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Chapter 9: Problem 93

An airplane flies at \(150 \mathrm{km} / \mathrm{hr}\). (a) The airplane is towing a banner that is \(b=0.8 \mathrm{m}\) tall and \(\ell=25 \mathrm{m}\) long. If the drag coefficient based on area \(b \ell\) is \(C_{D}=0.06,\) estimate the power required to tow the banner. (b) For comparison, determine the power required if the airplane was instead able to tow a rigid flat plate of the same size. (c) Explain why one had a larger power requirement (and larger drag) than the other. (d) Finally, determine the power required if the airplane was towing a smooth spherical balloon with a diameter of \(2 \mathrm{m}\)

Short Answer

Expert verified
The power required for the banner is 6500 W, for the rigid flat plate is 138780 W, and for the spherical balloon is 47980 W. The flat plate demands the highest power due to its higher resistance (drag), and the banner requires the least amount of power.

Step by step solution

01

Calculation for Banner

First, find the amount of power needed to tow the banner. Use the drag force formula: \(F = 0.5 \cdot C_D \cdot p \cdot A \cdot v^2\), where \(C_D\) is drag coefficient, \(p\) is air density, \(A\) is the reference area, and \(v\) is the velocity. This gives \(F = 0.5*0.06*1.184*(0.8*25)*(150/3.6)^2 = 156.34 \, \text{N}\). Remember that air density is approximately \(1.184 \, \text{kg/m}^3\) at sea level, and velocity needs to be converted to m/s. Then, calculate the power: \(P = F*v\), therefore \(P = 156.34*(150/3.6) = 6500 \, \text{W}\).
02

Calculation for Flat Plate

For a flat plate, the drag coefficient is higher, \(C_D = 1.28\). Repeating the calculation from step 1 with the new coefficient, \(F = 0.5*1.28*1.184*(0.8*25)*(150/3.6)^2 = 3340.47 \, \text{N}\). The power therefore is \(P = 3340.47*(150/3.6) = 138780 \, \text{W}\).
03

Explanation of Differences

The power required to tow the flat plate is greater than that needed for the banner due to the higher drag coefficient. This means that the flat plate creates more resistance against the air, requiring more force and energy to pull.
04

Calculation for Spherical Balloon

For a spherical balloon, the reference area becomes the cross-sectional area of the sphere, \(A = 0.25*\pi*d^2 = 0.25*\pi*(2^2) = \pi \, \text{m}^2\). The drag coefficient for a smooth sphere is approximately \(C_D = 0.47\). So, \(F = 0.5*0.47*1.184*\pi*(150/3.6)^2 = 1153.32 \, \text{N}\), and \(P = 1153.32*(150/3.6) = 47980 \, \text{W}\).

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