Question

If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level location? Assume an air temperature of 20∘C.

Answer 1

Complete question:

The classic Goodyear blimp is essentially a helium balloon— a big one, containing 5700 m³ of helium. If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level location? Assume an air temperature of 20°C.

Answer:

52.4 kN

Explanation:

The helium at 20°C has a density of 0.183 kg/m³, and the cargo load is the weight of the system, which consists of the envelope, the gondola, and the helium.

The helium mass is the volume multiplied by the density, thus:

mHe = 5700 * 0.183 = 1043.1 kg

The total mass is then 5343.1 kg. The weight is the mass multiplied by the gravity acceleration (9.8 m/s²), so:

W = 5343.1*9.8

W = 53362.38 N

W = 52.4 kN

Answer 2

QUESTION: The question is incomplete. See the complete question below and the answer.

The classic Goodyear blimp is essentially a helium balloon— a big one, containing 5700 m3 of helium. If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level location? Assume an air temperature of 20°C.

Answer:

Mass of the cargo load  = 1548.2kg

Explanation:

Calculating the weight of helium gas, we have

W = mg

   = (pv)g

   =(0.178 * 5700)g

   =(1014.6)g

Weight of envelope and gondola;

W = (4300kg)g

Weight of cargo load = mg (unknown)

The upthrust force on the balloon is calculated as;

F =pvg

  = (1.204 *5700)g

  =(6862.8)g

But;

Weight of helium gas + weight of envelope and gondola + weight of cargo = upthrust force on the balloon.

Substituting, we have;

(1.0146)g +  (4300kg)g + mg = (6862.8)g

mg =  (6862.8)g - (1014.6)g - (4300kg)g

 mg    = (1548.2kg)g

m =  (1548.2kg)g/g

m = 1548.2kg