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The drag force, \(F_{\mathrm{d}}\), imposed by the surrounding air on a vehicle moving with velocity \(\mathrm{V}\) is given by $$ F_{\mathrm{d}}=C_{\mathrm{d}} \mathrm{A}_{2}^{\frac{1}{2}} \rho \mathrm{V}^{2} $$ where \(C_{\mathrm{d}}\) is a constant called the drag coefficient, \(\mathrm{A}\) is the projected frontal area of the vehicle, and \(\rho\) is the air density. Determine the power, in \(\mathrm{kW}\), required to overcome aerodynamic drag for a truck moving at \(110 \mathrm{~km} / \mathrm{h}\), if \(C_{\mathrm{d}}=0.65, \mathrm{~A}=10 \mathrm{~m}^{2}\) and \(\rho=1.1 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
206.14 kW

Step by step solution

01

- Convert Velocity to Meters per Second

The given velocity is in \ (110 \ \mathrm{km/h}). Convert it to meters per second using the formula: \[ \mathrm{V} = 110 \ \mathrm{km/h} \ \times \ \frac{1000 \ \mathrm{m}}{1 \ \mathrm{km}} \ \times \ \frac{1 \ \mathrm{h}}{3600 \ \mathrm{s}} = 30.56 \ \mathrm{m/s} \] \
02

- Calculate Drag Force

Use the formula for drag force: \[ F_{\text{d}} = C_{\text{d}} \ \times \ \/\mathrm{A}^{1/2} \ \times \ \rho \ \times \ V^{2} \ \] \ Substitute the values: \[ F_{\text{d}} = 0.65 \ \times \ 10 \ \mathrm{m}^{2} \ \times \ 1.1 \ \mathrm{kg}/ \ \mathrm{m}^{3} \ \times \ (30.56 \ \mathrm{m/s})^{2} = 6745 \ \mathrm{N} \] \
03

- Calculate the Power Required in Watts

The power required to overcome the drag force is given by: \[ P = F_{\text{d}} \ \times \ V \] \ Substitute the values: \[ P = 6745 \ \mathrm{N} \ \times \ 30.56 \ \mathrm{m/s} = 206144.2 \ \mathrm{W} \]
04

- Convert Power to Kilowatts

We need the power in kilowatts: \[ \ \mathrm{P} = \ \frac{206144.2 \ \mathrm{W}}{1000} = 206.14 \ \mathrm{kW} \]

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Key Concepts

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

Drag Force
The drag force is a crucial concept in aerodynamics, representing the resistance a vehicle encounters as it moves through air. This force directly opposes the vehicle's motion. Drag force can be calculated with the formula: \[ F_{\text{d}} = C_{\text{d}} \times A \times \rho \times V^2 \]- **Drag Coefficient,** \( C_{\text{d}} \) is a dimensionless constant that depends on the shape of the object.- **Frontal Area,** \( A \) is the cross-sectional area of the vehicle that faces the air flow.- **Air Density,** \( \rho \) usually given in \( \text{kg/m}^3 \) depends on the altitude, humidity, and temperature of the environment.- **Velocity,** \( V \) is the speed of the vehicle through the air.Drag force increases with the square of the velocity and directly impacts factors such as fuel efficiency and maximum speed of the vehicle.
Drag Coefficient
The drag coefficient, denoted as \( C_{\text{d}} \), plays a significant role in determining the drag force on a vehicle. It is a dimensionless number that reflects the aerodynamic efficiency of the vehicle's shape. - Streamlined designs have lower drag coefficients, resulting in reduced drag forces.- Typical values for \( C_{\text{d}} \) range from 0.25 for a very aerodynamic car to 1.0 or higher for bluff objects like trucks.For a truck, a typical \( C_{\text{d}} \) might be around 0.65, as used in the given exercise. Reducing the drag coefficient via design improvements can substantially enhance a vehicle's performance and fuel efficiency.
Velocity to Meters per Second Conversion
Speed is often given in kilometers per hour (km/h), but for aerodynamic calculations, it is essential to convert it to meters per second (m/s). This ensures consistency within the metric system:\[ \text{Velocity} = 110 \times \frac{1000}{1} \times \frac{1}{3600} = 30.56 \text{ m/s} \]Here's the conversion process:- **Step 1:** Multiply by 1000 to convert kilometers to meters.- **Step 2:** Divide by 3600 to convert hours to seconds.Thus, a velocity of 110 km/h is equivalent to approximately 30.56 m/s. Correct units are critical for accurate calculations in physics and engineering disciplines.
Power Calculation
To determine the power needed to overcome aerodynamic drag, follow these steps:- First, find the drag force using the formula provided and the values for \( C_{\text{d}} \), \( A \), \( \rho \), and \( V \):\[ F_{\text{d}} = 0.65 \times 10 \times 1.1 \times (30.56)^2 = 6745 \text{ N} \]- Next, calculate the power, which is the product of the drag force and the vehicle's velocity:\[ P = F_{\text{d}} \times V = 6745 \text{ N} \times 30.56 \text{ m/s} = 206144.2 \text{ W} \]- Finally, to convert from watts to kilowatts:\[ P = \frac{206144.2}{1000} = 206.14 \text{ kW} \]This method helps determine the energy requirements for overcoming aerodynamic drag at a given speed. Knowing these values is crucial for vehicle design and energy efficiency optimization.

