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Hot Air Balloon Thremodynamics
Name: Luis V.
Status: Educator
Age: 40s
Location: N/A
Country: N/A
Date: June 2002
Question:
I made with my students a paper hot air balloon. If we
know the heating rate (watts) derived from rate at an alcohol mass is
burned, is there any simple equation about the loss of heat (radiation
and transmittance) if we know the balloon's surface and the ambient
temperature?
Replies:
This is a very difficult question. Radiation loss (assuming an ideal
radiator) is given by Planck's formula, which by itself is not a simple
equation, heat loss by conduction is proportional to the difference in
temperature between the balloon and the surrounding air and the constant of
proportionality is experimentally determined. There is no way that I am
aware of to calculate it from first principles, especially for something as
complex as paper. Convection is also going to be a major source of heat
loss. The hot air on the skin of the balloon heats the layer of air adjacent
to it decreasing the density of that air which causes this "skin of air" to
rise, which in turn sucks up some colder air to replace it, which then heats
up, decreases in density, rises, and again sucks up some more colder air,
and so on, and so on.
You might consider turning the experiment around. Knowing the rate of
combustion of the alcohol (Is it ethanol, isopropanol?? Whatever it is you
can look up the heat of combustion. If all that heat went into heating the
volume of air in your balloon, and assuming the products of combustion are
CO2, and H2O gases -- there is the difficulty that some of the H2O might
condense to liquid, but ignore that for the moment. These hot gases will
displace most of the air initially in the balloon. You can then calculate
the overall density of the balloon + hot gases. From the model of a standard
atmosphere (look up "standard atmosphere" in a handbook or on the Internet)
you can estimate how high the balloon should have risen IF THERE WERE NO
HEAT LOSSES. The actual height of rise of the balloon will be much less than
this ideal altitude. From the actual rise, you can calculate an equivalent
amount of heat loss from the "standard atmosphere" and convert it to a mass
of alcohol lost, which will in turn give you the percent efficiency of your
balloon. My guess is that it will be impressively inefficient.
Vince Calder
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Update: June 2012
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