In article ,
Edwin Pawlowski wrote:
dave martin wrote:

The propane burner requires about 10 cuft air for each 1000 btu
generated. Each 10 cuft of air burned generates about 1.6 cuft of
water vapor. That's a LOT of water.

How much water is in the water vapor? I'm not doubing the statistics, but
you are saying that 10 cu. ft of air reduces to 16% of vapor by volume.
What else is included here in the 1.6 cubic feet? That means that the
original atmosphere contain 16 percent vapor, but that must be reduced to a
given amount of liquid. How do we calculate the liquid content?

The answer to your last question is "absolute humidity". the math gets
somewhat messy. Google for a write-up.

Air consists partly of oxygen. the oxygen is the only part that is
involved in combustion.

The amount of oxygen in 10 cu. ft. of air will combine with a 'fuel' to
"make" about 1.6 cu. ft. of water vapor.

If you want to go through the actual numbers --
You start with the percentage of 'air' that is oxygen. the oxygen
molecule ("O2") has a molecular weight of 32 (roughly). thus 22.4 liters
of O2 will weight 32 grams.

Propane (C3H8, molecular weight 32) provides the hydrogen.

1 C3H8 + 5 O2 = 3 CO2 + 4 H20 + {heat}

CO2 weighs in at 40, and H20 at 18

In gaseous form (at standard temperature and pressure), 22.4 liters of
-anything- weighs, in grams, what the molecular weight of an individual
molecule is.

Have fun with the math. grin

One lb of propane, when burned, generates about 2-1/4 lbs of water. or about
35 fl. oz.