# Metered Fueling

Most people start off their first combustion cannon simply using aerosol fuels. While this may be perfect for somebody who's looking to just have fun, and not put a lot of investment into their launchers, if you're serious about performance, metered fueling is a must.

Basically, it creates the same air:fuel mixture in the combustion chamber, every shot, for consistent performance and ignition.

Metered fueling can be accomplished simply with a syringe, or a traditional fuel meter can be constructed which performs the same operation quicker and more conveniently. Both methods are outlined here.

## Note on Display

This document uses \KaTeX for rendering mathematics. Subsequently, this page may load a little slowly, and javaScript is obviously required to view it correctly.

Since Google Chrome has decided to gimp MathML, this is where we're at. Sorry.

## The Basics

You are looking to combine two chemicals in your chamber, and in the correct proportions, for optimum combustion. These two chemicals are oxygen and a gaseous fuel.

The fuel can be bought in pressurized tanks, and is usually either propane, MAPP gas, or propylene. The oxygen is present in fresh air already inside your combustion chamber, so the only reactant you will be handling and measuring out for each shot is the fuel.

## Fuel Types

Each fuel has its own chemical properties, and therefore takes a different amount of oxygen per unit volume. These values can be calculated using stoichiometry. Since propane is the most common fuel used for combustion cannons, we will go through this calculation for propane, and simply list the values for other common fuels.

It helps to identify your two reactants first. Propane's chemical formula is C3H8. Oxygen is of course O2, but it is mixed with other inert buffer gases like nitrogen in atmospheric air, with a normal concentration of 21%.

So, now that we have these values, we need to find the balanced chemical equation for the combustion of propane in air. This turns out to be:

\text{C}_{3} + 5\text{O}_{2} → 3\text{CO}_{2} + 4\text{H}_{2}\text{O}

What this equation is telling us is that one mole of propane reacts with 5 moles of oxygen to generate 3 moles of carbon dioxide and 4 moles of water, plus heat, which we will get to on another page...

A mole is a unit of measurement which is only marginally important in this discussion. For the same formula we could say that one molecule of propane reacts with... etc.

Using dimensional analysis, we can determine the stoichiometric ratio of propane to air.

\dfrac{1\text{ mol }\text{C}_{3}\text{H}_{8}}{5\sout{\text { mol }\text{O}_{2}}} × \dfrac{21\sout{\text{ mol }\text{O}_ {2}}}{100\text{ mol air}} = \dfrac{0.042\text{ mol } \text{C}_{3}\text{H}_{8}}{1\text{ mol air}}

Thus, we get the molar proportions of propane to air for a stoichiometric burn... but how the hell are we supposed to measure out moles?

The answer is we don't. Back in 1811 a guy named Avogadro hypothesized that two different samples of gas, at the same temperature, volume, and pressure, contained the same molar count.

What this means is that the molar proportion is exactly the same as the volumetric proportion, or, in other words, 4.2 cubic centimeters of propane will react with 100 cubic centimeters of air, and so on.

Phew! Just tell me the values!

Fuel Stoichiometric Ratio
Propane 0.042
Propylene 0.0467
MAPP See Below
MAP/Pro 0.0466 to 0.0467

MAPP gas is slowly being replaced by MAP/Pro, which is 99.5% - 100% propylene, and a very small amount of propane. If you still have an old bottle of MAPP gas, it is a proprietary blend of fuels which varies, but anything from 0.046 to 0.05 works very well. I calculated the proportion out to around 0.0489 using available MSDS sheets, but this isn't the most accurate, as the natural gas used in the mixture varies in composition.

## Syringe Fueling

Great! Now that we know the stoichiometric ratios of various fuels to atmospheric air, we can go about fueling a combustion cannon.

The simplest way to do this is to use a syringe, although it may not be the most user-friendly way. To increase your rate of fire, take a look at the traditional fuel meter setup outlined in the next section.

