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
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, [tex]\rho = 50 \;\rm lb/ft^{3}[/tex].
Let us consider a cylinder that has a height of dx, then the cylinder has a volume of:
[tex]V = \pi r^{2}dx\\\\V = \pi \times 5^{2}dx\\\\V =25 \pi dx[/tex]
The, the weight of oil (w) is calculated as,
[tex]w= \rho \times V\\\\w= 50 \times (25\pi dx)\\\\w = 1250 \pi \;dx[/tex]
Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have:
[tex]W = w \times x\\\\W = (1250 \pi \;dx) \times x\\\\W = 1250 \pi x\;dx[/tex]
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)
[tex]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[/tex]
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:
https://brainly.com/question/22599382
