The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkab

Question

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the building can cause windows to pop out. This happened with the original design of the John Hancock Building in Boston.
(a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m X 1.50 m. Assume the density of air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane?
(b) What force is experienced by the window pane from air if the airspeed outside is now 22.4m/s The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the building can cause windows to pop out. This happened with the original design of the John Hancock Building in Boston.

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Sapo 4 years 2021-08-26T06:01:31+00:00 1 Answers 104 views 0

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  1. thachthao
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    2021-08-26T06:03:24+00:00
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    Answer:

    a. 451.58 N b. 1806.34 N

    Explanation:

    (a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m X 1.50 m. Assume the density of air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane?

    Using Bernoulli’s equation

    P₁ + 1/2ρv₁² + ρgh₁ = P₂ + 1/2ρv₂² + ρgh₂ where P₁ = pressure of air in building = atmospheric pressure = 1.013 × 10⁵ N/m², ρ = density of air = 1.20 kg/m³, v₁ = speed of air in building = 0 m/s(since it is still), h₁ = h₂ = h = height of building, P₂ = pressure on the outside of window pane, v₂ = speed of air outside window pane = 11.2m/s and g = acceleration due to gravity

    So, since h₁ = h₂ = h

    P₁ + 1/2ρv₁² + ρgh = P₂ + 1/2ρv₂² + ρgh

    P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂²

    Also, v₁ = 0m/s

    So, P₁ + 1/2ρ(0 m/s)² = P₂ + 1/2ρv₂²

    P₁ + 0 = P₂ + 1/2ρv₂²

    P₁ = P₂ + 1/2ρv₂²

    P₁ – P₂ = 1/2ρv₂²

    So the net pressure on the window is ΔP = P₁ – P₂ = FA where F is the total force on the window pane and A is the area of the window pane = 4.00 m × 1.50 m = 6.00 m²

    So, P₁ – P₂ = 1/2ρv₂²

    ΔP = 1/2ρv₂²

    F/A = 1/2ρv₂²

    F = 1/2ρAv₂²

    Substituting the values of the variables into the equation, we have

    F = 1/2ρAv₂²

    F = 1/2 × 1.20 kg/m³× 6.00 m² × (11.2 m/s)²

    F = 1/2 × 1.20 kg/m³ × 6.00 m² × 125.44 m²/s²

    F = 451.584 N

    F ≅ 451.58 N

    (b) What force is experienced by the window pane from air if the airspeed outside is now 22.4m/s

    When v₂ = 22.4 m/s, F is

    F = 1/2ρAv₂²

    F = 1/2 × 1.20 kg/m³× 6.00 m² × (22.4 m/s)²

    F = 1/2 × 1.20 kg/m³ × 6.00 m² × 501.76 m²/s²

    F = 1806.336 N

    F ≅ 1806.34 N

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