Environmental Impacts of Energy Use: Air Pollutants, Greenhouse Gases, Emission Factors and Inventories

This chapter answers Section XE8.2 of XE8 Energy Science, the sub-heading “Environmental impacts of energy use”: “Air pollution (SOx, NOx, CO, particulates), greenhouse gas emissions and their sources, emission factors and inventories.” It belongs to the Engineering Sciences (XE) paper, as Section XE8.2 of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. The chapter teaches how each pollutant forms and what it does, the greenhouse gases and the way global warming potential converts them to a common scale, stoichiometric emission calculations, and the emission-factor method that builds an inventory. The numericals are CO₂ and SO₂ from fuel composition, dry flue-gas composition, ppm-to-mass conversion and oxygen correction, CO₂-equivalent, inventories from stated emission factors, and the efficiency of a precipitator; every factor is given in the question.

1. Air pollutants: SOx, NOx, CO and particulates

Combustion of a fossil fuel produces CO₂ and water in the ideal case, and a set of pollutants in the real one. Each has its own origin, and so its own remedy. The four named in the syllabus are best learned as a table of source, formation mechanism, effect and control.

The four pollutants
PollutantOriginEffectMain control
SOx (mainly SO₂, a little SO₃)Oxidation of sulphur in the fuel: S + O₂ → SO₂; coal and heavy oil are highRespiratory harm; acid rain (sulphuric acid); corrosionLow-sulphur fuel; flue-gas desulphurisation (limestone or lime scrubbing)
NOx (NO, NO₂)Thermal (N₂ + O at high flame temperature), fuel-bound nitrogen, and prompt NOx in the flame frontSmog and ground-level ozone with hydrocarbons and sunlight; acid rain; lung irritationLower peak flame temperature (staged air, flue-gas recirculation, low-NOx burners); SCR with ammonia
COIncomplete combustion: too little air, poor mixing, quenching on cold surfacesBinds to haemoglobin and reduces oxygen delivery; toxic indoorsAdequate air and mixing, good burner tuning
Particulates (ash, soot)Mineral ash in coal carried in flue gas; soot from incomplete combustion of oil and gasLung penetration; haze; PM₁₀ and PM₂.₅ are particles of aerodynamic diameter up to 10 µm and 2.5 µmCyclone, electrostatic precipitator, fabric filter
🎯 The combustion trade-off
More excess air burns CO out but cools the flame (less thermal NOx) and carries more heat up the stack; hotter, better-mixed flames burn out CO and soot but raise thermal NOx. No single setting minimises everything, so burner tuning balances CO, NOx and stack loss.

Two further points are needed for calculation. NOx is reported as the mass of NO₂ equivalent by convention (always check what the question states). Fuel-bound pollutants (SO₂ from sulphur, CO₂ from carbon) follow the fuel analysis exactly, whereas combustion-dependent ones (NOx, CO, soot) depend on the burner and its settings, so they cannot be derived from fuel composition alone.

2. Greenhouse gases, sources and CO₂-equivalent

Greenhouse gases absorb outgoing infrared radiation and warm the lower atmosphere. The main ones from human activity are carbon dioxide (CO₂), methane (CH₄), nitrous oxide (N₂O) and the fluorinated gases (hydrofluorocarbons, perfluorocarbons, sulphur hexafluoride). Water vapour is also a greenhouse gas, but its amount responds to temperature and is not set directly by emissions.

  • CO₂: burning coal, oil and gas (the largest source in energy systems), and the calcination of limestone in cement-making, which releases CO₂ from the raw material itself.
  • CH₄: livestock, rice paddies, landfills, leakage and venting in coal mining and in oil and gas systems, and incomplete combustion.
  • N₂O: fertilised soils, some industrial processes such as nitric-acid production, and some combustion processes.
  • Fluorinated gases: refrigerants and foam-blowing agents, and SF₆ as an insulating gas in electrical switchgear; present in tiny amounts but long-lived and strongly absorbing.

