Problem 36 - Thomas Calculus

Question

Water tower 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 300 \(\mathrm{ft}\) well through a vertical 4 in. pipe into the base of a cylindrical tank 20 \(\mathrm{ft}\) in diameter and 25 \(\mathrm{ft}\) high. The base of the tank will be 60 \(\mathrm{ft}\) aboveground. The pump is a 3 hp pump, rated at \(1650 \mathrm{ft} \cdot 1 \mathrm{b} / \mathrm{sec} .\) 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 \(62.4 \mathrm{lb} / \mathrm{ft}^{3} .\)

Step-by-Step Solution

Answer

92 hours.

Step 1: Calculate the Volume of the Tank

The tank is cylindrical, so we use the volume formula for a cylinder: \[ V = \pi r^2 h \]where \( r \) is the radius and \( h \) is the height. For this tank:- Diameter \( = 20 \text{ ft} \) implies radius \( r = 10 \text{ ft} \)- Height \( = 25 \text{ ft} \)Substitute these values into the formula:\[ V = \pi (10)^2 (25) = 2500\pi \text{ cubic feet}\]

Step 2: Calculate the Volume of the Pipe

The volume of a cylindrical pipe (excluding the bottom of the well) can be calculated using the same cylinder volume formula. The pipe is 300 feet and fills from the bottom:- Radius of the pipe \( = 2 \text{ in} \) = \( \frac{1}{6} \text{ ft} \)- Height \( = 300 \text{ ft} \)Substitute into the formula:\[ V = \pi \left(\frac{1}{6}\right)^2 (300) = \frac{300\pi}{36} = \frac{25\pi}{3} \text{ cubic feet}\]

Step 3: Calculate the Total Volume to be Filled

Add the volume of the tank and the volume of the pipe to find the total volume that needs to be filled:\[ \text{Total Volume} = 2500\pi + \frac{25\pi}{3} \]First, convert \( \frac{25\pi}{3} \) to a compatible form:\[ \frac{25\pi}{3} = \frac{75\pi}{9} \text{ cubic feet} \]Total volume:\[ \text{Total Volume} = 2500\pi + \frac{25\pi}{3} = \left(2500 + \frac{25}{3}\right)\pi \approx 2521\pi \text{ cubic feet}\]

Step 4: Use Pump Work Rate to Find Time

The pump is rated at 1650 ft-lb/sec. First, calculate the weight of the water using the weight per volume:\[ \text{Weight of Water} = 2521\pi \times 62.4 \text{ lb/ft}^3 \]Weight of the water:\[ \approx 157108\pi \text{ lb}\]Use the pump’s work rate in ft-lb/sec to find the time:\[ \text{Time in seconds} = \frac{\text{Total Weight}}{\text{Pump rate}} = \frac{157108\pi}{1650} \text{ seconds} \]Convert seconds to hours:\[ \text{Time in hours} = \frac{157108\pi}{1650 \times 3600} \approx 92.4 \text{ hours} \]

Step 5: Round the Time to the Nearest Hour

Rounding 92.4 to the nearest hour gives us 92 hours.

Key Concepts

Cylindrical Volume Calculation

Understanding the volume of a cylinder is crucial in many real-world scenarios, including calculating how much liquid a tank can hold. The formula used to find the volume of a cylindrical shape is: \[ V = \pi r^2 h \]where:

• \(V\) is the volume,
• \(\pi\) is a constant approximately equal to 3.14159,
• \(r\) is the radius, and
• \(h\) is the height of the cylinder.

In the water tower problem, we needed to calculate the volume of both the large tank and the pipes connecting to it. By substituting the dimensions of the tank into the formula, we get \( V = \pi (10)^2 (25) = 2500\pi \) cubic feet. This calculation tells us that the tank can hold about 2500\(\pi\) cubic feet of water.

Pump Work Rate

In this scenario, understanding the pump's work rate is key to solving how long it will take to fill the water tank. The pump here is rated at 1650 ft-lb/sec, meaning that it can move a certain amount of water weight within a given time frame.To calculate how long it takes the pump to fill the tank, we need to find out how much weight the pump will lift per unit time. With water weighing 62.4 lb/ft³, the total weight of water that needs moving is primarily dictated by both the total volume and the weight density of the water itself. Therefore, using the formula:\[ \text{Time in seconds} = \frac{\text{Total Weight}}{\text{Pump Rate}} \]Helps us find the time it takes for a specific pump to fill the water reservoir to its maximum capacity.

Fluid Mechanics

Fluid mechanics plays a significant role in understanding how water flows through the system, in this case, through a cylindrical pipe into a large tank. It involves the study of forces and motions in fluids, explaining how the water, driven by the pump, travels upward against gravity. The study of fluid mechanics considers factors like:

• The fluid's velocity,
• Viscosity (internal resistance of the fluid), and
• Pressure differences.

In engineering, understanding these concepts ensures that systems are efficiently designed, allowing for effective water distribution even over different elevations, such as using a pump to elevate water from the ground level to the top of a water tower.

Engineering Application

The engineering application of this water tower problem is a practical demonstration of several principles coming together to solve a community-based issue. Engineers combine principles from mathematics, physics, and engineering design to ensure that necessary infrastructure like this operates effectively. Considerations in this problem include:

• The design of the water tower to support its weight and height,
• The efficiency and capability of the pump to adequately fill the tank,
• Ensuring the steady supply of water to meet town needs.

Overall, this exercise underlines the importance of planning, proper system design, and maintenance to ensure that community water needs are met efficiently and sustainably. It showcases how interdisciplinary knowledge is crucial for practical problem-solving in real-life engineering contexts.