P6. Suppose a hot-air balloon is spherical, with a fixed diameter of 42 feet. Th

Question

P6. Suppose a hot-air balloon is spherical, with a fixed diameter of 42 feet. The amount of air inside the balloon is not fixed, but changes with temperature (holes in the bottom and the top allow air to flow into or out of the balloon so that the air pressure is the same inside and out). I is filled with air that is heated. As the air is heated, it expands, and thus reduces its density. If the mass of the balloon (and whatever is attached to it) is 310 kg (and the volume associated with this mass is small), to what temperature should the air inside the balloon be raised so that the balloon can float? Assume the surrounding air has a density of 1.10 kg/m', and a temperature of o C

Explanation / Answer

given
d = 42 ft = 12.8016 m
mass, m = 310 kg
hence

from equilibrium
for buoyant force B
B = mg = 3041.1 N

now, Pi = Po (pressure inside the ballon is outside the baloon)
hence
Po = rho*R*T
rho = 1.1 kg/m^3
R = 286.9 J/kg K
T = 273.16 K
hence
Po = 86206.5644 Pa

hence
Pi = 86206.5644 Pa = rho'*R*T ( where rho' is density of heloium, T is its temperature)
R = 2,077 for helium
now
B = (rho - rho')*g*pi*d^3/6
hence
rho' = 0.817791 kg/m^3
hence

T = 50.7529761349 K
so the helium needs to be cooled down to this temperature

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