In one line
Using balanced combustion equations, find the oxygen needed by C, H and S, subtract the oxygen already present in the fuel, and convert oxygen to air using 23% O2 by weight.
Explained simply
It is like a recipe. If 12 g of carbon always needs 32 g of oxygen, then from the coal's composition you can work out exactly how much oxygen the whole kilogram needs. Air is only about one-quarter oxygen by weight, so you need much more air than oxygen.
How it works
- Write the reactions with masses:
- C + O2 → CO2: 12 kg C needs 32 kg O2 (factor 32/12)
- H2 + ½O2 → H2O: 2 kg H needs 16 kg O2 (factor 8)
- S + O2 → SO2: 32 kg S needs 32 kg O2 (factor 1)
- Convert percentages to kg in 1 kg coal (80% C = 0.80 kg).
- Multiply each by its factor and add.
- Subtract O2 already present in the coal.
- Air by weight = O2 × 100/23.
- N, ash and moisture need no oxygen.
Key formula
\( \text{Min. O}_2 \text{ (kg)} = \dfrac{32}{12}C + 8H + S - O \)
\( \text{Min. air (kg)} = \text{Min. O}_2 \times \dfrac{100}{23} \)
C, H, S, O are masses in kg per kg of coal. Actual air supplied = min. air × (1 + excess fraction).
Quick example
Coal: C 80%, H 5%, O 6%, S 1%, N 2%, ash 6%. For 1 kg:
| Constituent | Mass (kg) | O2 needed (kg) |
|---|---|---|
| C | 0.80 | 0.80 × 32/12 = 2.1333 |
| H | 0.05 | 0.05 × 8 = 0.4000 |
| S | 0.01 | 0.01 × 1 = 0.0100 |
| Total | 2.5433 |
Less O2 in coal 0.06 kg → min. O2 = 2.4833 kg. Min. air = 2.4833 × 100/23 = 10.80 kg. With 20% excess air: 10.80 × 1.2 = 12.96 kg.
Common mistakes
- Using 21% (the volume figure) instead of 23% for weight of air.
- Adding the fuel's oxygen instead of subtracting it.
- Giving oxygen to nitrogen or ash.
How MU asks it
"Calculate the minimum quantity of air required for complete combustion of 1 kg of coal with composition..." (5 marks). Write reactions with molar masses, a table like the one above, the subtraction of fuel oxygen, and the conversion to air with the 23% basis. If a mass other than 1 kg is given (e.g., 2 kg or 5 kg), multiply at the end.