The air drag force on a car is 0.225 ApV^2. Assume air at 290 K, 100 kPa and a car frontal area

Name: Problem 29, Chapter 4: Fundamentals of Thermodynamics
Uploaded: 2020-07-21T17:41:18-08:00
Duration: 2 min 15 s
Channel: Sanjeev Kumar
Description: The air drag force on a car is 0.225 ApV^2. Assume air at 290 K, 100 kPa and a car frontal area of 4 m^2 driving at 90 km / h. How much energy is used to overcome the air drag driving for 30 min?

step 1 Rho is equal to 1 upon V, which is equal to p upon RT, which is equal to 100 pressure P. R is 0

Question

The air drag force on a car is $0.225 \mathrm{ApV}^{2}$. Assume air at $290 \mathrm{K}, 100 \mathrm{kPa}$ and a car frontal area of $4 \mathrm{m}^{2}$ driving at $90 \mathrm{km} / \mathrm{h}$. How much energy is used to overcome the air drag driving for 30 min?

   The air drag force on a car is $0.225 \mathrm{ApV}^{2}$. Assume air at $290 \mathrm{K}, 100 \mathrm{kPa}$ and a car frontal area of $4 \mathrm{m}^{2}$ driving at $90 \mathrm{km} / \mathrm{h}$. How much energy is used to overcome the air drag driving for 30 min?

Fundamentals of Thermodynamics

Richard E. Sonntag,… 6th Edition

Chapter 4, Problem 29

Instant Answer

Solved by Expert Sanjeev Kumar

07/21/2020

Step 1

The formula for air density is given by $\rho = \frac{p}{RT}$, where p is the pressure, R is the specific gas constant for air, and T is the temperature. Substituting the given values, we get \[\rho = \frac{100}{0.287 \times 290} = 1.2015 \, \text{kg/m}^3\] Show more…

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The air drag force on a car is $0.225 \mathrm{ApV}^{2}$. Assume air at $290 \mathrm{K}, 100 \mathrm{kPa}$ and a car frontal area of $4 \mathrm{m}^{2}$ driving at $90 \mathrm{km} / \mathrm{h}$. How much energy is used to overcome the air drag driving for 30 min?

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Key Concepts

Air Drag Force

This concept refers to the resistive force encountered by an object moving through air. It is a fundamental topic in fluid dynamics and shows how forces opposing motion depend on factors like the speed of the object (often a square dependence), the cross?sectional area exposed to the flow, and properties of the air.

Work-Energy Principle

This principle links force and energy by stating that the work done against a force (in this case, air drag) over a distance is equal to the energy expended. In problems involving drag, calculating the work done involves integrating the force along the path over which it acts.

Unit Conversion

Converting between different units of measurement, such as transforming speeds from kilometers per hour to meters per second and time from minutes to seconds, is crucial. This ensures that all quantities are in compatible units for correct calculation of forces, distances, and ultimately, energy.

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