Home / physics / hot air balloon physics calculations Hot air balloon physics calculations 09/08/2021 The mass of a hot air balloon and also its cargo (not including the air inside) is 200 kg.You watching: Hot air balloon physics calculations The air external is at 10 oC and also 101 kPa. The volume of the balloon is 400 m3. To what temperature must the air in the balloon be heated prior to the balloon will lift off. (Air density at 10 oC is 1.25 kg/m3.) Solution:Concepts:The buoyant forceReasoning:For the balloon to lift off, the buoyant force B have to be greater than its weight.Details of the calculation:The buoyant pressure is equal to the weight of the disinserted air at 10 oC = 283 K. B = (1.25 kg/m3)(400 m3)(9.8 m/s2) = 4900 N. The weight of the balloon is 200 kg(9.8 m/s2) + weight of hot air. The hot air therefore need to weigh much less than 4900 N - 1960 N = 2940 N. Its mass should be much less than 2940 N/(9.8 m/s2) = 300 kg. Its thickness need to be less than r = 300 kg/(400 m3) = 0.75 kg/m3.At consistent pressure, the volume of a gas is proportional to the absolute temperature. (Law of Gay-Lussac) The pressures on the inside and also exterior of the inflated balloon are virtually equal. The press on the external is the constant atmospheric pressure. The Law of Gay-Lussac therefore uses.Because the volume of a gas at consistent pressure is proportional to its temperature, its thickness ρ = m/V is proportional to 1/T.See more: Part A The Pictorial Representation Of A Physics Problem Consists Of A Physics We have actually ρ1/ρ2 = T2/T1. ρ1T1/ρ2 = T2. (1.25 kg/m3)(283 K)/(0.75 kg/m3) = 472 K = T2. The air in the balloon should be heated to even more than 472 K = 199 oC.Problem:Two balloons have been filled up via air under atmospheric press to quantities V1 and also V2, respectively. They are currently submerged under water. A thin string of length L, which is run through a pulley at a addressed depth H, connects the balloons. (The radii of the pulley-block and also the balloons are much smaller sized than the size of the string.) By setting the initial positions of the balloons, one can accomplish a state of equilibrium. Neither balloon is climbing or going dvery own. Determine the difference in the depth of the balloons (in terms of H and L) under those problems. The mass of the balloon skins, of the string, and also of the air is negligible. The temperature of the water is constant and also equal to the temperature of the air.Solution:Concepts:The buoyant pressure, the ideal gas law, pressure at depth h, Ph = Poptimal + ρghThe right gas law: PV = NkTReasoning:The buoyant pressure is equal to the weight of the disinserted water. To suffer the exact same buoyant pressure, the 2 balloons need to have actually the exact same volume under water.Details of the calculation:At atmospheric pressure and at the exact same temperature we have actually V1/V2 = N1/N2 from the best gas legislation.To have actually the exact same volume under water at the very same temperature we need P1/P2 = N1/N2 according to the right gas law. We therefore needP1/P2 = V1/V2.(Poptimal + ρgh1)/( Poptimal + ρgh2) = V1/V2.Pheight + ρgh1 = (V1/V2)( Ppeak + ρgh2).(1 - V1/V2) Ptop + ρgh1= (V1/V2)ρgh2.h2 = ((V2/V1) - 1)(Ptop/ρg) + (V2/V1)h1h2 - h1 = ((V2/V1) - 1)<(Ptop/ρg) + h1)>h1 + h2 = 2H - Lh2 - h1 = <(1 - (V1/V2))/ <(1 + (V1/V2))><2(Ptop/ρg) + 2H - L>Here Pheight = 101 kPa and also ρ = 1000kg/m3.