1 Combustion Analysis
Learning Objectives
Operate the Plant at 80% capacity burning coal to
- Perform combustion analyses for two types of coal,
- Compare results.
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:
| Element | Symbol | Molecular Weight |
| Carbon | C | 12 |
| Sulphur | S2 | 32 |
| Hydrogen | H2 | 2 |
| Oxygen | O2 | 32 |
| Nitrogen | N2 | 28 |
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 CnHm, 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, CnHm, becomes
[latex]C_{n}H_{m}+y(O_{2}+\frac{79}{21}N_{2}\rightarrow aCO_{2}+bH_{