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Chapter 1: Problem 108

The "power available in the wind" of velocity \(V\) through an area \(A\) is \\[ \dot{W}=\frac{1}{2} \rho A V^{3} \\] where \(\rho\) is the air density \(\left(0.075 \mathrm{lbm} / \mathrm{ft}^{3}\right) .\) For an 18 -mph wind, find the wind area \(A\) that will supply a power of 4 hp.

Short Answer

Expert verified
The wind area 'A' that will supply a power of 4 hp for an 18-mph wind is approximately 350.17 ft^2.

Step by step solution

01

Convert units

To start with, convert the wind velocity, 'V' from mph to ft/s using the conversion factor 1 mile = 5280 ft and 1 hour = 3600 s. Similarly, convert the power from hp to ft.lb/s using the conversion factor 1 hp = 550 ft.lb/s. After conversion, velocity 'V' is 18*5280/3600 = 26.4 ft/s and power '\(\dot{W}\)' is 4*550 = 2200 ft.lb/s.
02

Substitute the values into the formula

Substitute the values of power '\(\dot{W}\)', velocity 'V' and air density 'ρ' into the formula \(\dot{W} = 0.5 * ρ * A * V^3\) and solve for area 'A'. The formula becomes 2200 = 0.5 * 0.075 * A * (26.4)^3.
03

Calculate the wind area 'A'

Calculate the wind area 'A' by rearranging the formula to solve for 'A': \(A = 2*\dot{W} / (ρ * V^3)\). Substituting the given values gives \(A = 2*2200 / (0.075 * (26.4)^3)\). Solving this gives \(A\) approximately 350.17 ft^2.

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