Theory

In the Boiler Efficiency lab, we stated that Combustion Efficiency is defined as the ratio of the burner’s capability to burn fuel completely to the unburned fuel and excess air in the exhaust. In this lab, we will perform a combustion analysis.

Fossil fuels may be classified into solid, liquid and gaseous fuels. The vast majority of fuels are based on carbon (C), hydrogen (H₂) or some combination of carbon and hydrogen called hydrocarbons.

During combustion, oxygen (O₂) combines rapidly with C, H₂, sulphur (S₂) and their compounds in solid, liquid and gaseous fuels and results in the liberation of energy. Except for special applications such as oxyacetylene welding, in which a high-temperature flame is required, the O₂ necessary for combustion is obtained from air. Air contains O₂ and nitrogen (N₂), plus negligible amounts of other gasses and for engineering purposes, may be considered to have the following percentage composition by mass:

O₂: 23%
N₂: 77%

The proportions in which the elements enter into the combustion reaction by mass are dependent upon the relative molecular weights as shown below:

Stoichiometric Combustion Theory

Complete combustion of simple hydrocarbon fuels forms carbon dioxide (C0₂) from the carbon and water (H₂0) from the hydrogen, so for a hydrocarbon fuel with the general composition C_nH_m, the combustion equation on a molar basis is as flows:

[latex]C_{n}H_{m}+yO_{2}\rightarrow aCO_{2}+bH_{2}O[/latex]

Where the balance should be satisfied following the moles for any mathematcial  equation:

Carbon balance:

[latex]a=n[/latex] kmol CO₂/ kmol fuel

Hydrogen balance:

[latex]2b=m[/latex] 

[latex]b = \frac{m}{2}[/latex] kmol H₂O/ kmol fuel

Oxygen balance:

[latex]2y = 2a+b[/latex]

[latex]y = a+\frac{b}{2}[/latex] kmol O₂/ kmol fuel 

Considering that combustion occurs in air rather than in pure oxygen, the nitrogen in the air may react in the combustion process to produce nitrogen oxides. Beside, some fuels contain elements other than carbon, and these elements may react with oxygen during combustion. Also, combustion is not always complete, and the exhaust gases contain unburned and partially burned products in addition to C0₂ and H₂O.

Air is composed of oxygen, nitrogen, and small amounts of carbon dioxide, argon, and other trace components. For the purposes of the further calculation it is perfectly reasonable to consider air as a mixture of 21% (mole basis) 02 and 79 % (mole basis) N₂.  Nitrogen will be considered as an “inert” gas in the combustion calculations. The stoichiometric relation for complete combustion of a hydrocarbon fuel, C_nH_m, becomes

[latex]C_{n}H_{m}+y(O_{2}+\frac{79}{21}N_{2}\rightarrow aCO_{2}+bH_{2}O+cN_{2}[/latex]

The balance equations are:

Carbon balance:

[latex]a=n[/latex] kmol CO₂/ kmol fuel

Hydrogen balance:

[latex]2b=m[/latex] 

[latex]b = \frac{m}{2}[/latex] kmol H₂O/ kmol fuel

Oxygen balance:

[latex]2y = 2a+b[/latex]

[latex]y = a+\frac{b}{2}[/latex] kmol O₂/ kmol fuel 

Nitrogen balance:

[latex]y\frac{79}{21}=c[/latex] kmol N₂/kmol fuel

Flue gas compositions are presented in terms of mole fractions as kmol of product per kmol of fuel.

Example 1.  Combustion of Octane in Air

Determine the stoichiometric air/ fuel mass ratio and product gas composition for combustion of octane ( C₈H₁₈)  in air.