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Most popular questions from this chapter

A heat pump cycle whose coefficient of performance is \(2.5\) delivers energy by heat transfer to a dwelling at a rate of \(20 \mathrm{~kW}\). (a) Determine the net power required to operate the heat pump, in \(\mathrm{kW}\). (b) Evaluating electricity at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), determine the cost of electricity in a month when the heat pump operates for 200 hours.

A gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of \(1000 \mathrm{~W}\). Heat transfer between the gas and the surroundings occurs at a rate of \(\dot{Q}=-50 t\), where \(\dot{Q}\) is in watts, and \(t\) is time, in min. (a) Plot the time rate of change of energy of the gas for \(0 \leq t \leq 20 \mathrm{~min}\), in watts. (b) Determine the net change in energy of the gas after 20 min, in kJ. (c) If electricity is valued at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), what is the cost of the electrical input to the resistor for \(20 \mathrm{~min}\) of operation?

A wire of cross-sectional area \(A\) and initial length \(x_{0}\) is stretched. The normal stress \(\sigma\) acting in the wire varies linearly with strain, \(\varepsilon\), where $$ \varepsilon=\left(x-x_{0}\right) / x_{0} $$ and \(x\) is the length of the wire. Assuming the cross-sectional area remains constant, derive an expression for the work done on the wire as a function of strain.

An electric motor draws a current of 10 amp with a voltage of \(110 \mathrm{~V}\). The output shaft develops a torque of \(10.2 \mathrm{~N} \cdot \mathrm{m}\) and a rotational speed of 1000 RPM. For operation at steady state, determine (a) the electric power required by the motor and the power developed by the output shaft, each in \(\mathrm{kW}\). (b) the net power input to the motor, in \(\mathrm{kW}\). (c) the amount of energy transferred to the motor by electrical work and the amount of energy transferred out of the motor by the shaft, in \(\mathrm{kW} \cdot \mathrm{h}\) during \(2 \mathrm{~h}\) of operation.

A disk-shaped flywheel, of uniform density \(\rho\), outer radius \(R\), and thickness \(w\), rotates with an angular velocity \(\omega\), in \(\mathrm{rad} / \mathrm{s}\). (a) Show that the moment of inertia, \(I=\int_{\text {vol }} \rho r^{2} d V\), can be expressed as \(I=\pi \rho w R^{4} / 2\) and the kinetic energy can be expressed as \(\mathrm{KE}=I \omega^{2} / 2\). (b) For a steel flywheel rotating at 3000 RPM, determine the kinetic energy, in \(\mathrm{N} \cdot \mathrm{m}\), and the mass, in \(\mathrm{kg}\), if \(R=0.38 \mathrm{~m}\) and \(w=0.025 \mathrm{~m}\). (c) Determine the radius, in \(\mathrm{m}\), and the mass, in \(\mathrm{kg}\), of an aluminum flywheel having the same width, angular velocity, and kinetic energy as in part (b).

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