of the cylinder, with the formula being Density · Volume of slice · Displacement of slice. There is also a mention of an incorrect volume calculation and a desire to stand the cylinder upright.
  • #1
marc017
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0

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:

[tex]
x^2 + y^2 = 3\\
x = \sqrt{3-y^2}\\
= 14\sqrt{3-y^2}\\

\int 42(16-h)*14\sqrt{3-y^2}\,dy\\
[/tex]
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.
[tex]

= 14\left( \sqrt{3-y^2} \right)^2dy\\
[/tex]This would be so easy if they would let me stand the cylinder upright...
 
Last edited:
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  • #2
What's the equation for a circle with a radius of 3 feet?
 

1. What is the purpose of pumping oil from a horizontal tank?

Pumping oil from a horizontal tank is necessary in order to extract the oil from the tank and transport it to its intended destination, such as a refinery or storage facility.

2. How is work done on a horizontal tank pumping oil?

Work is done on a horizontal tank pumping oil when a force is applied to the oil in order to move it from the tank. This force can come from a pump or other mechanical means.

3. How is the amount of work done on a horizontal tank determined?

The amount of work done on a horizontal tank pumping oil is determined by calculating the force applied to the oil multiplied by the distance it is moved. This is known as the work-energy principle.

4. What factors affect the amount of work done on a horizontal tank pumping oil?

The amount of work done on a horizontal tank pumping oil can be affected by factors such as the viscosity of the oil, the efficiency of the pumping mechanism, and the distance the oil needs to be moved.

5. Can work done on a horizontal tank pumping oil be converted into other forms of energy?

Yes, work done on a horizontal tank pumping oil can be converted into other forms of energy, such as heat or electricity, through the use of a generator or other devices. However, there will always be some energy lost in the conversion process.

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