So, for syringe fueling, you will obviously need a syringe. I find the best way to go about this is to use a large, unmarked syringe, with no needle. Depending on the size of your cannon, a smaller or larger syringe may/should be used.

Now it's time to see how much fuel to measure out and inject into your cannon. But let's take a look at how we need to calculate this value.

Assuming your combustion chamber is filled with fresh air, simply multiply the chamber volume by the fuel:air ratio for the fuel you're using. For these examples, we will use propane.

\dfrac{0.042\text{ mL }\text{C}_{3}\text{H}_{8}}{1\sout{\text { mL air}}} × \dfrac{1200\sout{\text{ mL air}}}{1} = 50.4\text{ mL }\text{C}_{3}\text{H}_{8}

What this means is that for 1200 mL of fresh air (for a freshly ventilated 1200 mL combustion chamber), you would need 50.4 mL of propane to react completely.

However, think of how you inject fuel into a combustion chamber.

When injecting fuel with a syringe, you are often injecting into a hole or cracked endcap, which means air will be displaced as you inject fuel, which means there will be slightly less than 1200 mL of air to react with. If you are injecting into a closed chamber (i.e. by injecting through some type of grommet with check valve), you do not have to worry about this. If the small amount of error doesn't bug you either, you don't have to worry about this as well. If the last two conditions apply, simply use the value calculated like above. Otherwise, read on.

To know how much fuel you should inject given the displacement of air during injection, it's as simple as solving a system of equations, like below.

V_{\text{chamber}} - V_{\text{fuel}} = V_ {\text{air}} \quad , \qquad V_{\text{fuel}} = 0.042 V_{\text{air}}

The first equation is simply stating that the volume of fresh air in the chamber after displacement is equal to the chamber volume minus the volume of the fuel. The second equation is stating that 0.042 times the volume of the air left after displacement is equal to the fuel volume. This seems redundant, but solving is simple. Simply substitute Vair (the left side of the first equation) into the second and solve for the correct fuel ratio for this situation.

\begin{aligned} V_{\text{fuel}} &= 0.042V_{\text{air}}\\ V_{\text{fuel}} &= 0.042(V_{\text{chamber}} - V_{\text{fuel}})\\ 1.042V_{\text{fuel}} &= 0.042V_{\text{chamber} }\\ V_{\text{fuel}} &≈ 0.0403V_{\text{chamber}} \end{aligned}

Taking the coefficient from the right-hand term, we can deduce that the modified "ratio" for injection when air is displaced is about 0.0403. So by repeating the simple method of multiplying the chamber volume by the ratio using 0.0403 instead of 0.042, you can adjust for air displacement. The adjusted ratios for this situation are listed below.

Fuel Modified Ratio
Propane 0.0403
Butane 0.0313
Propylene 0.0446
MAP/Pro 0.0445 to 0.0446

Now, as far as how to get the fuel into the syringe, here's what I do.

Take your unmarked syringe and a graduated cylinder of sufficient accuracy. Measure out your required fuel volume in water with the graduated cylinder, and fill the (dry) syringe with all of this water. Avoid air bubbles. Mark the location of the plunger on the outside of syringe with a permanent marker.

To fill with propane, use a standard torch head, and remove the air mixer on the end of the shaft. Press the plastic around the needle attachment of the syringe over the orifice in the end of the shaft, and use the needle valve to fill the syringe until the plunger is at the mark you've made. That's it.

For filling, I usually drill and tap a hole into the chamber at 1/4" NPT threads, and install a small brass pipe plug. Remove to inject, and replace to fire. Simple.

Now that we know the simplest way to fuel a launcher, it's time to learn the bulky, overcomplicated, but convenient way.

The traditional fuel meter works by trapping a known volume of gas at a known pressure between two valves, and then using one valve to inject this gas into the combustion chamber.

By using a simple gas law and some logic, we can develop a simple equation for the pressure required for a known volume meter and chamber, or, the meter volume required for a known volume chamber and meter pressure, etc.

We start by using Boyle's Law.

P_{1}V_{1} = P_{2}V_{2}

This law relates a sample of gas at different pressure and volumes. While it is not chemically the most accurate equation to be using, it will work just fine for a normal combustion cannon. Eventually, I will get up a page on advanced fuel metering (for hybrids) which will go a little more in depth.

So, how does this help us? Well, let's consider our first scenario: we know the chamber volume, the meter pipe volume, and we want to know what pressure we should use in our meter.

So, by assigning the correct variables, we can begin the process. P1 is what we are solving for, so therefore V1 is set to the volume of the meter pipe. P2 would be the pressure at equilibrium, which for these purposes we will set to 1 atmosphere (14.7 PSIA). V2 would be the volume of the propane gas at equilibrium, which would be the fuel:air ratio multiplied by the volume of the combustion chamber. So, let's work it out.

The following example is for propane, like before.

\begin{aligned} P_{m}V_{m} &= (14.7)(0.042)V_{c}\\ P_{m} &= \frac{0.6174V_{c}}{V_{m}} \end{aligned}

Now, you may be thinking that it's over, that's the answer, but you'd be wrong. While, for all reasons, it should work in theory, in practice, it is too simplified. Let's discuss.

Using the above equation on small to medium sized launcher, or a launcher with bad fuel mixing, you will have a significant amount of misfires. The reason is that the mixture is too lean.

Why? Because, when opening the valve from the meter to the combustion cannon, the propane will flow through the injection orifice until the pressures are at equilibrium, for our purposes, one atmosphere.

This means that, barring diffusion of propane through the orifice, which most shooters are not patient enough to wait for, there is still one atmosphere of propane left in the meter.

The simplest way to fix this, of course, is to just add another atmosphere of pressure to the results from the above equation, like so (in PSIA).

P_{m} = \dfrac{0.6174V_{c}}{V_{m}} + 14.7

People have been fueling cannons for years now without even knowing this, though, by simply not knowing the fact that the output of the equation is meant to be PSIA (absolute pressure). So, when they use the very first equation we derived, or Boyle's law, they fill in 14.7 PSIA for the chamber pressure, but when they obtain their answer, they treat it as gauge PSI (PSIG, which is the absolute pressure minus one atmosphere). So, by simply not knowing your math and disregarding unit matching, you can obtain the correct answer. However, when working on more advanced metering systems, you may pull your hair out.

### Fixed Pressure

Another common method people use for propane meters is to use a fixed pressure and adjust the volume of their meter pipe to obtain the correct mixture.

I don't like this method as the adjustment is not as stepless as adjusting pressure. Pipe and fittings come in fixed sizes and the best you can do is to try and add up the right combination to get the volume you want.

I will provide the equation for an approximation to the correct length of a pipe nipple used in a fuel meter.

We start off with the formula for the volume of a right cylinder i.e. the meter pipe.

V = πr^{2}l

Then, by substitution into our first equation:

\begin{aligned} P_{m} &= \frac{0.6174V_{c}}{(πr^{2}l)} + 14.7\\ P_{m} &= \frac{0.6174V_{c} + 14.7(π r^{2}l)}{(πr^{2}l)}\\ P_{m}(πr^{2}l) &= 0.6174V_{c} + 14.7( πr^{2}l)\\ (P_{m} - 14.7)πr^{2}l &= 0.6174V_{c} \\ l &= \frac{0.6174V_{c}}{(P_{m} - 14.7) πr^{2}} \end{aligned}

Then, by plugging in chamber volume, meter pressure, and the internal radius of the pipe you are using, you can obtain an approximate length to ensure correct fueling.

## Conclusion

While metered fueling may seem complicated, it actually is not. It all boils down to getting the correct amount of fuel into a fixed amount of air. While I would recommend syringe fueling to most beginners, a traditional propane meter is not complicated (it's just a piece of pipe between two ball valves and a way to control the pressure of propane inside them), and can provide more convenience than a syringe.

Hybrid and other advanced fueling topics will be covered in the future on another page.