Respuesta :
Answer:
Explanation:
Work in pumping water from the tank is given as
W = ∫ y dF. From a to b
Where dF is the differential weight of the thin layer of liquid in the tank, y is the height of the differential layer
a is the lower limit of the height
b is the upper limit of the height.
We know that, .
F = ρVg
Where F is the weight
ρ is the density of water
V is the volume of water in tank
g is the acceleration due to gravity
Then,
dF = ρg ( Ady)
We know that the density and the acceleration due to gravity is constant, also the base area of the tank is constant, only the height that changes.
Then,
ρg = 62.4 lbs/ft³
Area = L×B = 3 × 9 = 27ft²
dF = ρg ( Ady)
dF = 1684.8dy
The height reduces from 12ft to 0ft
Then,
W = ∫ y dF. From a to b
W = ∫ 1684.8y dy From 0 to 12
W = 1684.8y²/2 from 0 to 12
W = 842.4 [y²] from y = 0 to y = 12
W = 842.4 (12²-0²)
W = 121,305.6 lb-ft
2 .095 × 10⁶ J of work will be done by pumping all of the liquid out over the top of the tank.
Calculating the work required:
Let us denote the dimension of the tank as:
depth h = 12 feet
length l = 3 feet
width b = 9 feet
As we pump out the liquid from the top of the tank, the height of the surface level of the liquid in the tank decreases. So we will have to calculate the amount of work required to pump out the liquid for a small amount of decrease in the height of the surface level, then integrate it over the depth of the tank.
The density of the liquid is ρ = 110 pounds/ foot³
Area of the surface, A = l×b = 3×9 ⇒ A = 27 foot²
So the gravitational force or the weight of the liquid in a small depth element dh will be :
dF = ρgAdh ( where ρAdh gives the mass contained in dh depth)
The work done is given by:
dW = hdF = ρgAhdh
dW = 110×9.8×27×hdh
dW = 29106hdh
[tex]W=29106\int\limits^{12}_0 {h} \, dh\\\\W=29106[h^2/2]_0^{12}\\\\[/tex]
W = 2 .095 × 10⁶ J
Learn more about work done:
https://brainly.com/question/16976412?referrer=searchResults
