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Chapter 2: Problem 21

The drag force, \(F_{\mathrm{d}}\), imposed by the surrounding air on a vehicle moving with velocity \(\mathrm{V}\) is given by $$ F_{\mathrm{d}}=C_{\mathrm{d}} \mathrm{A}_{2} \rho \mathrm{V}^{2} $$ where \(C_{\mathrm{d}}\) is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \(\rho\) is the air density. Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at (a) 25 miles per hour, (b) 70 miles per hour. Assume \(C_{\mathrm{d}}=0.28, \mathrm{~A}=25 \mathrm{ft}^{2}\), and \(\rho=0.075 \mathrm{lb} / \mathrm{ft}^{2}\).

Short Answer

Expert verified
The power required is 47.38 hp at 25 mph and 1060.31 hp at 70 mph.

Step by step solution

01

- Convert Velocity to Feet per Second

First, convert the given velocities from miles per hour to feet per second. There are 5280 feet in a mile and 3600 seconds in an hour. Thus, to convert a velocity V from miles per hour (mph) to feet per second (ft/s), use the formula: \[ V_{\text{ft/s}} = V_{\text{mph}} \times \frac{5280}{3600} = V_{\text{mph}} \times \frac{22}{15} \]Let's calculate this for both given velocities:
02

- Calculate Drag Force

Next, calculate the drag force using the formula: \[ F_{\text{d}} = C_{\text{d}} \times A \times \rho \times V^2 \]Here, the values given are: \[ C_{\text{d}} = 0.28 \]\[ A = 25 \text{ ft}^2 \]\[ \rho = 0.075 \text{ lb/ft}^3 \]Calculate the drag force for each velocity.
03

- Calculate Power in ft-lb/s

Power required to overcome drag force can be calculated by multiplying the drag force with velocity: \[ \text{Power}_{\text{ft-lb/s}} = F_{\text{d}} \times V \]Use the velocities converted to feet per second from Step 1.
04

- Convert Power to Horsepower

Convert the power from ft-lb/s to horsepower (hp). There are 550 ft-lb/s in one horsepower. Thus, the formula to convert from ft-lb/s to hp is: \[ \text{Power}_{\text{hp}} = \frac{\text{Power}_{\text{ft-lb/s}}}{550} \]Calculate this for each power value from Step 3.
05

Example Calculations for (a) 25 mph

1. Convert 25 mph to ft/s: \[ V_{\text{ft/s}} = 25 \times \frac{22}{15} = 36.67 \text{ ft/s} \]2. Calculate drag force: \[ F_{\text{d}} = 0.28 \times 25 \times 0.075 \times (36.67)^2 = 710.47 \text{ lb} \]3. Calculate power in ft-lb/s: \[ \text{Power}_{\text{ft-lb/s}} = 710.47 \times 36.67 = 26060.76 \text{ ft-lb/s} \]4. Convert to horsepower: \[ \text{Power}_{\text{hp}} = \frac{26060.76}{550} = 47.38 \text{ hp} \]
06

Example Calculations for (b) 70 mph

1. Convert 70 mph to ft/s: \[ V_{\text{ft/s}} = 70 \times \frac{22}{15} = 102.67 \text{ ft/s} \]2. Calculate drag force: \[ F_{\text{d}} = 0.28 \times 25 \times 0.075 \times (102.67)^2 = 5683.67 \text{ lb} \]3. Calculate power in ft-lb/s: \[ \text{Power}_{\text{ft-lb/s}} = 5683.67 \times 102.67 = 583668.15 \text{ ft-lb/s} \]4. Convert to horsepower: \[ \text{Power}_{\text{hp}} = \frac{583668.15}{550} = 1060.31 \text{ hp} \]

