Question

Consider a spherical tank full of water with radius 3 m (plus a spout on top 1m high). Set up an integral expression for the work required to pump all of the water out of the spout. You can assume the density of water is 1000 kg/m3 , and g ≈ 9.81 m/s2 is gravitational acceleration.

Answer 1

Answer:

The answer to the question is

4433.416 kJ

See explanation below

(3-y)²+r² = 3² or

6y-y² = r²

r =√(6y-y²)

The volume of a small section of height Δy =  Δy ×(√(6y-y²))²×π

For water with density of 1000 kg/m³, the mass of the slice

= 1000×Δy ×(√(6y-y²))²×π and since force = mass × acceleration we have

1000×Δy ×(√(6y-y²))²×π ×g = 1000×Δy ×(√(6y-y²))²×π ×9.81

The work done to move a unit height of y+1 = Force × Distance

W  = 1000×Δy ×(√(6y-y²))²×π ×9.81 × (y+1)

Integrating the entire section of the sphere that is 2×r high, or from 0 to 6 we get

W =$\int\limits^6_0 {1000*(6y-y^{2} ) *\pi *9.81 * (y+1)} \, dy$

= $9810*\pi \int\limits^6_0 {5y^{2} -y^{3} +6y \, dy$

$= 9810*\pi *[\frac{5y^{3} }{3} -\frac{y^{4} }{4} +3y^{2} ]^{6} _{0}$

=9810×π×144 =4433416 J