Find the work required to pump the water out of the spout.

Question

A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft³. (Assume r = 4 ft, R = 8 ft, and h = 12 ft.)

Answer 1

To find the work required to pump the water out of the spout, you can calculate the potential energy of the water in the tank and then convert it into work. The potential energy of an object is given by the formula:

Potential Energy (PE) = mass (m) * gravitational acceleration (g) * height (h)

Given:

- Water density (ρ) = 62.5 lb/ft³

- Inner radius of the tank (r) = 4 ft

- Outer radius of the tank (R) = 8 ft

- Height of the water column (h) = 12 ft

- Acceleration due to gravity (g) = 32.2 ft/s² (approximately)

First, let's calculate the volume of water in the tank. To do this, we'll find the volume of the larger cylinder (with radius R) and subtract the volume of the smaller cylinder (with radius r) that is not filled with water.

Volume of the larger cylinder = π * R² * h

Volume of the smaller cylinder = π * r² * h

Now, calculate the volume of water in the tank:

Volume of water = Volume of the larger cylinder - Volume of the smaller cylinder

Volume of water = π * (8 ft)² * 12 ft - π * (4 ft)² * 12 ft

Next, calculate the mass of the water by multiplying the volume by the water density:

Mass of water (m) = Volume of water * Water density

Mass of water = [π * (8 ft)² * 12 ft - π * (4 ft)² * 12 ft] * 62.5 lb/ft³

Now, calculate the potential energy (work done) required to pump the water out of the spout:

Potential Energy (PE) = Mass of water * Gravitational acceleration * Height

Potential Energy = [π * (8 ft)² * 12 ft - π * (4 ft)² * 12 ft] * 62.5 lb/ft³ * 32.2 ft/s² * 12 ft

Calculate this expression to find the potential energy in foot-pounds. This represents the work required to pump the water out of the spout.