Three men in a balloon (to say nothing of the dog). It is a late fall, slightly

Question

Three men in a balloon (to say nothing of the dog). It is a late fall, slightly chilly morning, t = 10.0 degree C. Three men and a dog drive a couple of miles until they reach an empty farmland area where a large balloon is anchored. They climb into a sturdy wicker gondola equipped with propane cylinders and a twin gas burner, all attached to a large spherical balloon, and decide to take a flight. They slide their aviator goggles and put leather gloves on, ignite the burners, and get ready for a lift off (no goggles for the dog). The passengers, dog, basket, ropes, burners, gas cylinders, and the nylon-Nomex envelope of the balloon weigh 678 kg. (Typical weight of a mid-size balloon, a basket with -three to five passengers, a twin burner, and -two to four gas cylinders is 650-750 kg.) As the air inside the balloon becomes warmer it becomes less dense and when its buoyancy exceeds the net weight of the balloon and the cargo, the balloon will take off. Given that the fully inflated balloon is a nearly perfect sphere of 9.0 m radius what should be the average temperature of the air inside the balloon [degree C] needed for a liftoff? Keep in mind that the balloon is open and the pressure of the air inside the balloon equals the pressure of the air outside the balloon. A good estimate to use when comparing tire pressure to temperature is for every 10 degrees F, tire pressure will adjust by 1 psi. For example, if the temperature increases 10 degrees, the tire pressure will increase by 1 psi. (1 psi= 2.21 bar; 1 K =1.8 degrees F). A formula one car was inflated to 32 psi initially at 95 degrees F. The car operates at a tire temperature of ~ 230 degrees F. (a) Estimate the pressure of the tire pressure at operation, (b) In actual operation, the tire has an undetectable leak, making the tire pressure only reaches 36 psi. How much air was leaked?

Explanation / Answer

So this is an example of buoyancy

The difference in density (hot air inside the balloon ) vs. the density of outside air will make the balloon rise

We need to make the next balance:

Fdown = Fupward

Total Weight = Buoyancy force

in order to go up, we need at least

Total Weight < Buoyancy force

Total W = total mass * gravity

total mass * g = Buoyancy force

Calculate Total W = (total mass) * g = 678 kg * 9.8 m/s2 = 6644.4 N

Buoyancy force = (difference of density in air)*gravity*(Volume of air)

Volume of air = Volume of balloon

Vballoon = 4/3*PI*(R^3)

Vballoon = 4/3*3.14159*(9m)^3 = 3053.62548 m3

g = 9.8 m/s2

Air density at normal conditions = 1.19 kg/m3

Air density needed at T = ?

Solve for Air density @ T

Buoyancy force = (difference of density in air)*gravity*(Volume of air)

6644.4 N = (D@T - 1.19 kg/m3)*(9.8 m/s2)*(3053.62548 m3)

solve for D@T

6644.4/(9.8*3053.62548) kg/m3+1.19 kg/m3 = D@T

1.41203 kg/m3 = D@T

Now, we can find T using Ideal gas Law

but we can model it as follows:

PV = nRT

n = mol

mol = mass/MW, where MW is the molar weight of air, which is 29 g/mol

substitute data

PV = m/MW*RT

force "density" to appear in the formula, D = m/V

T = P(V/m)*MW/R

T = P*D*MW/R

P = atmospheric pressure = 101325 Pa

D = 1.41203 kg/m3

MW = 29 g/mol

R = 8.314 Pa*m3 / (mol*K)

T = P*D*MW/R

T = (101325 Pa)(1.41203 kg/m3*(29 g/mol) / (8.314 Pa*m3 / mol K)

note that we need to change g to kg

T = (101325 Pa)(1.41203 kg/m3*(0.029 kg/mol) / (8.314 Pa*m3 / mol K)

T = (101325*1.41203*0.029)/(8.314) = 499.05 K

T = 499.05-273 = 226.05 C

Temperature of balloon must be 226.05 C

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