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Chapter 10: Problem 51

A school bus that is used on a long-distance route has a frontal area of \(9 \mathrm{~m}^{2}\) and an estimated drag coefficient of \(0.85 .\) The bus regularly travels on the highway at the speed limit of \(90 \mathrm{~km} / \mathrm{h}\). A design modification of the shape of the bus can reduce the drag coefficient to \(0.75 .\) Determine the percentage reduction in the engine power that the bus will need to use to maintain a speed of \(90 \mathrm{~km} / \mathrm{h}\).

Short Answer

Expert verified
The percentage reduction in engine power required is approximately 11.73%.

Step by step solution

01

Convert Speed Units

The speed of the bus is given in kilometers per hour (km/h), but for aerodynamic calculations, it is generally more convenient to work in meters per second (m/s). Use the conversion: \[ 1 \, \text{km/h} = \frac{1000}{3600} \, \text{m/s} \] Thus, the speed of the bus is:\[ 90 \, \text{km/h} \times \frac{1000}{3600} = 25 \, \text{m/s} \]
02

Calculate the Initial Drag Force

The drag force experienced by the bus is calculated using:\[ F_{\text{drag}} = \frac{1}{2} \cdot \rho \cdot C_d \cdot A \cdot v^2 \]where:- \(\rho \) is the air density (assumed to be \(1.225 \, \text{kg/m}^3\)).- \(C_d = 0.85\) is the current drag coefficient.- \(A = 9 \, \text{m}^2\) is the frontal area.- \(v = 25 \, \text{m/s}\) is the bus speed.Substitute the values:\[ F_{\text{drag, initial}} = \frac{1}{2} \times 1.225 \times 0.85 \times 9 \times 25^2 \approx 1170.56 \, \text{N} \]
03

Calculate the Modified Drag Force

Repeat the drag force calculation with the modified drag coefficient:\[ C_d = 0.75 \]\[ F_{\text{drag, modified}} = \frac{1}{2} \times 1.225 \times 0.75 \times 9 \times 25^2 \approx 1033.12 \, \text{N} \]
04

Calculate the Initial and Modified Power Requirements

The power to overcome aerodynamic drag is given by:\[ P = F_{\text{drag}} \times v \]For the initial scenario:\[ P_{\text{initial}} = 1170.56 \times 25 \approx 29264 \, \text{W} \]For the modified scenario:\[ P_{\text{modified}} = 1033.12 \times 25 \approx 25828 \, \text{W} \]
05

Calculate the Percentage Reduction in Power

The percentage reduction in power is calculated as:\[ \text{Percentage Reduction} = \left(\frac{P_{\text{initial}} - P_{\text{modified}}}{P_{\text{initial}}}\right) \times 100\% \]Substitute the power values:\[ \text{Percentage Reduction} = \left(\frac{29264 - 25828}{29264}\right) \times 100\% \approx 11.73\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Drag Coefficient
In aerodynamics, the drag coefficient (\(C_d\)) is a measure of an object's resistance to air flow. It plays a crucial role in determining how easily a vehicle can move through the air.
  • A lower drag coefficient means less resistance and usually higher efficiency when moving through the air.
  • This coefficient is influenced by the shape of the vehicle, its surface roughness, and how air flows over it.
Considering our school bus from the exercise, the initial drag coefficient was 0.85. By altering the shape of the bus, it's possible to reduce this coefficient to 0.75.
Reducing the drag coefficient directly impacts energy consumption because the bus will face less aerodynamic drag, thereby lowering the power requirement for maintaining speed. In this context, even small changes to the drag coefficient can result in significant improvements in fuel efficiency.
Calculating Power Requirement for Vehicles
Power requirement refers to the amount of energy needed to keep a vehicle moving at a particular speed, especially when combating aerodynamic forces.The equation to calculate this power due to drag is:\[ P = F_{\text{drag}} \times v \]Where:
  • \(F_{\text{drag}}\) is the drag force experienced by the vehicle.
  • \(v\) is the velocity of the vehicle.
In our example, initially, the bus had a power requirement of approximately 29264 Watts due to a drag force of around 1170.56 Newtons. With better aerodynamics and a reduced drag coefficient, this requirement dropped to about 25828 Watts.
Reducing the power requirement is crucial for energy conservation and can lead to economic benefits as well, such as lower fuel costs.
The Role of Frontal Area in Aerodynamics
Frontal area (\(A\)) is the portion of a vehicle that is directly facing forward, exposed to oncoming air. In the context of aerodynamics, a larger frontal area usually increases air resistance.
  • It is measured in square meters (m虏).
  • It directly affects the drag force; a greater area leads to higher drag.
For the school bus example, the frontal area is 9 m虏. This area greatly affects the overall drag force.
Therefore, for aerodynamic efficiency, it's often advantageous to minimize the frontal area as much as possible without compromising the vehicle's functionality or safety.
Streamlining the shape and reducing unnecessary protrusions are common strategies to lower the effective frontal area.
Aerodynamics in Transportation
Aerodynamics is the study of how air interacts with moving objects. In transportation, understanding these interactions allows for the design of more energy-efficient and safer vehicles.
  • Good aerodynamics lead to vehicles that have reduced air resistance, translating to lower fuel or energy consumption.
  • Improved aerodynamic design contributes to stability at high speeds and reduces noise levels.
In practical terms, the role of aerodynamics encompasses everything from the shape of a car to the surface texture of aircraft hulls. In the exercise, by focusing on reducing the drag coefficient through better aerodynamics, the bus achieves a notable reduction in energy needs.
By improving aerodynamics, transportation systems can significantly decrease their environmental footprint while also cutting costs, crucial benefits for long-distance vehicles.

