find the total amount of work needed to pump the gasoline out of the tank.

Question

A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 2 meters, its length is 2 meters, and its top is 2 meters under the ground, find the total amount of work needed to pump the gasoline out of the tank. (The density of gasoline is 673 kilograms per cubic meter; use=9.8m/s^2 .)

Work=_______

(include units)

Answer 1

You actually don't need calculus for this problem for a full tank. Because the potential energy is linear, you can argue by symmetry that the average height that you lift the gasoline is simply r + depth to top of the tank on its side = 2 + 2 = 4 m (The depth of the centerline of the tank)

M = ρV = πr²hρ where ρ is density

W = ΔE_pot. = mg*(average height raised) units of W are kgm²/s² which are Joules.

Let's say you were given that the tank is filled to the level Y meters from the bottom in order to motivate the more general calculus solution:

W = ∫ dF(y) (2+2r -y) from 0 to Y dF will be the weight of rectangular horizontal slabs through the cylinder of height dy and area 2x by h and (2+2r-y) is how high each slab volume has to be lifted.

Let's find the chord across the cross-section of the circular tank by using the circle equation solved for x centered at (0,2) with (0,0) being the bottom of the tank: 2x = 2sqrt(r² - (y-2)²)

dF = gdm = ρgdV = ρg(2x)hdy

So, the W = 2ρgh∫ sqrt(r² - (y-2)²)(2+2r-y)dy from 0 to Y This could be simplified by using y = 0 at the centerline, but I wanted to solve generally for a given Y.

Actually integrable: If you integrate from 0 to 4, you get 16πρgh which is what we get from the average calculation.

Good luck!