Physics, 22.09.2019 07:30 Reddolls

Afuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius of 5 feet and is 15 feet high, although the current oil level is only 14 feet deep. calculate the work required to pump all of the oil to the surface. oil weighs 50lb/ft350lb/ft3.

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Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of:

Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is:

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.

Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)

W = ∫ 1250π x dx
W = 1250π ∫ x dx
W = 625π * (x2 – x1)

x2 = 14 ft + 15 ft = 29 ft

x1 = 14 ft + 1 ft = 15 ft

W = 625π * (29^2 - 15^2)
W = 385,000π ft-lbs = 1,209,513.17 ft-lbs

Answer from: munziruddin204

The work required to pump all of the oil to the surface is of 1209513.17 ft-lbs.

Given data:

The height of circular top from the ground is, h = 14 feet.

The height of tank is, H = 15 feet.

The radius of tank is, r = 5 feet.

The level of oil inside the tank is, h' = 14 feet.

The density of oil is, $\rho = 50 \;\rm lb/ft^{3}$ .

Let us consider a cylinder that has a height of dx, then the cylinder has a volume of:

$V = \pi r^{2}dx\\\\V = \pi \times 5^{2}dx\\\\V =25 \pi dx$

The, the weight of oil (w) is calculated as,

$w= \rho \times V\\\\w= 50 \times (25\pi dx)\\\\w = 1250 \pi \;dx$

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:

$W = w \times x\\\\W = (1250 \pi \;dx) \times x\\\\W = 1250 \pi x\;dx$

Now, Integrating from X' to X" ft gives the total work done. Where,

X' = distance from top liquid level to ground level ( X' = H = 15 feet)

X" = distance from bottom liquid level to ground level ( X" = H +h = 15 +14 = 29 feet)

$W = \int \limits^{X"}_{X'} { 1250 \pi x} \, dx\\\\W = 1250 \pi \times \dfrac{X"^{2}-X'^{2}}{2} \\\\W = 1250 \pi \times \dfrac{29^{2}-15^{2}}{2}\\\\W = 1250 \pi \times \dfrac{29^{2}-15^{2}}{2}\\\\W = 1209513.17 \;\rm ft-lbs$

Thus, we can conclude that the work required to pump all of the oil to the surface is of 1209513.17 ft-lbs.

Learn more about the concept of work done here:

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Afuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground lev...