A spherical tank with radius 4 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top.

Question

Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)

Answer 1

Calculate the increase in the potential energy of the liquid. The height of the liquid increases from the location of its center of mass to 9 meters. The center of mass of the lquid initially ( filling a hemisphere) is 5/8 r heigh.

Answer 2

To find the weight of the liquid in the tank, we first need to find the volume of liquid in the tank. Since the tank is half full, the volume of liquid is half the volume of the sphere. The volume of a spherical cap is given by the formula V = πh^2(3R - h)/3, where R is the radius of the sphere (4 m), and h is the height of the liquid in the tank.

To find h, we need to consider the geometry of the situation. Since the tank is half full, the liquid level is at a distance of 2 m from the bottom of the tank. Therefore, the height of the liquid h is R - 2 m = 4 m - 2 m = 2 m.

Plugging in the values, we get the volume of liquid:

V = π(2)^2(3(4) - 2)/3

V = π(4)(10)/3

V = 40π/3 ≈ 41.89 m^3

Next, we can find the weight of the liquid by multiplying the volume by the density:

Weight = Volume x Density

Weight = 41.89 m^3 x 900 kg/m^3

Weight = 37,701 kg

Now, considering the spout at the top, there will be a pressure difference at the base of the tank due to the liquid column in the spout. This is represented by the formula: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the liquid, g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the liquid column in the spout (1 m).

Substitute the values:

ΔP = 900 kg/m^3 x 9.81 m/s^2 x 1 m

ΔP = 8829 Pa

Therefore, the pressure at the base of the tank will be the atmospheric pressure plus the pressure due to the liquid column in the spout.