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30% Propane, 70% Butane Mix, Any good?

Post questions and info about combustion (flammable vapor) powered cannons here. This includes discussion about fuels, ratios, ignition systems, safety, and anything else relevant.
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30% Propane, 70% Butane Mix, Any good?

Unread postAuthor: athoul » Sun Jul 15, 2007 12:14 pm

Taymar

A woman i do gardening for has a can of this in her shed for a small flamethrower thing that is meant to burn small path weeds.

Its mix is
30% propane
70% butane


Will this be any good to use as fuel?
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Unread postAuthor: noname » Sun Jul 15, 2007 12:20 pm

Hell yes it will! It'll be hard to get right, but once you get the fueling down, you'll have the best spray and pray fuel there is!
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Unread postAuthor: athoul » Sun Jul 15, 2007 12:21 pm

Hard to get right?
Do you mean the right amount of spray?
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Unread postAuthor: th3p0p0 » Sun Jul 15, 2007 12:23 pm

yes iv tride it and it and it works very good but hard to get the right amont.
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Unread postAuthor: Binder17 » Mon Jul 16, 2007 12:02 pm

GEt the chemical equation for the combustion of propane and butane with oxygen. Make sure there are no other chemicals in the can. Stoichiometry may be difficult to use in figuring this out so maybe just experiment.
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Unread postAuthor: jimmy101 » Tue Jul 17, 2007 3:48 pm

If the can really only contains propane and butane then the stoichiometry is pretty simple.

Method 1: Just take the weighted average of the fuel ratios for the two fuels;
30% butane which is used at 3.12% (squirt-and-screw method)
70% propane which is used at 4.02% (squirt-and-screw method)
Volume percent of fuel is;
(0.30)(3.12%)+(0.70)(4.02%) = 3.75%
(this is only approximately but close enough for most purposes)

Method 2: The balanced chemical equation is;
3C<sub>4</sub>H<sub>10</sub> + 7C<sub>3</sub>H<sub>8</sub> + (109/2)O<sub>2</sub> = 33CO<sub>2</sub> + 43H<sub>2</sub>O
or
6C<sub>4</sub>H<sub>10</sub> + 14C<sub>3</sub>H<sub>8</sub> + 109O<sub>2</sub> = 66CO<sub>2</sub> + 86H<sub>2</sub>O
For every 6+14=20 volumes of fuel we need 109 volumes of oxygen.
Since air is only 21% oxygen;
(20/109)*(0.21) = 3.85% fuel in air.
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