Free Image from Pixabay QuestionIn a hurricane, the air (density 1.2 kg/m^3) is blowing over the roof of a house at a speed of 110 km/h. (a) What is the pressure difference between inside and outside that tends to lift the roof? (b) What would be the lifting force on a roof of area 93 m^2? SolutionN.B. We use the law of the conservation of energy to solve this problem.
(a) We divide the problem into two parts. First, we look at air flow outside the house. The total energy of the air flow per unit volume outside the house is: > p1 + (1/2) ro . v1^2 + ro . g . h1 = constant  (1) Where: The pressure of the flow of air about the roof is p1 Speed of the air flow about the roof is v1 Height of the air flow above the roof from the ground is h1 Inside the house, we have a similar equation as: > p2 + (1/2) ro . v2^2 + ro . g . h2 = constant  (2) Where: The pressure of the flow of air below the roof is p2 Speed of the air flow below the roof is v2 Height of the air flow below the roof from the ground is h2 We have to assume that inside the house the velocity of air is zero. So the above equation reduces to: > p2 + ro . g . h2 = constant  (3) From the conservation of energy principle, the total energy of air flow per unit volume inside and outside the house is the same. So, we have: > p1 + (1/2) ro . v1^2 + ro . g . h1 = p2 + ro . g . h2 The pressure difference, p2  p1, is given by: > p2  p1 = (1/2) ro . v1^2 + ro . g . (h1  h2) Now, the height inside and outside the roof is the same, i.e., h1 = h2. So, > p2  p1 = (1/2) ro . v1^2  (4) The speed blowing above the roof, v1, is 110 km/h So, v1 = 110 km/h = 110 x 1000 m/h = 110,000/3600 m/s or, v1 = 30.55 or 30.6 m/s The density of air, ro, is give as 1.2 kg/m^3. Substituting values of v1 and ro in equation (4) gives us: p2  p1 = (1/2) ro . v1^2 = 0.5 x 1.2 x (30.6)^2 = 561.816 N/m So the pressure difference inside and outside the roof is: 561.816 N/m (b) The lift force is equal to the pressure difference between the inside and outside of the house times the area of the roof. Or, in other words, FL, the lift force, = (p2 = p1) x Ar, the area of the roof. The area of the roof is 93m^2 and the pressure difference i 561.816 N/m. Substituting the values, we get: FL = 561.816 x 93 = 52,248.888 N. Rounding up, we get: The lift force, FL, = 52,250N.
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