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The Question
(Submitted January 06, 1997)
My son, as a requirement for his 7th grade science class, is to
have a working model science fair project. While going through some
magazines he came upon the idea to build a "Trash Bag Hot-Air
Balloon", which would show and demonstrate one type of lighter
than air vehicle.
While this seemed to be a great idea for his project, his teacher
is not too enthused as "everyone knows how they work." As a
result of
his comments we have been trying to find the explanation of the
scientific principles that are being demonstrated in this project and,
hopefully, prove his project more "science project" worthy to his
teacher.
Is there any help, information, or advice you can possibly offer
that will help him with this project?
Thank You for your time.
The Answer
Our expertise is in designing and building detectors
to collect X-rays and gamma-rays from astrophysical objects, and then
to interpret the data. It might seem like hot air balloons would be
a little outside of our area of interest, but actually, since the radiation
we are most interested in observing is absorbed by the Earth's atmosphere,
many of our high energy astrophysics experiments are flown on balloons.
This way they can get above a substantial fraction of the absorbing
atmosphere. The balloons used for scientific payloads are helium filled,
but the principle of employing a balloon filled with a lighter gas to get
and remain airborne is basically the same.
The basic physics behind hot air balloon travel is the effect of
increased temperature on the motions of molecules of a gas, and thereby
on the density of the gas. In order to understand this, you'll need a little
algebra, and one of the basic ideas of thermodynamics, called the Ideal
Gas Law.
A hot air balloon stays afloat in the cooler air surrounding it
due to the buoyant force on it. This is the same force that acts on you
when you are in a pool of water. You have probably noticed it is much easier
to lift someone if you are both in a pool of water, and this is due to the
partial support the water is offering: the buoyant force. This force was
studied by the Greeks before 200 B.C. and can be understood by Archimedes'
principle: any body completely or partially submerged in a fluid is buoyed
up by a force equal to the weight of the fluid displaced by the body.
If B represents the buoyant force and W the weight of the displaced fluid,
then B=W.
We need to consider the sum of the forces acting on the balloon,
which is totally submerged in the air around it. The buoyant force B
acts upward on the balloon, and gravity acts downward. The weight of the
balloon, w, is the same as the gravitational force downward. Since these
two forces act in opposite directions (buoyant force up, gravity down), the
total force on the balloon is F(total)=B-w.
Now we need to represent each force in terms of things we can measure.
These are: the density of the fluid and the volume of the balloon. Remember
that density is defined as the mass of an object divided by its volume. The
weight of the displaced air equals the buoyant force, and weight is always
equal to the mass times the acceleration due to gravity (W=Mg). The mass of
the displaced air, M, is just the density of the displaced air multiplied
by the volume of the balloon (that's the volume being displaced). So,
if D=density of the cool, surrounding air, then W=D*V*g, where V is the
volume of the balloon. And since B=W, B=D*V*g. For the weight of the
balloon, w=mg. Assuming the basket attached to the balloon can be ignored
for now, we need only get the mass of the air inside the balloon. That will
be equal to the density of the air inside, d, times the volume. w=d*V*g.
Then F(total)=B-w=D*V*g-d*V*g=(D-d)*V*g. When this is a positive number,
the force is in the upward direction. That occurs when the density of the
air inside the balloon, d is less than the density of the surrounding, cooler
air, D.
How can the density of the air inside become less that the air outside?
Since the gas inside is the same as outside the balloon, we can use the
Ideal Gas Law to study what happens to the density as the temperature is
increased. One way of stating this law is: for a gas with a constant
molecular weight, the pressure is proportional to product of the density
and the temperature (P=K*D*T). Here, K is just a constant, and T is the
temperature. That means for a gas at constant temperature, the density
is given by: D=P/(K*T). As the temperature increases, the density decreases.
At this point the story is almost complete. If you and your son are still
interested in following this line for a science project, then I will leave it
to you to discover what is happening to the molecules of gas inside the
balloon that causes the density to drop. You can find a discussion of the
ideas of buoyancy, and how gases are affected by temperature, in any
introductory high school or college physics book. The books will have lots of
pictures and worked-out examples, and will give you a more detailed description
than the one here.
I spoke with a high school physics teacher this weekend about your
question, and she told me that there is a kit available to students for
making a hot air balloon. Perhaps your son's teacher was concerned that
if he used a kit to build the balloon he might not learn as much as if he
started a project from scratch. I would encourage you to talk to the teacher
some more, and find out what they want the kids to get out of their project.
If your son comes away from this project with a better understanding of
the forces acting on a balloon, and on the effects of heat on a gas, then
it sounds like a good learning experience. It the teacher is aiming for
something different in the class, perhaps you and your son can build the
balloon on your own in your spare time!
I hope this helps. Good luck with the balloon and the science fair.
Regards,
Padi Boyd
for Imagine the Universe!
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