A hot air balloon takes off from the airport. The airport is at a geometric alti

Question

A hot air balloon takes off from the airport. The airport is at a geometric altitude 900ft on a standard day. The burner in the hot air balloon maintains the air inside the balloon at a density of 0.95kg/m^3 The balloon has a volume of 70,000ft^3 and a weight of 380lbf. Attached to the balloon is basket that weighs 40 lbf and has one passenger that has a mass of 180 lbm. The reference area for this balloon is the area of a circle at the equator of the balloon where the diameter is 50 ft. Assume the acceleration due to gravity is 32.17 ft/s^2 and the drag coefficient for the balloon is 0.55. The balloon is traveling upward in equilibrium; determine the ascent rate for this hot air balloon.

Explanation / Answer

given the baloon is ascending in equilibrium, so net force on the baloon is 0
now from the given data
geometric altitude = h = 900 ft
density of air inside the baloon, rho1 = 0.95 kg/m^3
volume of baloon, V = 70,000 ft^3
weight W = 380 + 40 + 180*32.17 lbf
diameter of baloon, D = 50 ft
Cd = 0.55
density of air at that altitude be rho
then from the chart, rho = 1.1 kg/m^3

so, buoyant force = (rho - rho1)*V = 122.54 lbf upwards
weight = 6180 lbf
the baloon should not b e travelling upwards, it should be travelling downwardxs weith velocity v
so that for equilibrium Drag + buoyancy = weight
drag = 6180 - 122.54 = 6057.46 lbf = 26944.92441 N
but Drag = Cd*0.5*rho*v^2 = 26944.92441 = 0.55*0.5*1.2*v^2
v = 285.74 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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