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Chapter 9: Problem 75

A full-sized automobile has a frontal area of \(24 \mathrm{ft}^{2},\) and a compact car has a frontal area of \(13 \mathrm{ft}^{2}\). Both have a drag coefficient of 0.5 based on the frontal area. Find the horsepower required to move each automobile along a level road in still air at 55 mph. Assume that the power required to deform the tires continuously at this speed (called rolling resistance) is equal to the power to overcome the air resistance. Estimate the gas mileage of both automobiles if the energy supplied to the drive wheels is \(\frac{1}{4}\) that available in the fuel. A gallon of fuel has \(1.0 \times 10^{8} \mathrm{ft} \cdot 1 \mathrm{b}\) available energy.

Short Answer

Expert verified
The horsepower required and gas mileage for the full-sized automobile are approximately 86.04 hp and 14.59 MPG respectively. For the compact car, these values are approximately 46.60 hp and 26.92 MPG.

Step by step solution

01

Calculate drag force

Drag force on a moving object can be calculated using the formula \( F_{d} = 0.5 * C_{d} * A * ρ * V^{2} \), where \( C_{d} \) is the drag coefficient, \( A \) is the frontal area, \( ρ \) is the air density, and \( V \) is the velocity. Here, both cars have a drag coefficient of 0.5, the frontal area is given, and we can consider air density to be about \( 0.002377 \, slug/ft^{3} \) and velocity is \( 55 \, mph (=80.67 \, fps) \). By plugging these values into the formula, we can calculate the drag force for both cars.
02

Calculate the power required

The power to overcome the air and tire resistance each is equal to the drag force times the speed. Therefore, total power required is double that amount. Power can be calculated as \( P = 2 * F_{d} * V \). After calculating the drag force for both cars in the previous step, we can just plug in these values with the velocity to find the power required for each car in foot-pounds per second (ft-lbf/s), which can then be converted to horsepower (1 horsepower = 550 ft-lbf/s).
03

Calculate the fuel consumed

Given that \( \frac{1}{4} \) of the energy in fuel is supplied to the drive wheels to generate this power, rest is wasted as heat, inefficiencies, etc. So we can calculate the energy consumed from the fuel by \( E = P * 4 \). This will give the energy consumed in \( ft \cdot lbf \).
04

Calculate gas mileage

Finally, given that a gallon of fuel has \( 1.0 \times 10^{8} ft \cdot lbf \) energy, we can calculate the gas mileage or how many miles can the car run per gallon (MGP) of fuel by dividing the energy per gallon by the energy consumed per second by speed in \( mph \). In other words, \( MGP = \frac{1.0 \times 10^{8}}{E * 3600} * V_{mph} \). This process needs to be repeated for both cars.

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Most popular questions from this chapter

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