How do you find the volume of air at STP needed to burn coal, including excess air?

In one line

Convert the minimum mass of oxygen into moles, multiply by 22.4 to get its volume at STP, then divide by 0.21 because air contains 21% oxygen by volume.

Explained simply

Blowers and fans in a boiler house are rated in cubic metres of air per minute, not kilograms. So engineers need the air in volume. At STP, one mole of any gas occupies 22.4 L, and one kilomole occupies 22.4 m³. That is the bridge between mass and volume.

How it works

1. Find minimum O₂ by weight: \( \frac{32}{12}C + 8H + S - O \) (kg per kg of coal).
2. Moles of O₂ = mass/32 (kmol when mass is in kg).
3. Volume of O₂ at STP = kmol × 22.4 m³.
4. Volume of air = volume of O₂ × 100/21.
5. If excess air is asked, multiply by (1 + % excess/100).

Excess air is supplied in practice because mixing of fuel and air is never perfect; without it some carbon burns only to CO (wasting heat and causing pollution). Too much excess air, however, carries heat away up the chimney.

Key formula

\( V_{O_2} = \dfrac{\text{Min. O}_2\ (\text{kg})}{32} \times 22.4\ \text{m}^3 \)

\( V_{\text{air}} = V_{O_2} \times \dfrac{100}{21} \)

• 22.4 m³ = volume of 1 kmol of gas at STP (0 °C, 1 atm).
• 21 = volume % of O₂ in air.

Quick example

Coal: C 80%, H 5%, O 6%, S 1%. Min. O₂ = 2.4833 kg per kg (from the weight method).

• kmol O₂ = 2.4833/32 = 0.07760
• V_O2 = 0.07760 × 22.4 = 1.738 m³
• V_air = 1.738 × 100/21 = 8.28 m³
• With 40% excess air: 8.28 × 1.4 = 11.59 m³

For gaseous fuels the same idea is simpler: volumes react in mole ratio, e.g., CH₄ + 2O₂ → CO₂ + 2H₂O, so 1 m³ CH₄ needs 2 m³ O₂ = 2 × 100/21 = 9.52 m³ air.

Common mistakes

• Using 23% (the weight figure) for volume; use 21%.
• Dividing by 12 or 16 instead of 32 for O₂ moles.
• Applying excess air to the oxygen but forgetting it in the final air figure (or applying it twice).

How MU asks it

Usually combined with the weight calculation: "Calculate the weight and volume of air required for complete combustion of 1 kg coal... if 30% excess air is used" (5–6 marks). For full marks show both bases clearly (23% by weight, 21% by volume), the 22.4 m³ conversion, and the excess-air step with final units.