Gases are put on a common scale with the global warming potential (GWP): the warming caused by 1 kg of the gas over a stated time horizon (commonly 100 years) relative to 1 kg of CO₂, so GWP(CO₂) = 1. The CO₂-equivalent of an emission is CO₂e = Σ mᵢ × GWPᵢ. GWP values depend on the horizon chosen and on which scientific assessment they are taken from, so a question must state them. Worked, with GWP(CH₄) = 28 and GWP(N₂O) = 265 as given: 1000 kg CH₄ and 10 kg N₂O are 28 000 + 2650 = 30 650 kg CO₂e.

⚠️ A big GWP does not mean a big contribution
GWP is per kilogram. A gas with a huge GWP but a tiny emitted mass can matter less than CO₂, whose GWP is 1 but whose mass is billions of tonnes. Always multiply by the mass emitted before comparing.

3. Stoichiometric emission calculations

Take atomic masses C = 12, H = 1, O = 16, S = 32 (kg/kmol). Each complete reaction fixes the emission per unit of fuel element: C + O₂ → CO₂ gives 44/12 = 3.667 kg CO₂ per kg carbon; S + O₂ → SO₂ gives 64/32 = 2 kg SO₂ per kg sulphur (if all sulphur is oxidised and none is retained in ash); CH₄ + 2O₂ → CO₂ + 2H₂O gives 44/16 = 2.75 kg CO₂ per kg methane. For a coal with 72 % carbon, burning 200 kg/h emits 200 × 0.72 × 44/12 = 528 kg CO₂/h.

Flue-gas composition. Take air as 21 % O₂ and 79 % N₂ by volume (3.76 kmol N₂ per kmol O₂). For methane with theoretical air: CH₄ + 2O₂ + 7.52N₂ → CO₂ + 2H₂O + 7.52N₂. Dry products are 1 + 7.52 = 8.52 kmol, so dry CO₂ = 1/8.52 = 11.74 %. With 20 % excess air the extra 0.4 kmol O₂ (and 1.504 kmol N₂) appear unreacted: dry products = 1 + 0.4 + 9.024 = 10.424 kmol, CO₂ = 9.59 % and O₂ = 3.84 %. Excess air dilutes every pollutant, so measured concentrations must be corrected to a common oxygen level.

  • ppm to mass: for a gas of molar mass M, C (mg/Nm³) = ppm × M/22.414, where 22.414 L/mol is the molar volume at 0 °C and 1 atm (“normal” conditions). 500 ppm SO₂ (M = 64) is 500 × 64/22.414 = 1428 mg/Nm³.
  • Oxygen correction: C_ref = C_meas × (21 − O₂,ref)/(21 − O₂,meas), with oxygen in % by volume on the same (dry) basis. 300 mg/Nm³ NOx measured at 6 % O₂ is 300 × 18/15 = 360 mg/Nm³ at a 3 % O₂ reference, since the dilution by excess air is removed.
⚠️ Do not shrink the number by diluting the stack
Adding air before the sampling point lowers the measured concentration but not the mass emitted. This is why standards quote concentrations at a reference oxygen and why the correction multiplies by (21 − O₂,ref)/(21 − O₂,meas), which is above 1 when the sample was diluted.

4. Emission factors and inventories

An emission factor (EF) is the mass of a pollutant released per unit of an activity: kg CO₂ per GJ of fuel, g NOx per kWh of electricity, kg SO₂ per tonne of coal. An inventory multiplies each activity (the quantity of fuel burned, electricity bought, product made) by its EF and adds up: E = Σ Aᵢ × EFᵢ. If a control device of removal efficiency η is fitted, E = A × EF × (1 − η).

For greenhouse-gas inventories of an organisation, emissions are grouped by the widely used scope convention: Scope 1 direct emissions from sources the organisation owns (boilers, furnaces, vehicles, process emissions); Scope 2 indirect emissions from the generation of the electricity, steam or heat it buys; Scope 3 all other indirect emissions in the value chain (purchased materials, transport, use of products). A purchased kWh carries the grid emission factor, the average kg CO₂ per kWh of the supply, so cutting purchased electricity cuts Scope 2 by that factor per kWh.

