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2 years ago

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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 addi

tional 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.

1 answer:

RSB [31]2 years ago

3 0

Answer:

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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