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