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15 March, 12:36

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the additional weight of the balloon material itself. The balloon material has a mass of 60 g/m2. Ambient air is at 25 °C and 1 atm. The hot air inside the balloon is at 70 °C and 1 atm. Assuming that the balloon has a perfectly spherical shape, what balloon diameter will just support the total weight? Neglect the size of the hot air inlet vent and assume an ideal gas law (P = rhoRT) for the air.

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  1. 15 March, 14:26
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    r = 4.44 m

    Explanation:

    For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

    B = ρ g V

    Now let's use Newton's equilibrium relationship

    B - W = 0

    B = W

    The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

    σ = W / A

    W = σ A

    The area of a sphere is

    A = 4π r²

    W = W₁ + σ 4π r²

    The volume of a sphere is

    V = 4/3 π r³

    Let's replace

    ρ g 4/3 π r³ = W₁ + σ 4π r²

    If we use the ideal gas equation

    P V = n RT

    P = ρ RT

    ρ = P / RT

    P / RT g 4/3 π r³ - σ 4 π r² = W₁

    r² 4π (P/3RT r - σ) = W₁

    Let's replace the values

    r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

    r² (11.81 r - 0.060) = 13000 / 4pi

    r² (11.81 r - 0.060) = 1034.51

    As the independent term is very small we can despise it, to find the solution

    r = 4.44 m
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