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Most popular questions from this chapter

A baseball with a mass of \(145 \mathrm{~g}\) and a diameter of \(71.6 \mathrm{~mm}\) is dropped from a height of \(1 \mathrm{~km}\) in a standard atmosphere. (a) Estimate the terminal velocity attained by the baseball, assuming that atmospheric conditions remain constant and equal to those at the release height. (b) Determine the time it takes the baseball to attain \(90 \%\) of its terminal velocity. (c) Determine the distance traveled for the baseball to attain \(90 \%\) of its terminal velocity. Based on your results, comment on whether the assumption of constant atmospheric conditions is reasonable.

A prototype sports car has an engine that can deliver \(360 \mathrm{~kW}\) of power. The shape of the car is such that is has an estimated drag coefficient of 0.20 and a frontal area of \(2.50 \mathrm{~m}^{2}\). If the car is to be tested on a track at sea level under standard conditions, estimate the maximum possible speed the car can attain.

A small military helicopter has four blades with a rotor diameter of \(10.70 \mathrm{~m},\) and each blade has a width of \(0.73 \mathrm{~m}\). When the helicopter is flying at normal speed, the blades rotate at \(360 \mathrm{rpm} .\) The standard engine on the helicopter has a rated power of \(606 \mathrm{~kW}\). Assuming that the blades can be treated as flat plates for the purpose of estimating the frictional force that must be overcome in turning the blades, estimate the power required to turn the blades. What percentage of the engine power is used to turn the blades? Assume standard air in your analysis.

Experiments on a car in a wind tunnel indicate that the drag coefficient of the car with all the windows up is 0.35 and that the drag coefficient with all the windows down is 0.45 . The car has a frontal area of \(3.0 \mathrm{~m}^{2}\). Consider the case where the car is driven at \(90 \mathrm{~km} / \mathrm{h}\) for a distance of \(205 \mathrm{~km}\), gasoline costs \(\$ 0.6604 / \mathrm{L}\), and the available energy from gasoline for overcoming aerodynamic drag is \(4.50 \mathrm{MJ} / \mathrm{kg}\). Estimate the additional gasoline cost incurred as a result of driving with the windows down. Assume standard air.

Consider the general case in which a body of mass \(m\) and frontal area \(A\) free-falls in an environment where the effective gravity is \(g^{\prime},\) the density is \(\rho,\) and the drag coefficient of the body is constant and equal to \(C_{\mathrm{D}}\). Show that the time interval, \(\Delta t,\) for the velocity to change from \(V_{1}\) to \(V_{2}\) is given by $$ \Delta t=\frac{m}{2 \sqrt{a b}}\left[\ln \left|\frac{\sqrt{a}+\sqrt{b} V}{\sqrt{a}-\sqrt{b} V}\right|\right]_{V_{1}}^{V_{2}}, \quad \text { where } \quad a=m g^{\prime} \quad \text { and } \quad b=\frac{1}{2} \rho A C_{\mathrm{D}} $$ Explain how this formula could be used to calculate the time it takes a body dropped in the atmosphere to attain a speed equal to \(90 \%\) of its terminal speed.
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