Problem 30 - Thomas Calculus in SI Units

Question

Your town has decided to drill a well to increase its water supply. As the town engineer, you have determined that a water tower will be necessary to provide the pressure needed for distribution, and you have designed the system shown here. The water is to be pumped from a \(90-\mathrm{m}\) well through a vertical \(10 \mathrm{cm}\) pipe into the base of a cylindrical tank \(6 \mathrm{m}\) in diameter and \(7.5 \mathrm{m}\) high. The base of the tank will be \(18 \mathrm{m}\) above ground. The pump is a 3 -hp pump, rated at \(2000 \mathrm{W}\) ( \(\mathrm{J} / \mathrm{s}\) ). To the nearest hour, how long will it take to fill the tank the first time? (Include the time it takes to fill the pipe.) Assume that water weighs \(9800 \mathrm{N} / \mathrm{m}^{3}\).

Step-by-Step Solution

Answer

It will take approximately 3 hours to fill the tank.

Step 1: Calculate the Volume of the Cylindrical Tank

The volume of a cylinder is determined by the formula \( V = \pi r^2 h \). The diameter of the tank is 6 m, so the radius \( r \) is 3 m. The height \( h \) is 7.5 m. Substitute these values to find the volume: \( V = \pi (3)^2 (7.5) = 67.5\pi \approx 212.06 \text{ m}^3 \).

Step 2: Calculate the Total Mass of Water in the Tank

The weight density of water is given as 9800 N/m\(^3\). To find the total mass \( m \) of the water, we use \( m = \frac{Weight}{g} = \frac{9800 \times 212.06}{9.8} \approx 21206 \text{ kg} \), where \( g = 9.8 \text{ m/s}^2 \).

Step 3: Calculate the Work Done to Lift Water into the Tank

The work done \( W \) is given by \( W = mgh \), where \( h \) is the total elevation gain from the bottom of the well to the tank top (90 m + 18 m + 7.5 m = 115.5 m). Thus, \( W = 21206 \times 9.8 \times 115.5 \approx 2.392 \times 10^7 \text{ J} \).

Step 4: Calculate the Time to Fill the Tank

The power of the pump is rated at 2000 W, which implies that it does 2000 J of work per second. The time \( t \) taken to fill the tank is given by \( t = \frac{W}{P} \), where \( P = 2000 \text{ W} \). Substitute to find \( t = \frac{2.392 \times 10^7}{2000} \approx 11960 \text{ seconds} \).

Step 5: Convert Time from Seconds to Hours

There are 3600 seconds in an hour. Divide the total seconds by 3600 to convert to hours: \( 11960 \div 3600 \approx 3.32 \text{ hours} \). Hence, to the nearest hour, it will take 3 hours to fill the tank.

Key Concepts

Volume of a Cylinder

The volume of a cylinder is an important concept in physics and engineering, especially when dealing with problems involving storage tanks, such as a water tower. The formula to calculate the volume of a cylinder is given by: \[ V = \pi r^2 h \] where:

• \( V \) is the volume,
• \( r \) is the radius of the base of the cylinder, and
• \( h \) is the height of the cylinder.

In the context of a water tank, knowing the volume is essential for determining how much water it can hold. For instance, if the diameter of the tank is 6 meters, the radius will be half of that, which is 3 meters. The height of the tank is specified as 7.5 meters. Substituting these into the formula gives us the volume of the water tank:\[ V = \pi (3)^2 (7.5) = 67.5\pi \approx 212.06 \text{ m}^3\] Understanding and calculating the volume correctly is crucial for determining the amount of water that can be stored, which impacts further calculations such as the time required to fill the tank.

Weight Density of Water

The weight density of water is another key concept in physics, particularly when calculating the forces in fluid dynamics. Weight density is defined as the weight per unit volume, and for water, it is often given as 9800 \( \text{N/m}^3 \). This value is essential for determining other physical parameters, like the mass of water in a container.To find the mass of the water in a tank, one can use the relationship between weight and mass, which is governed by gravity. The formula is:\[ m = \frac{\text{Weight}}{g} \] where:

• \( m \) is the mass,
• \( \text{Weight} \) is the weight of the water, and
• \( g \) is the acceleration due to gravity, approximately 9.8 \( \text{m/s}^2 \).

Applying this to our example, if the water in the tank has a volume of 212.06 cubic meters, then its total weight is 9800 times that volume. Dividing this weight by the acceleration due to gravity will give us the mass of the water:\[ m = \frac{9800 \times 212.06}{9.8} \approx 21206 \text{ kg}\] This calculation is vital for determining how much force is required to transport the water to the tower's height.

Power and Work in Physics

Power and work are fundamental concepts in physics that are interrelated. Understanding them is crucial when dealing with systems like pumps that move water. **Work**Work (\( W \)) is defined as the energy transferred by a force acting over a distance. It's calculated using the formula:\[ W = mgh \] where:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity, and
• \( h \) is the height the object is lifted.

In the water tank example, the energy or work needed to lift water from the well to the tank involves moving the mass of water a total elevation gain. This gain is the sum of the height of the well, the height of the tank above the ground, and the height of water within the tank, giving a total height of 115.5 meters.**Power**Power (\( P \)) measures the rate at which work is done. It's expressed in watts (W), where one watt is equivalent to one joule per second. The relationship is given by:\[ t = \frac{W}{P} \] where:

• \( t \) is the time taken,
• \( W \) is the work done, and
• \( P \) is the power.

For a pump rated at 2000 watts, this means it can perform 2000 joules of work per second. To find the time required to fill the tank, calculate the total work involved and divide by the pump's power:\[ t = \frac{2.392 \times 10^7}{2000} \approx 11960 \text{ seconds}\] Converting seconds to hours shows it takes approximately 3.32 hours, which rounds to 3 hours. Understanding these concepts helps explain how energy and time are related and guides the practical aspect of running equipment like pumps efficiently.