Question 2.6: A particular coal has a mass analysis of 81% C, 5% H, 5% O, ...
A particular coal has a mass analysis of 81% C, 5% H, 5% O, 9% ash. The volumetric analysis of the dry products is: 10% CO_{2}, 1% CO, 8% O_{2}, 81% N_{2}
Determine the minimum air required for complete combustion of 1 kg of coal and the percentage of excess air required to produce the product gas proportions.
Aims:
To realize the ash content effect on the air–fuel ratio.
To determine the stoichiometric reaction.
To realize the effect of oxygen in coal on the stoichiometric reaction.
To calculate excess air supplied.
In this case it is useful to determine the required oxygen by working out for each component how much oxygen is required to burn it individually. So instead of a representative reaction equation, it is more obvious to create a table:
| Constituent | Reaction | Mass per kg of coal | Mass of O_{2} per kg | Oxygen required |
| carbon | C + O_{2} \rightarrow CO_{2} | 0.81 | 32/ 12 = 2.67 | 2.16 |
| hydrogen | H_{2} + ^{1}/_{2}O_{2} \rightarrow H_{2}O | 0.05 | 16/2 = 8 | 0.4 |
| oxygen | 0.05 | -1 | – 0.05 | |
| ash | 0.09 | 0 | 0 |
Adding the oxygen required in the last column, per kg of coal there is 2.51 kg of O_{2} required. For complete combustion, the air required for 1 kg of coal is this mass multiplied by the air:oxygen ratio by mass in the atmosphere, which is 1:0.233 = 4.29.
Therefore complete combustion requires 4.29 × 2.51 = 10.77 kg.
In order to find the actual air supplied, we have to take account of the information presented by the constituents of the product gas stream. This contains nitrogen that can only have come from the air supplied. The carbon came in a fixed proportion of the coal, and if we can work out the nitrogen to carbon ratio, we can infer the air:carbon ratio, and hence the air:coal ratio. The other useful information is that the ratio of products is given by volume, and we therefore know the molar proportions, which then give not only the mass proportions of each species molecule, but also the mass proportions of the species atoms.
For 1 kmol of product gases, there is 0.81 kmol of N_{2}, with mass 0.81 × 28 = 22.68 kg.
This is carried in with a proportion of air by mass of 1:0.767 or \frac{22.68}{0.767} = 29.57 kg
There is 0.1 kmol of CO_{2}, containing 0.1 kmol of carbon, with mass 0.1 × 12 = 1.2 kg, and there is 0.01 kmol of CO, containing 0.01 kmol of carbon, with mass 0.01 × 12 = 0.12 kg. The total amount of carbon in 1 kmol of product gas is therefore 1.32 kg, which is carried in the proportion 1:0.81 of coal to carbon. Therefore the mass of coal to carry 1.32 kg of carbon is \frac{1.32}{0.81} = 1.63 kg
The air to coal ratio is therefore 29.57:1.63 = 18.14 by mass.
Therefore percentage excess air is:
excess air % = \frac{18.14 − 10.77}{10.77}
or 68.4%.