Combustion calculations, excess air and flue gas analysis

Theoretical oxygen and air, percent excess air, wet and dry (Orsat) flue-gas analysis, and back-calculating excess air and fuel composition from flue gas.

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Why it matters

Furnaces, boilers, incinerators and gas turbines burn fuel in air, and their efficiency, safety and emissions depend on how much air is supplied. Combustion balances give the air requirement and flue-gas flow for fan and stack design, and working backwards from a flue-gas (Orsat) analysis tells an operator the excess air actually being used.

Key ideas

Reactions. Complete combustion turns C → CO₂, H → H₂O and S → SO₂. Incomplete combustion leaves CO (and possibly unburnt fuel or soot). Nitrogen in air is normally treated as inert.

Air. Dry air is taken as 21 mol % O₂ and 79 mol % N₂ (N₂/O₂ = 3.76), mean molar mass ≈ 29 kg/kmol (28.84 with this two-component model; 28.97 for real air). On a mass basis air is about 23.2 % O₂.

Theoretical (stoichiometric) oxygen and air. The O₂ needed to burn all the fuel completely to CO₂, H₂O and SO₂ – even if in practice some forms CO. Theoretical air = theoretical O₂ / 0.21. Oxygen already present in the fuel reduces the O₂ that air must supply.

Excess air. Percent excess air = (air fed − theoretical air) / theoretical air × 100 = (O₂ fed − theoretical O₂) / theoretical O₂ × 100. It is the same on a mole or mass basis. It is defined from complete combustion, so it does not depend on how much CO actually forms. Typical values: 5–20 % for gas, 15–30 % for oil, 20–50 % for coal. Too little air gives CO and soot; too much wastes heat in the hot flue gas.

Flue (stack) gas analysis. Reported either on a wet basis (including H₂O) or a dry basis (the Orsat analysis, which excludes water because the sample is cooled). Convert between them using the moles of water.

Back-calculation from an Orsat analysis. When the fuel contains no nitrogen, all N₂ in the flue gas came from air, so O₂ supplied = N₂ × 21/79. The free O₂ in the gas, corrected for CO (which would need ½ mol O₂ more to become CO₂), is the excess O₂.

Fuels. Gaseous fuels are given by mole fractions; liquid and solid fuels (coal, oil) by an ultimate analysis in mass % C, H, O, N, S, ash and moisture. Convert masses to kmol of atoms (C/12, H/1 as H atoms or H₂/2, S/32, O/32 as O₂) before finding O₂.

Formulas

  • C + O₂ → CO₂ ; H₂ + ½O₂ → H₂O ; S + O₂ → SO₂ ; C + ½O₂ → CO
  • CₓHᵧ + (x + y/4)·O₂ → x·CO₂ + (y/2)·H₂O
  • Theoretical O₂ = n_C + n_H/4 + n_S − n_O₂,fuel – kmol per basis of fuel; n_H in kmol of H atoms.
  • Theoretical air = theoretical O₂ / 0.21 ; N₂ with air = 3.76 × O₂ fed
  • % excess air = (O₂ fed − O₂ theoretical) / O₂ theoretical × 100
  • % excess air (from Orsat, N-free fuel) = (O₂ − 0.5·CO) / (0.266·N₂ − (O₂ − 0.5·CO)) × 100 – O₂, CO, N₂ in mol % of dry flue gas; 0.266 = 21/79.
  • Air/fuel ratio (mass) = (moles air × 28.84) / (mass of fuel)

Worked examples

Example 1 (standard). Methane is burned completely with 20 % excess air. Basis 100 mol CH₄. Find the flue-gas composition on a wet and a dry basis.

  1. CH₄ + 2O₂ → CO₂ + 2H₂O; theoretical O₂ = 200 mol.
  2. O₂ fed = 1.20 × 200 = 240 mol; N₂ = 240 × 79/21 = 902.9 mol.
  3. Products: CO₂ = 100, H₂O = 200, O₂ left = 240 − 200 = 40, N₂ = 902.9 mol; total 1242.9 mol.
  4. Wet basis: CO₂ 8.05 %, H₂O 16.09 %, O₂ 3.22 %, N₂ 72.64 %.
  5. Dry basis (remove 200 mol H₂O, total 1042.9 mol): CO₂ 9.59 %, O₂ 3.84 %, N₂ 86.58 %.

Answer: Dry (Orsat) analysis ≈ 9.6 % CO₂, 3.8 % O₂, 86.6 % N₂

Example 2 (GATE level). A hydrocarbon fuel (no N, O or S) burns in air. The Orsat analysis of the flue gas is CO₂ 10.0 %, CO 1.0 %, O₂ 4.5 %, N₂ 84.5 %. Find the H/C atomic ratio of the fuel and the percent excess air.

  1. Basis 100 mol dry flue gas. O₂ from air = 84.5 × 21/79 = 22.46 mol.
  2. Carbon in fuel = CO₂ + CO = 10.0 + 1.0 = 11.0 mol C.
  3. O₂ appearing in dry gas = CO₂ (10.0) + ½CO (0.5) + free O₂ (4.5) = 15.0 mol. The rest went to water: 22.46 − 15.0 = 7.46 mol O₂ → 14.92 mol H₂O → 29.85 mol H.
  4. H/C = 29.85 / 11.0 = 2.71.
  5. Theoretical O₂ (complete combustion) = C + H/4 = 11.0 + 7.46 = 18.46 mol.
  6. % excess air = (22.46 − 18.46) / 18.46 × 100 = 21.7 %. Check with the Orsat formula: (4.5 − 0.5) / (22.46 − 4.0) × 100 = 21.7 % ✓.