Worked example: annual CO₂ inventory with stated factors
SourceActivity × factorEmission
Scope 1: gas boiler10 000 GJ × 56 kg/GJ560 t CO₂
Scope 2: grid electricity2.0 × 10⁶ kWh × 0.82 kg/kWh1640 t CO₂
Total560 + 16402200 t CO₂
After saving 10 % of grid electricity0.10 × 1640164 t saved
ℹ️ Why CO₂ factors are firm and NOx factors are not
The CO₂ factor of a fuel follows from its carbon content and calorific value, so it is nearly the same wherever the fuel is burned. A NOx or CO factor depends on the burner type, load, excess air and maintenance, so published factors are averages with wide scatter, and a measured value at the plant is better when it is available.

5. Control efficiency and the precipitator

The removal efficiency of a control device is η = (C_in − C_out)/C_in = 1 − C_out/C_in, or equivalently by mass flow. Devices in series combine through their penetrations: the fraction passing is the product of the individual (1 − η), so η_total = 1 − (1 − η₁)(1 − η₂). A cyclone of 70 % followed by a fabric filter of 99 % passes 0.30 × 0.01 = 0.003, so η_total = 99.7 %. A scrubber taking 90 % of 12 t/h of SO₂ leaves 12 × 0.10 = 1.2 t/h to the stack.

In an electrostatic precipitator the particles are charged and drawn to collecting plates. For ideal plug flow with migration velocity w towards the plates, collecting area A and gas flow Q, the Deutsch–Anderson equation gives η = 1 − exp(−wA/Q). The exponent wA/Q = 4 gives η = 1 − e⁻⁴ = 98.2 %; doubling the plate area to wA/Q = 8 gives 99.97 %. Efficiency rises with area and falls with gas flow, and fine particles with low w need large plate area.

⚠️ Control device and pollutant must match
A precipitator or fabric filter removes particles, not gases. SO₂ needs a scrubber or sorbent, and NOx needs combustion modification or selective catalytic reduction with ammonia. Limestone scrubbing takes 100 kg CaCO₃ per 64 kg SO₂ at a molar ratio of 1, since CaCO₃ + SO₂ → CaSO₃ + CO₂.

Key takeaways

  • SOx is fuel-bound (S + O₂ → SO₂, 2 kg SO₂ per kg S); NOx is mostly thermal and falls with lower flame temperature; CO signals incomplete combustion; particulates come from ash and soot.
  • Greenhouse gases are CO₂, CH₄, N₂O and fluorinated gases; GWP puts them on a CO₂ scale per kilogram, and CO₂e = Σ m × GWP with the GWP values stated.
  • Stoichiometry fixes CO₂ at 44/12 kg per kg carbon; dry CO₂ in flue gas falls as excess air rises, and concentration is corrected to a reference O₂ by (21 − O₂,ref)/(21 − O₂,meas).
  • An inventory is E = Σ activity × emission factor, with (1 − η) for control; CO₂ factors are firm, NOx and CO factors depend on combustion conditions.
  • Scope 1 is direct, Scope 2 is purchased electricity, steam or heat, and Scope 3 is the rest of the value chain.
  • Control efficiencies in series combine through penetrations; the precipitator obeys η = 1 − exp(−wA/Q); each device removes only the pollutant class it is built for.

Practice questions (16)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A gas-fired furnace is modified by recirculating part of the flue gas into the combustion air, which lowers the peak flame temperature. The pollutant most directly reduced is

    1. SO₂, because its formation is governed mostly by the flame temperature
    2. unburnt ash, because a cooler flame leaves less mineral matter to be carried along
    3. CO₂, because at the lower temperature less of the fuel carbon is oxidised
    4. thermal NOx, because N₂ oxidation rises steeply with flame temperature
    Show answer

    Answer: D — thermal NOx, because N₂ oxidation rises steeply with flame temperature

    Thermal NOx forms from atmospheric N₂ and O at high temperature and is steeply temperature dependent, so cooling the flame cuts it. SO₂ is fixed by fuel sulphur, and CO₂ by fuel carbon, whatever the flame temperature; gas fuel has no mineral ash to reduce.

