Designing a High Altitude Balloon

Problem:
Design a gas balloon to carry an instrument package in the earth's atmosphere at an altitude of 90 km. Specify how large the balloon will have to be and the mass of gas it will require to maintain this altitude if:

Assume that the ideal gas law holds at the desired altitude, that the inflated balloon is spherical, and that the total payload mass (balloon + instrument package) is 10 kg.

Solution:
Let's assume that:

1. The buoyant force on the balloon is equal to the weight of the (fluid) atmosphere displaced.
2. The ideal gas law holds in the form p = nkT
3. where n is the number density in m^-3, k is Boltzmann's constant (k = 1.38 X 10^-23 j/K), T is the absolute temperature, and p is the pressure.
4. Atmospheric conditions at an altitude of 90 km are:

1. Temperature:	T = 187 K
   Pressure:	p = 1.38 x 10-3 Torr
   Density:	A = 3.42 x 10-6 kg/m3

   (ref.: U. S. Standard Atmosphere, 1976).

   Then:

   1. In order for the balloon to stay aloft, we know that the buoyant force must equal the weight of the balloon and instrument package plus the weight of the (hydrogen or helium) gas inside the balloon. If _AV_Bg is the weight of displaced atmosphere, then
   2. _AV_Bg = (10 kg)g + _GV_Bg
      

2. or

   AVB = 10 kg + GVB

   1.

    

where _G is the gas density inside the balloon, V_B is volume of the balloon, and g is the local gravitational acceleration.

1. We use the ideal gas law to obtain the gas density inside the balloon. For helium: p = nkT = (_G/m)kT

   with m = 6.68 x 10^-27 kg. Set p = 1.38 X 10^-3 torr = 1.82 x 10^-6 atm = 0.18 nt/m², and T = 187 K, then solve for _G:

G = 4.66 x 10-7 kg/m3

2.

This number represents the density of helium at the pressure and temperature conditions stated for 90 km altitude. It is assumed to be the helium gas density inside the balloon as well.⁸

1. We now use eq. 2 in eq. 1 to calculate the volume of the balloon:

VB = 3.39 x 106 m3

3.

and we use eq. 3 with eq. 2 to calculate the mass of helium required:

mHe = 1.58 kg

4.

If the balloon is assumed to be spherical, then

V_B = (4/3) r_B³.

where r_B is the balloon radius. Using the value of V_B in eq. 3, we find that the radius of the balloon is

rB = 93 m

5.

If hydrogen is used instead of helium, then we set m_H2 = 1.67 x 10^-27 kg in the ideal gas law, and repeat the above procedure to find

G = 1.16 x 10-7 kg/m3

2a.

VB = 3.03 x 106 m3

3a.

mH2 = 0.35 kg

4a.

rB = 90 m

5a.

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⁸ Thus, the interior and exterior pressures are equal. This assumption is equivalent to saying that the balloon is inflated without stretching; i.e., that there is no tension in its material.