Common Liquid Fuels for Combustion Spud Guns

Occasionally, people decide to use liquid fuels, like gasoline, in place of the more common gaseous fuels like propane, in a combustion spud gun. This page contains information on some of the common liquid fuels that might be used in a spud gun.

Combustion

In order for combustion to occur you need two things, fuel and oxygen. In a typical spudgun, air is the source of the oxygen and the fuel is a gaseous hydrocarbon such as propane. Here, we will be looking at the use of liquid fuels such as gasoline. Regardless of whether you use a gaseous or liquid fuel, the ratio of fuel to oxygen must be correct to get the maximum amount of power out of your gun. Furthermore, to get the fuel to even ignite is must be within the fuel's combustibility limits.

Combustion Equation

The complete combustion of hydrocarbon fuels using oxygen (O2) produces carbon dioxide (CO2), water and heat. As an example, the balanced (stoichiometric) chemical equation for the complete combustion of octane (C8H18, a major component of gasoline) in oxygen is;

2C8H18 + 25O2 = 16CO2 + 18H20 + Heat

This equation tells us that for each molecule of octane we need 25/2=12.5 molecules of oxygen. If we have either too much or too little fuel, the reaction does not go to completion and the energy released by the combustion is decreased.

Optimal Fuel Ratio

Most fuels only burn over a limited range of concentrations in air. For octane, that range is about 1% to 6.5%. If we have less than 1% octane, or more than 6.5% octane, the octane will not burn. So, too much fuel, or too little fuel, will simply not ignite. The optimal amount of fuel is what is predicted by the combustion equation, the volume of the chamber, and the percentage of oxygen in air. Too much fuel will never cause a spudgun to fail since with too much fuel the gun won't fire.

For each fuel the maximum energy will be produced when the ratio of fuel to oxygen is correct for the particular fuel. However, most spudguns use air as the oxygen source instead of pure oxygen. Air is about 21% oxygen (by volume), 88% nitrogen and about 1% other gases. We have to take this into account when calculating the amount of fuel to use in a gun. The chemical equation for the combustion of octane indicates that we need 12.5 volumes of oxygen per (gaseous) volume of octane. Since air is 21% oxygen then we need about (1/12.5)(21%) = 1.7% by volume gaseous octane in the chamber with air for maximum power.

With liquid fuels we have a couple of problems not present with gaseous fuels, for example;

  1. Exactly what volume of the liquid is needed to give the correct percentage
    of fuel in the chamber? 
  2. How do we measure the fairly small amounts of liquid fuel with sufficient accuracy?
  3. How do we get the liquid fuel to vaporize completely in the chamber?

One thing that you should keep in mind with liquid fuels is the potential for weakening the PVC. Take a look at the ingredients list on your cans of PVC cleaner and glue. Anything listed on those cans probably should be avoided as fuels. Acetone and tetrahydrofuran (THF) in particular are probably not the best idea for fuel since they are present in PVC glue and cleaner and will soften and/or be absorbed by the PVC. PVC should be fine with small amounts of gasoline.

Calculating the volume needed of a liquid fuel

In order to calculate the volume of a liquid fuel for a particular sized gun we need several pieces of information; the volume of the chamber (from which the volume of available oxygen can be calculated), the molecular weight (MW, in grams/mol) of the fuel, the density (in g/ml) of the liquid fuel and the amount of oxygen required per mole of fuel. The volume of liquid fuel required is given by;

(Vchamber)(0.01639L/in3)(273.15/Temp)*(1mol/22.4L)*(MW in g/mol)(mol fuel/mol oxygen)(0.2095)

(density of liquid fuel in g/ml)

Where, Vchamber is the volume of the chamber (in in3), and Temp is the ambient temperature in degrees kelvin (72F=295K). The ratio "mol fuel/mol oxygen" is the number of molecules of fuel per molecule of oxygen from the balanced chemical equation for the combustion. The factor of 0.2095 is the mole fraction of oxygen in air. The 0.0164L/in3 factor converts the chamber volume in in3 to liters.

As an example lets do octane (basically gasoline). The "mol fuel/mol oxygen" is 0.08 (2 molecules of octane for 25 molecules of oxygen), the density is 0.70g/ml, the MW is 114g/mol and we will assume a temperature of 72F (295K) and a chamber volume of 100in3 (1.64L). Plugging these values into the equation we get;

(100in3)(0.0164L/in3)(273K/295K)*(1mol/22.4L)*(114g/mol oct.)(0.08 mol oct./mol O2)(0.2095)

(0.70g oct./ml oct.)

= 0.185ml octane

So, for a 100in3 chamber filled with air we need to add 0.185ml of liquid octane. The parameters of several other liquid fuels are given in Table I.

