


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:
 It uses
helium (He).
 It uses
hydrogen (H_{2}).
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:
 The buoyant force
on the balloon is equal to the weight of the (fluid) atmosphere displaced.
 The ideal gas
law holds in the form p
= nkT

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.

Atmospheric
conditions at an altitude of 90 km are:
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:
 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 _{A}V_{B}g
is the weight of displaced atmosphere, then

_{A}V_{B}g
= (10 kg)g + _{G}V_{B}g
or

_{A}V_{B}
= 10 kg + _{G}V_{B}


where
_{G}
is the gas density inside the balloon, V_{B} is volume of
the balloon, and g is the local gravitational acceleration.

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/m^{2}, and T = 187
K, then solve for _{G}:
_{G}
= 4.66 x 10^{7} kg/m^{3}


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}

We
now use eq. 2 in eq. 1 to calculate
the volume of the balloon:
V_{B}
= 3.39 x 10^{6} m^{3}


and
we use eq. 3 with eq. 2 to calculate
the mass of helium required:
If
the balloon is assumed to be spherical, then
V_{B}
= (4/3) r_{B}^{3}.
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
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/m^{3}


V_{B}
= 3.03 x 10^{6} m^{3}


^{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.
