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Name: Luis V.
Status: Educator
Age: 40s
Location: N/A
Country: N/A
Date: June 2002

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?

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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