Table I: Common Liquid Fuels for Combustion Spud Guns

Fuel Boiling Point,
C (F)
MW, g/mol Density, g/ml Heat of Combust.,
KJ/mol
(note 1)
Flammability Limits in Air Stoich. % Fuel in Air Energy Relative to Propane Combustion
Formula
Moles fuel to moles oxygen ml per 100in3 Chamber Volume
(note 2)
Approx. drops per 100in3 Chamber Volume
(note 3)
Methanol
(wood alcohol)
65
(149)
32 0.79 725 6%–36% 14.0% 1.09 2CH4O + 3O2 =
2CO
2 + 4H2O
0.667 0.384 4.7
Ethanol
(grain alcohol)
78
(172)
46 0.79 1378 3.3%–24.5% 7.0% 1.04 1C2H6O + 3O2 =
2CO
2 + 3H2O
0.333 0.277 3.4
Isopropanol
(rubbing alcohol)
83
(181)
60 0.79 2030 2%–12% 4.7% 1.02 2C3H8O + 9O2 =
6CO
2 + 8H2O
0.222 0.241 2.9
Acetone 57
(134)
58 0.78 1761 2.6%–12.8% 5.2% 1.00 1C3H6O + 4O2 =
3CO
2 + 3H2O
0.250 0.262 3.2
Diethyl Ether 34.6
(94)
74 0.71 2683 1.7%–48% 3.5% 1.01 1C4H10O + 6O2 =
4CO
2 + 5H2O
0.167 0.245 3.0
Hexane 69
(156)
86 0.65 4167 1.2%–7.7% 2.2% 0.99 2C6H14 + 19O2 =
12CO
2 + 14H2O
0.105 0.196 2.4
Octane
(gasoline)
126
(258)
114 0.70 5473 1%–6.5% 1.7% 0.99 2C8H18 + 25O2 =
16CO
2 + 18H2O
0.080 0.184 2.2
Toluene
(methyl benzene)
111
(231)
92 0.86 4234 1.1%–7.1% 2.3% 1.07 1C7H8 + 9O2 =
7CO
2 + 4H2O
0.111 0.168 2.0
Propane
(for reference)
-42
(-44)
44 0.51 2208 2.37%–9.5% 4.2% 1 1C3H8 + 5O2 =
3CO
2 + 4H2O
0.200    

Table Notes:

  1. Calculated using the method at http://home.fuse.net/clymer/rq.
  2. Based on a temperature of 72F (22C) and air containing 20.95% (volume) oxygen.
  3. Based on 12.2 drops = 1 ml.

How do we measure the fairly small amounts of liquid fuel
with sufficient accuracy?

For our 100in3 chamber we determined we needed 0.185ml (about 1/26th of a teaspoon) of liquid octane (gasoline). How exactly do you measure a volume that small? Well, 1 drop of a liquid is generally about 0.08ml (this is very approximate and depends on the shape of the dropper and the viscosity of the liquid). So 0.185ml would be about 2.3 drops.

If you try to draw a volatile (low boiling) liquid, like gasoline, up into an eye dropper the liquid will piddle back out of the dropper almost immediately. This is caused by the evaporation of the volatile liquid which raises the pressure in the dropper. To minimize this affect you can draw the liquid up into the dropper, allow most of the liquid to be pushed out of the dropper, then refill the dropper. If you repeat this process several times eventually the liquid will, more or less, stay in the dropper for the length of time needed to measure it into your chamber. By repeatedly filling and draining the dropper you are replacing the air in the dropper with fuel vapor. Once the dropper is full of fuel vapor the liquid will stop evaporating.

You may want to do a rough calibration of your eye dropper with your chosen fuel. Measure how many drops it takes to fill a 1/4 teaspoon measure (1/4 teaspoon = 1.225ml), it should be somewhere around 15 +/- 5 drops. Divide the actual number of drops you measured 15 by and use this number as a scaling factor for the values in the column "Approx. drops per 100in3 Chamber Volume" in Table I .



 (measured number drops in 1/4 teaspoon)( Approx. drops per 100in3 Chamber Volume) 
corrected number drops =



15

How do we get the liquid fuel to vaporize completely in the chamber?

Now that we know how many drops of fuel are required, and we can measure the number of drops with reasonable accuracy, we need to make sure that the fuel actually evaporates once it is inside the chamber. For best results, this is going to require a chamber fan. I suggest adding the appropriate number of drops of fuel to the chamber, closing the chamber and running the fan for ten seconds or so. Open the chamber up and see if the drops of fuel are still present. If they are then you need to run the fan longer. Do this a couple of times until you have a good idea of how long the particular fuel takes to evaporate with your fan setup.

When you are ready to actually fire the gun you should thoroughly air out the test fuel load(s) from the chamber, measure in new fuel, close the chamber and run the fan. Do not open the chamber to check for complete evaporation since you will loose some of your fuel.

Heats of Combustion and the "Power" of a Fuel

The "heat of combustion" of a fuel is a measure of the amount of energy released when the fuel is burned. This is an important, but not the only factor affecting the performance of a fuel. The table above lists the heats of combustion along with other parameters for a variety of liquid fuels.

The key value for comparing two fuels is not the heats of combustion. Instead, you need to take into account both the heat of combustion and the amount of oxygen required to burn the fuel. The amount of fuel that should be used is limited by the amount of oxygen present in the chamber. Fuel is added to match that amount of oxygen. In the table, the "Energy Relative to Propane" column gives the theoretical amount of energy that can be obtained for a particular fuel, relative to the energy with propane as the fuel. A graph of the relative energies is shown below.

Relative energy of liquid fuels

As you can see from the graph, there is relatively little energy difference between the various fuels. The best fuel in the set, by this measure, is methanol which gives 9% more energy than propane. The worst fuel is octane (gasoline) which gives about 1% less energy than propane.


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Modified: 25 June 2007
©2007 James Sluka