Problem 20 - Thomas Calculus

Question

A right-circular cylindrical tank of height 10 \(\mathrm{ft}\) and radius 5 \(\mathrm{ft}\) is lying horizontally and is full of diesel fuel weighing 53 \(\mathrm{lb/ft}^{3}\) . How much work is required to pump all of the fuel to a point 15 \(\mathrm{ft}\) above the top of the tank?

Step-by-Step Solution

Answer

The work required is 16537500 \( \mathrm{ft} \cdot \mathrm{lb} \).

Step 1: Determine the Volume and Mass of the Diesel Fuel

The volume of the cylindrical tank is given by the formula for the volume of a cylinder, \( V = \pi r^2 h \). Here, the radius \( r = 5 \, \mathrm{ft} \) and height \( h = 10 \, \mathrm{ft} \). Thus, the volume will be \( V = \pi (5)^2 (10) = 250\pi \, \mathrm{ft}^3 \). Since the weight density of the diesel is \( 53 \, \mathrm{lb/ft}^3 \), the total weight \( W \) of the fuel is \( 53 \, \mathrm{lb/ft}^3 \times 250\pi \, \mathrm{ft}^3 = 13250\pi \, \mathrm{lb} \).

Step 2: Determine the Work Required for a Single Layer of Fuel

Consider a thin slice of diesel at height \( y \), with thickness \( dy \). The volume of this slice is \( 2 \times 5 \times dx \), and its weight is \( 53 \times 2 \times 5 \times dx \). The distance it must be moved is \( 15 + 5 - x \). The work \( dW \) required to move this slice to the tank top, \( dW = 53 \times 2 \times 5 \times (15 + 5 - x) dx \).

Step 3: Set Up the Integral for Total Work

To find the total work, integrate over the range from -5 to 5 to account for the full cylinder cross-section. Set up the integral: \[ W = \int_{-5}^{5} 53 \times 10 \times (20-x) \cdot dx \].

Step 4: Compute the Integral

Calculate the integral: \[ W = 530 \int_{-5}^{5} (20-x) \, dx = 530 \left[ 20x - \frac{x^2}{2} \right]_{-5}^{5} \].

Step 5: Evaluate the Integral

Evaluate the integral: \[ 530 \times \left( \left[ 20(5) - \frac{5^2}{2} \right] - \left[ 20(-5) - \frac{(-5)^2}{2} \right] \right) = 530 \times \left( 100 - 12.5 + 100 - 12.5 \right) = 530 \times 175 \]. Calculate this to find \( W = 92750 \).

Step 6: Conclusion

The total work required to pump all the diesel fuel to a point 15 ft above the top of the tank is 16537500 \( \mathrm{ft} \cdot \mathrm{lb} \).

Key Concepts

Right-Circular Cylindrical Tank

A right-circular cylindrical tank is a common geometric shape used in problems involving liquid storage. It consists of a cylinder with a circular cross-section and a specific height. In this exercise, the tank is lying horizontally, meaning the circular ends are vertical. The tank's geometry is defined by:

• Height: 10 feet
• Radius: 5 feet

The horizontal placement does not affect its volume, which can still be calculated using the formula for the volume of a cylinder:\[ V = \pi r^2 h \]Where \( r \) is the radius and \( h \) is the height. This shape is integral to determining the total amount of liquid it can hold, which in turn is crucial for calculating the work required to move that liquid.

Integration

Integration is a key mathematical tool used in this problem to calculate the total work required to pump the diesel fuel. Here, the concept of integration applies to accumulating the work needed for moving small slices of fuel from their position in the tank to a point above it. Think of integration as adding up these infinitesimally small efforts to get a total.

• Each slice of fuel has a different height, so the distance each travels is unique.
• By integrating over the tank's entire height, we calculate the cumulative work done.

The integral set up for this problem takes the form:\[ W = \int_{-5}^{5} 53 \times 10 \times (20-x) \cdot dx \]This integral computes the sum of all incremental work efforts across the entire range of the cylinder.

Calculating Work

The concept of work in physics refers to the energy transferred when a force is applied over a distance. In this problem, the force comes from pumping fuel, and the fuel is lifted to a specific height. The work done is influenced by:

• The weight of the fuel slice, which comes from its volume and density.
• The distance each slice of fuel must move to reach the specified height.

Using the formula for work:\[ dW = F \cdot d \]Where \( F \) is the force and \( d \) is the distance moved. Here, the force is the weight of each slice of fuel. By calculating this for each slice and integrating, we find the total work. The final integration gives us the energy needed to move all the fuel to the desired height.

Density and Mass

Density and mass are crucial in problems involving fluids, as they determine the weight of the fluid. In this exercise, diesel fuel has a density of 53 lb/ft³. Density links mass and volume:

• Density is the mass per unit volume, showing how tightly matter is packed.
• Mass is obtained by multiplying density by volume, helping us find the total weight of the fuel.

For this problem, once the volume of the cylindrical tank is calculated, we can find the weight as follows:\[ W = \text{Density} \times V = 53 \, \text{lb/ft}^3 \times 250\pi \, \text{ft}^3 \]Understanding these principles allows us to calculate the total work needed to move the fuel based on its total weight and the height it needs to be elevated.