Question

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

206.14 kW

Step-by-step solution

- Convert Velocity to Meters per Second

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

- 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: \[ F_{\text{d}} = 0.65 \ \times \ 10 \ \mathrm{m}^{2} \ \times \ 1.1 \ \mathrm{kg}/ \ \mathrm{m}^{3} \ \times \ (30.56 \ \mathrm{m/s})^{2} = 6745 \ \mathrm{N} \] \

- 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: \[ P = 6745 \ \mathrm{N} \ \times \ 30.56 \ \mathrm{m/s} = 206144.2 \ \mathrm{W} \]

- Convert Power to Kilowatts

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

Key concepts

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.