Combustion (2)_Air and Fuel

Combustion (2)_Air and Fuel (A/F Ratio)

 

As written in part (1) of combustion, the composition of fuel is hydrocarbon (HC), but the chemical bonding of every kind of HC is different from each other. Depending on the type of bondage of HC, the type of combustion inside combustion chamber will also be decided. The picture below shows a couple of different kind of HC chemical structure.

 

     

                    

 

       Paraffinic HC (CnH2n+2)         Naphthenic hydrocarbon (CnH2n)

 

The “n” in the equation means the ration of hydrogen and carbon in the HC. There are several other types of HC and the fuel for our engine is mixed with several different types of hydro carbons.

 

It is not possible to count number of hydrogen molecules and carbon molecules, instead we can measure the volume or mass of the fuel and calculate statistically the number of molecules of hydrogen and carbon involves in combustion process by applying the ratio of hydrogen and carbon in the fuel. In case of gasoline fuel the composition ration of hydrogen and carbon is 0.856 (C ) and 0.144 (H) and 0,875 (C ) and 0.125 (H) in case of diesel fuel.

 

While we are filling fuel in the tank at a gas station, we do not fill oxygen for combustion. We use the oxygen which exists in the air. Now let’s have a look on the “air” the counter partner of fuel for combustion process a little bit more precisely. As seen in the combustion reaction equation in the combustion (1), only oxygen is involved (or useful) in combustion process, and all the other components of air are just for combustion point of view useless. If we get gram of air and analyze the components, the list will be like the table below.

 

Component	Symbol	Volume Ratio (%)	Mass Ratio (%)
Nitrogen	N2	78.03	75.47
Oxygen	O2	20.99	23.20
Argon	Ar	0.933	1.280
Carbon-dioxide	CO2	0.030	0.046
Hydrogen	H2	0.010	0.001
Neon	Ne	0.0018	0.0012
Helium	He	0.0005	0.00007
Krypton	Kr	0.0001	0.0003
Xenon	Xe	0.000009	0.00004

 

As seen in the table, 3/4 of the air composed of nitrogen and the other is oxygen. As written as an equation form, the combustion process has an exact rule, meant the mass of each material necessary for combustion is should be exactly correct for an optimal combustion. Now, let’s calculate the amount of air which is needed for a given amount of fuel.

 

As in the combustion equation, one oxygen molecule need one carbon molecule for carbon combustion, and molecular weight ration of oxygen and carbon is 8:3. Therefore, for an optimal conversion of carbon into carbon dioxide, the mass ratio of carbon and oxygen should be 8:3. With the same theory the ratio of hydrogen and oxygen for an optimal combustion should be 8:1. This relationship can be expressed in the equation below.

 

Oxygen  Mass   = (8/3) x Carbon Mass + 8 x Hydrogen Mass

for Combustion

 

The equation above expressed the mathematical relationship of carbon, hydrogen and oxygen for an optimal combustion.  Using this mass relationship between the three components, we can drive the equation between air and fuel, because we know that the 23.2% of air mass is oxygen. If we express this in equation form, this will be like below.

 

Air Mass for = Mass of Oxygen/0.232 x{(8/3) x C-Mass + 8 x H-Mass}

Combustion

 

If we can apply ratio of carbon and hydrogen composition in the fuel, we can calculate the necessary amount of air for an optimal combustion process. Let’s take the composition ration of gasoline fuel and calculate the necessary amount of air.

 

Air  Mass = 4.31 x { (8/3) x 0.856 + 8 x 0.144 }

              = 14.82

 

 ** 4.31 : Inverse Number of 0.232 (Oxygen rate in air)

   0.856 : Composite rate of carbon in gasoline fuel

   0.144 : Composite rate of hydrogen in gasoline fuel

 

We can interpret the calculation result in this way. To make optimal combustion of one kilogram of fuel, we need to supply 14.8 kilogram of air into the combustion chamber. The ratio of 14.8:1, we call Air-Fuel ratio for gasoline engine.

 

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