Answer: H/C ≈ 2.71; excess air ≈ 21.7 %

Common mistakes

  • Basing theoretical O₂ on the actual products (with CO) instead of complete combustion.
  • Forgetting the N₂ that comes with the air, or using 21/79 the wrong way round.
  • Mixing wet and dry flue-gas analyses.
  • Ignoring oxygen already in the fuel, or counting H atoms as H₂ molecules (n_H/4 vs n_H₂/2).
  • Using mass % of a solid fuel as if it were mole %.
  • Assuming excess air must be on a mass basis – the percentage is the same on any basis.

For GATE CH

Expect NAT questions on theoretical air for a gas or coal analysis, flue-gas composition at a given excess air, and percent excess air or fuel composition back-calculated from an Orsat analysis. Practise the N₂-tie-element method and keeping wet and dry bases separate.

Quick check

  1. What is the theoretical O₂ for 1 mol of propane (C₃H₈)?
  2. Air fed is 110 % of theoretical. What is the percent excess air?
  3. Why is water absent from an Orsat analysis?
  4. In a flue gas with 79 mol N₂ from air, how much O₂ was supplied?
  5. Does CO formation change the theoretical air?

Answers: 1. 5 mol; 2. 10 %; 3. the sample is cooled and water condenses; 4. 21 mol; 5. no – theoretical air assumes complete combustion.

Interview questions

All Process Calculations interview questions

Try answering each one aloud before you open it.

  1. 1.What is combustion in the context of chemical engineering?Concept

    Combustion is a chemical process where a fuel reacts with an oxidant, typically oxygen, to produce heat and often light. It is an exothermic reaction, meaning it releases energy. In chemical engineering, combustion is used to generate energy for various processes, such as power generation and heating.

  2. 2.Explain the concept of excess air in combustion processes.Concept

    Excess air refers to the additional air supplied to the combustion process beyond the stoichiometric requirement. It ensures complete combustion of the fuel, minimizing the production of carbon monoxide and unburned hydrocarbons. However, too much excess air can lead to energy losses as it carries away heat in the form of hot flue gases.

  3. 3.What is flue gas analysis and why is it important?Concept

    Flue gas analysis involves measuring the composition of gases emitted from combustion processes. It is important for optimizing combustion efficiency, reducing emissions, and ensuring compliance with environmental regulations. By analyzing flue gases, engineers can adjust the air-to-fuel ratio to improve performance and reduce pollutants.

  4. 4.Why is oxygen typically used as the oxidant in combustion processes?Application

    Oxygen is used as the oxidant in combustion processes because it is abundant in the atmosphere and highly reactive, making it effective for supporting combustion. Its high reactivity ensures that it can readily combine with most fuels to release energy efficiently.

  5. 5.What happens if there is insufficient air in a combustion process?Application

    If there is insufficient air in a combustion process, incomplete combustion occurs. This can lead to the production of carbon monoxide, soot, and unburned hydrocarbons, which are harmful pollutants. Incomplete combustion also results in lower energy efficiency as not all the fuel is converted into energy.

  6. 6.How does excess air affect the efficiency of a combustion process?Application

    Excess air can improve combustion efficiency by ensuring complete combustion of the fuel. However, too much excess air can reduce efficiency because it increases the volume of flue gases, which carry away heat. The key is to find an optimal balance where combustion is complete without excessive heat loss.

  7. 7.What are the typical components of flue gas from a combustion process?Concept

    Typical components of flue gas include carbon dioxide (CO₂), water vapor (H₂O), nitrogen (N₂), oxygen (O₂), and trace amounts of pollutants such as carbon monoxide (CO), sulfur dioxide (SO₂), and nitrogen oxides (NOx). The exact composition depends on the fuel used and the combustion conditions.

  8. 8.Calculate the mass of air required for the complete (stoichiometric) combustion of 1 kg of methane (CH₄). Take air as 21 mol % O₂.Numerical

    CH₄ + 2O₂ → CO₂ + 2H₂O. 1 kg CH₄ = 1000 / 16.04 = 62.34 mol, needing 2 × 62.34 = 124.7 mol O₂. Air = 124.7 / 0.21 = 593.7 mol, and with M_air ≈ 28.97 g/mol that is 17.2 kg of air per kg of CH₄ (about 13.3 Nm³). Working via volume needs the density at the same reference state as the molar volume (1.293 kg/m³ at 0 °C with 22.4 L/mol).

  9. 9.A combustion process uses 10% excess air. If the stoichiometric air requirement is 15 kg, how much air is actually used?Numerical
    1. Calculate the excess air: 10% of 15 kg = 1.5 kg.
    2. Total air used = Stoichiometric air + Excess air = 15 kg + 1.5 kg = 16.5 kg.
  10. 10.Explain why flue gas analysis is crucial for environmental compliance.Application

    Flue gas analysis is crucial for environmental compliance because it helps identify and quantify pollutants emitted from combustion processes. By monitoring emissions such as CO, SO₂, and NOx, companies can ensure they meet regulatory limits and avoid penalties. It also aids in optimizing combustion to reduce emissions and improve air quality.

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