The mass of a hot-air balloon and its cargo (not including the air inside) is 20

Question

Explanation / Answer

Draw a Free Body diagram around the balloon. There is the weight of the cargo (Wc) and balloon pulling down. There is the weight of the air (Wi) inside the balloon pulling down. There is a buoyant force (B) pushing up. So for the balloon to start to lift, B >= Wc + Wi B is equal to the weight of the air displaced by the balloon. Before the air inside is heated, Wi = B, the weight of the displaced air is the same as the weight of the air inside. We know Wc and we can easily calculate B from the air density (Da) and the volume (Vb) of the balloon. (B = Da*Vb*g) To lift off, Wi = B - Wc Wi = (Da*Vb*g) - Wc Wi = (Da*Vb*g) - Mc*g Wi = g((Da*Vb) - Mc) Wi = 9.81 m/s^2((1.25 kg/m^3*500 m^3) - 200 kg) Wi = 4170 N or Mi = 425 kg Hmmm, now it seems to be getting tricky.....the balloon will take off when the mass of air inside is less than 425 kg. So the air density inside the balloon must be 425 kg / 500 m^3 = 0.85 kg/m^3. There is not a fixed amount of air inside the balloon. The bottom of the balloon is open, so as the air is heated, the higher pressure inside will force some of the air to leave the balloon until the inside and outside pressure equalize. If you have a chart with air density versus temperature, then this would be the time to use it. At what temp will the density equal 0.85 kg.m^3? If you have no such chart, try PV = nRT. You would have to calculate the number of moles needed to make 425 kg. Then use T = PV/nR to calculate Temp. (P is atm, and V is 500 m^3). Sorry, that is too much for me to handle tonight.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.