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Chapter 3: Problem 28

A 40 -mph wind blowing past your house speeds up as it flows up and over the roof. If elevation effects are negligible, determine (a) the pressure at the point on the roof where the speed is 60 mph if the pressure in the free stream blowing toward your house is 14.7 psia. Would this effect tend to push the roof down against the house, or would it tend to lift the roof? (b) Determine the pressure on a window facing the wind if the window is assumed to be a stagnation point.

Short Answer

Expert verified
The pressure at the point on the roof where the speed is 60 mph is less than the initial pressure. This effect tends to lift the roof due to lower pressure on top. The pressure on a window facing the wind is more than the initial pressure which results in pushing against the window.

Step by step solution

01

Determine the pressure at the point on the roof

Firstly, convert the speeds from mph to m/s: 40 mph = 17.88 m/s, 60 mph = 26.82 m/s. Now, use the Bernoulli's equation to find the pressure at the point on the roof where the wind speed is 60 mph. Air density (\(\rho\)) is 1.184 kg/m^3, the given initial pressure \(P_1\) is 14.7 psia which converted to pascal is 101353 Pascal. Substitute these values into the Bernoulli's equation: \(P_2 = P_1 + \frac{1}{2} \cdot \rho \cdot (v_1^2 - v_2^2) \)
02

Calculate the pressure and analyze its effect on the roof

After calculation, the pressure \(P_2\) will be found to be less than \(P_1\). The lower pressure on top means that there is a net upward pressure, which tends to lift the roof.
03

Determine the pressure on a window facing the wind

For the window assumed to be at the stagnation point, the velocity of the wind (\(v_2\)) becomes zero. Hence \(P_{window} = P_1 + \frac{1}{2} \cdot \rho \cdot (v_1^2) \). Calculate \(P_{window}\) to get the pressure on the window.

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