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Key Concepts

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

drag coefficient
The drag coefficient, denoted as \(C_{\mathrm{d}}\), is a dimensionless number that quantifies the drag or resistance of an object in a fluid environment such as air or water. It essentially represents how aerodynamic an object is. The lower the drag coefficient, the more streamlined the vehicle, and the less force it will take to move through the air. For vehicles, having a lower drag coefficient is crucial as it improves fuel efficiency and speed. In the example, a drag coefficient of 0.28 indicates a reasonably aerodynamic vehicle.
frontal area
The frontal area, often denoted as \(A\), refers to the size of the front of a vehicle as it faces the airflow. It is typically measured in square feet (\(\text{ft}^2\)). A larger frontal area results in more air resistance since more surface area is being pushed against the air. In this exercise, the given frontal area is 25 \(\text{ft}^2\). This relatively large area means that the vehicle will encounter a significant amount of drag as it moves, which directly affects the amount of power required to overcome this resistance. To minimize drag, vehicle designers try to streamline this area as much as possible.
power conversion
Power conversion involves converting the units of power from one system to another. In this problem, we need to convert power from foot-pounds per second (\(\text{ft-lb/s}\)) to horsepower (hp). This conversion is necessary because horsepower is a more familiar unit when discussing car engines and performance. The conversion factor is: \[ \text{Power}_{\text{hp}} = \frac{\text{Power}_{\text{ft-lb/s}}}{550} \] This means that one horsepower is equivalent to 550 foot-pounds per second. Thorough understanding of this conversion helps in better comprehending how much power a car engine needs to overcome aerodynamic drag.
velocity conversion
Velocity conversion is crucial when dealing with different units of speed. Typically, vehicle speeds are given in miles per hour (mph) but for calculations involving forces and power, it's more useful to convert these speeds to feet per second (\(\text{ft/s}\)). The conversion formula used is: \[ V_{\text{ft/s}} = V_{\text{mph}} \times \frac{5280}{3600} = V_{\text{mph}} \times \frac{22}{15} \] This converts mph to \(\text{ft/s}\), allowing us to perform more accurate calculations in the given equations. In our step-by-step solution, converting 25 mph and 70 mph to \(\text{ft/s}\) helped in accurately determining the aerodynamic drag force and the power required to overcome this drag for the vehicle.

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Most popular questions from this chapter

A gas contained within a piston-cylinder assembly undergoes two processes, \(A\) and \(B\), between the same end states, 1 and 2 , where \(p_{1}=10\) bar, \(V_{1}=0.1 \mathrm{~m}^{3}, U_{1}=400 \mathrm{~kJ}\) and \(p_{2}=1\) bar, \(V_{2}=1.0 \mathrm{~m}^{3}, U_{2}=200 \mathrm{~kJ}\) : Process A: Process from 1 to 2 during which the pressurevolume relation is \(p V=\) constant. Process B: Constant-volume process from state 1 to a pressure of 2 bar, followed by a linear pressure-volume process to state 2 . Kinetic and potential energy effects can be ignored. For each of the processes \(A\) and \(B\), (a) sketch the process on \(p-V\) coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.

At steady state, a spherical interplanetary electronicsladen probe having a diameter of \(0.5 \mathrm{~m}\) transfers energy by radiation from its outer surface at a rate of \(150 \mathrm{~W}\). If the probe does not receive radiation from the sun or deep space, what is the surface temperature, in \(\mathrm{K}\) ? Let \(\varepsilon=0.8\).

A gas undergoes a process in a piston-cylinder assembly during which the pressure-specific volume relation is \(p v^{1.2}=\) constant. The mass of the gas is \(0.4 \mathrm{lb}\) and the following data are known: \(p_{1}=160 \mathrm{lbf} / \mathrm{in}^{2}, V_{1}=1 \mathrm{ft}^{3}\), and \(p_{2}=\) \(390 \mathrm{lbf} /\) in. \({ }^{2}\) During the process, heat transfer from the gas is \(2.1\) Btu. Kinetic and potential energy effects are negligible. Determine the change in specific internal energy of the gas, in Btu/lb.

An object of mass \(1000 \mathrm{~kg}\), initially having a velocity of \(100 \mathrm{~m} / \mathrm{s}\), decelerates to a final velocity of \(20 \mathrm{~m} / \mathrm{s}\). What is the change in kinetic energy of the object, in \(\mathrm{kJ}\) ?

An object of mass \(10 \mathrm{~kg}\), initially at rest, experiences a constant horizontal acceleration of \(4 \mathrm{~m} / \mathrm{s}^{2}\) due to the action of a resultant force applied for \(20 \mathrm{~s}\). Determine the total amount of energy transfer by work, in kJ.
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