[TheForge] allstate oxy-propane torch

[Rich Maynard]

Some observations;

Your table talks about molecules of gas combining in a chemical reaction.

You take no account of the volumes of gas this represents (which is probably
closest to what we would measure in practice).

You take no account of the fact that burning one molecule of methane will
produce less heat than burning one molecule of gasoline (for example).

Also, the heat needed for the cutting of steel comes about largely from the
reaction between oxygen and the iron in the steel, and not from burning the
fuel.

Cheers,

Rich.

>
> methane   == CH4
> acetylene == C2H2
> ethylene  == C2H4
> ethane    == C2H6
> propylene == C3H6
> propane   == C3H8
> butylene  == C4H8
> butane    == C4H10
> gasoline  == C8H18
> hydrogen  == H2
> oxygen    == O2
>
>           fuel        oxygen         carbon      water
>           gas                        dioxide     vapor
> methane   CH4(g)   + ( 2.0)O2(g) = (1)CO2(g) + (2)H20(g)
> acetylene C2H2(g)  + ( 2.5)O2(g) = (2)C02(g) + (1)H20(g)
> ethylene  C2H4(g)  + ( 3.0)O2(g) = (2)CO2(g) + (2)H20(g)
> ethane    C2H6(g)  + ( 3.5)O2(g) = (2)CO2(g) + (3)H20(g)
> propylene C3H6(g)  + ( 4.5)O2(g) = (3)CO2(g) + (3)H20(g)
> propane   C3H8(g)  + ( 5.0)O2(g) = (3)CO2(g) + (4)H20(g)
> butylene  C4H8(g)  + ( 6.0)O2(g) = (4)CO2(g) + (4)H20(g)
> butane    C4H10(g) + ( 6.5)O2(g) = (4)CO2(g) + (5)H20(g)
> gasoline  C8H18(g) + (12.5)O2(g) = (8)CO2(g) + (9)H20(g)
> hydrogen  H2(g)    + ( 0.5)O2(g) =             (1)H2O(g)
>
>           oxygen to fuel gas ratios
>