[latex]C_{8}H_{18}+y(O_{2}+\frac{79}{21}N_{2}\rightarrow aCO_{2}+bH_{2}O+cN_{2}[/latex]

Carbon balance:

[latex]a=n=8[/latex] kmol CO₂/ kmol fuel

Hydrogen balance:

[latex]2b=18[/latex] 

[latex]b = \frac{18}{2}=9[/latex] kmol H₂O/ kmol fuel

Oxygen balance:

[latex]2y = 2(8)+b[/latex]

[latex]y = 8+\frac{9}{2}=12.5[/latex] kmol O₂/ kmol fuel 

Nitrogen balance:

[latex]y\frac{79}{21}=c=12.5\frac{79}{21}=47[/latex] kmol N₂/kmol fuel

The combustion equation becomes:

[latex]C_{8}H_{18}+12.5(O_{2}+\frac{79}{21}N_{2}\rightarrow 8CO_{2}+9H_{2}O+47N_{2}[/latex]

Air/ fuel mass based ratio considering that 1 kmol fuel is 114 kg of fuel ( 8*12 + 18 *(1) = 114):

[latex]12.5\frac{kmol\ O_{2}}{kmol\ fuel}\frac{1\ kmol\ fuel}{114\ kg\ fuel}32\frac{kg\ O_{2}}{kmol\ O_{2}}\frac{100\ kg\ air}{23\ kg\ O_{2}}=15.25\frac{kg\ air}{kg\ fuel}[/latex]

Flue gas composition on molar basis is:

Total number of kmol of flue gasses = kmol CO₂  + kmol H₂O + kmol N₂ = 8 +9+ 47 = 64 kmol flue gasses/ kmol fuel

CO₂  = 8/64 = 12.5 %

H₂O = 9/64 = 14 %

N₂ = 47/64 = 73.5%

Other components and impurities in the fuel make the calculation process more complicated. For example if sulfur exists in fuel it is usually combust into sulfur dioxide (SO₂). Ash, the noncombustible inorganic (mineral) impurities in the fuel, undergoes a number of transformations at combustion temperatures, will be neglected in the further calculation (ash will be assumed to be inert).

For most solid and liquid fuels, the chemical composition is on a mass basis, as determined in the ultimate analysis.

Mass Based Chemistry of Combustion

The combustion reactions are written following the stoichiometric rules as defined above. The quantity of matter entering into a reaction is equal to the quantity of matter in the products of the reaction.

The reaction for the complete combustion of C may be written as follows:

C+O₂=CO₂

or, if molecular weights are used,

12+32=44

1 kg C + 2⅔ kg O₂ = 3⅔ kg CO₂

The complete combustion of H₂ occurs as follows:

2H₂ +O₂ =2H₂O

4+32=36

1 kg H₂ + 8 kg O₂ = 9 kg H₂O

Sulphur burns as follows:

S +O₂ =SO₂

32+32=64

1 kg S + 1 kg O₂ = 2 kg SO₂

In addition, 1 kg of O₂ (stoichiometric mass of O₂) is contained in 1/0.232=4.3 kg air which is the stoichiometric mass of air. This air will contain 4.3-1=3.3 kg N₂. Therefore we can write:

1 kg S + 4.3 kg air = 2 kg SO₂+3.3 kg N₂

Procedure

If the analysis of fuel is given by mass, follow the steps below:

1. Total O₂ required: Determine the mass of O₂ required for each constituent and find the total mass of O₂ (Subtract any O₂ which may be in the fuel)
2. Stoichiometric air: Stoichiometric mass of air = O₂ required/0.232
3. Total mass of combustion products: Determine the mass of each combustion product. For example, given C content of 84.9%, CO₂=84.9/100*3⅔=3.11% and find the total mass of combustion products.
4. Analysis of combustion products by mass: Suppose the total mass of combustion products in step 3 has been found as 12.09 kg/kg fuel, then CO₂=3.11/12.09*100=25.74. This means that 25.74% of the flue gas is CO₂. Repeat this calculation for each constituent.

Hints & Tips

In this lab, you are carrying out two combustion analyses. For data collection, set up your trends, a sample trend plot is shown below (Make sure your trend printouts are labeled properly otherwise, data analysis will be very confusing):

To change the fuel composition use the Variable List Page#: 0111 on MD180: