When you have a fuel ratio of 14 to 1 does tha mean 1lbs of fuel to 14lbs of air or 14lbs of O2?

 

Mass of Air to Mass of Fuel

 

Its not that easy 1 gallon of fuel wieghs about 6 lbs and air wieghs about .010026lbs per gallon and say you get 10 miles per gallon that says you use 648 gallons of air for 1 gallon of gas. A 351 and 1800rpm doing a mile a minute would use .1gallons of fuel while using 2844 gallons of air. Somethings not right.

Air = .075#/cubic ft
Gas = 6#/gal

 

We are dealing with molecular MASS the fuel and air. Here is a quick example.

The stoichiometric air-fuel ratio is the ratio at which all oxygen is used up and all fuel is completely burnt. This ratio is a basic property of a fuel and is the result of its chemical composition. Let’s for example look at natural gas (methane). When burning any carbon-based fuel, carbon dioxide and hydrogen are formed. Going back to the methane example, the following reaction equation describes the oxidation of the fuel:

CH4 + 2O2 -> CO2 + 2H20

Following from the equation, we need 2 molecules of oxygen for every octane molecule.

If we look up the atomic weights of the atoms that make up octane and oxygen, we get the following numbers:

Carbon (C): 12,01

Oxygen (O): 16

Hydrogen (H): 1,008

So 1 molecule of methane has a molecular weight of: 1 * 12,01 + 4 * 1,008 = 16,042
One oxygen molecule weighs: 2 * 16 = 32.
The oxygen-fuel mass ratio is then: 2 * 32 / 1 * 16,042 = 64 / 16,042 = .
So we need 3,99 kg of oxygen for every 1 kg of fuel.
Since 23,2 mass-percent of air is actually oxygen, we need : 3,99 * 100/23,2 = 17,2 kg air for every 1 kg of methane.
So the stoichiometric air-fuel ratio of methane is 17.2.

Common fuels

When the composition of a fuel is known, this method can be used to derive the stoichiometric air-fuel ratio. For the most common fuels, this, however, is not necessary because the ratios are known:

Gasoline: 14.7

Natural gas: 17.2

Propane: 15.5

Ethanol: 9

Methanol: 6.4

Hydrogen: 34

Diesel: 14,6