Asked by Anonymous
A 650 kg weather balloon is designed to lift a 4600 kg package. What volume should the balloon have after being inflated with helium at 0 degrees celcius and 1 atm pressure to lift the total load?
The buoyancy force must equal the weight lifted. Let V be the volume.
g*[(He density)* V + 650 kg] = (Air density)*V*g
Cancel out the g's. Assume the outside air density correesponds to the same 1 atm and 0 C. He density is 4 kg/22.4 m^3 and the air density is 29/22.4 kg/m^3
0.179 V + 650 = 1.295 V
Solve for V (in cubic meters)
The buoyancy force must equal the weight lifted. Let V be the volume.
g*[(He density)* V + 650 kg] = (Air density)*V*g
Cancel out the g's. Assume the outside air density correesponds to the same 1 atm and 0 C. He density is 4 kg/22.4 m^3 and the air density is 29/22.4 kg/m^3
0.179 V + 650 = 1.295 V
Solve for V (in cubic meters)
Answers
Answered by
Salah
Lifted weight= Bouyant force
g*[(He density)*V+650+4600]=Air density *V*g
g= gravity= 9.8 N/Kg
Air Density= 1.29 Kg/m^3
He Density= 0.179 Kg/ m^3
Ballon Voulme= Gas Volume= Air volume approximately
Solving the equation
V= 4227 m^3
g*[(He density)*V+650+4600]=Air density *V*g
g= gravity= 9.8 N/Kg
Air Density= 1.29 Kg/m^3
He Density= 0.179 Kg/ m^3
Ballon Voulme= Gas Volume= Air volume approximately
Solving the equation
V= 4227 m^3
Answered by
possi
funny how you helped me a decade later
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