\'\'A typical balloon system - envelope, gondola, fuel tanks, and 40 gallons of
ID: 1279698 • Letter: #
Question
''A typical balloon system - envelope, gondola, fuel tanks, and 40 gallons of fuel - will weigh about 600 pounds, deflated on the ground. In the air, the complete system, including the weight of the air inside the envelope, will weigh about 2Y2 tons.'' Assume the information stated in the quote, and that additionally, there are three passengers in the balloon with a total mass of 200 kg. Also assume that the air surrounding the balloon is at temperature 10 degree C and pressure 0.90 atm. If the balloon is in static equilibrium, what is the temperature of the air inside the balloon? Neglect the volume of all parts of the balloon system other than the hot air inside the balloon envelope.Explanation / Answer
2.5 tons = 2500 kg
600 pounds = 272.15 kg
0.9 atm = 0.9*101325 Pascals = 91192.5 Pa
10 deg C = 10 +273 K = 283 K
Density of surrounding air = P/RT = 91192.5 / (287*283) = 1.122 kg/m^3
Volume of balloon = V
Weight of diplaced fluid = 1.122*V*g
Weight of balloon + passengers = (2500 + 3*200)*g = 3100*g
Equating both
1.122*V*g = 3100*g
V = 2761 m^3
Mass of air = 2500 - 272.15 = 2227.85 kg
m = PV/(RT)
T = PV / (mR)
T = 91192.5*2760 / (2227.85*287)
T = 393.6 K
T = 393.6 - 273 deg C = 120.6 deg C
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.