  2. On a boiler, as the excess air is reduced below a certain level the CO in the flue gas rises sharply. The most direct explanation is that

    1. the available oxygen and mixing are no longer enough to oxidise all the CO to CO₂
    2. the flame temperature rises so much that CO₂ dissociates entirely into solid carbon and oxygen
    3. the sulphur in the fuel is converted to CO instead of SO₂ when air is short
    4. nitrogen in the air reacts with the fuel carbon directly to form CO
    Show answer

    Answer: A — the available oxygen and mixing are no longer enough to oxidise all the CO to CO₂

    Below the needed air the carbon cannot all reach CO₂, so CO (and soot) appear; this is why oxygen trim stops short of zero excess air. The other mechanisms do not occur: sulphur gives SO₂, and N₂ does not form CO.

  3. A coal contains 2.0 % sulphur by mass, all of which is oxidised to SO₂ with none retained in the ash. Atomic masses: S = 32, O = 16. The SO₂ released per tonne of coal burned is

    1. 20 kg, taking the SO₂ mass equal to the sulphur mass in the coal
    2. 40 kg, since SO₂ is twice as heavy as its sulphur
    3. 32 kg, from using the molar mass of sulphur in place of the coal fraction
    4. 64 kg, from using the molar mass of SO₂ in place of the coal fraction
    Show answer

    Answer: B — 40 kg, since SO₂ is twice as heavy as its sulphur

    Sulphur in 1 t is 20 kg. S + O₂ → SO₂ gives 64/32 = 2 kg SO₂ per kg S, so 40 kg. Quoting the sulphur mass itself forgets the added oxygen.

  4. An analyser reads 300 mg/Nm³ of NOx in dry flue gas containing 6 % O₂ by volume. A regulation quotes NOx at a 3 % O₂ reference, using C_ref = C_meas × (21 − O₂,ref)/(21 − O₂,meas). The reference-basis concentration is

    1. 250 mg/Nm³, from inverting the ratio of the two oxygen terms
    2. 360 mg/Nm³, from 300 × 18/15
    3. 150 mg/Nm³, from the direct ratio of the oxygen readings, 300 × 3/6
    4. 600 mg/Nm³, from the direct ratio of the oxygen readings, 300 × 6/3
    Show answer

    Answer: B — 360 mg/Nm³, from 300 × 18/15

    (21 − 3)/(21 − 6) = 18/15 = 1.2, so 300 × 1.2 = 360 mg/Nm³. The sample was diluted by excess air (6 % O₂ against 3 %), so removing the dilution raises the number; the inverted ratio would lower it, and the simple oxygen ratios have no basis in the correction.

  5. The CO₂-equivalent of a greenhouse-gas emission is obtained by

    1. multiplying each gas’s emitted mass by its GWP and summing
    2. adding the emitted masses of all the gases together without any weighting factor
    3. dividing each gas’s emitted mass by its molar mass and summing
    4. multiplying the emitted masses by the fraction of carbon in each gas
    Show answer

    Answer: A — multiplying each gas’s emitted mass by its GWP and summing

    CO₂e = Σ m × GWP expresses each gas as the mass of CO₂ with the same warming over the stated horizon. Unweighted addition ignores the large differences in warming per kg, and molar or carbon fractions measure composition, not warming.

  6. Under the scope convention for greenhouse-gas inventories, the emissions arising at a power station from the electricity a factory buys from the grid are counted in the factory’s inventory as

    1. Scope 1, as direct emissions from sources that the factory owns and controls
    2. Scope 2, as indirect emissions from purchased electricity
    3. Scope 3, as emissions from the use of the products that the factory sells
    4. not counted at all, since the factory itself has no stack for them
    Show answer

    Answer: B — Scope 2, as indirect emissions from purchased electricity

    Purchased electricity, steam and heat form Scope 2; direct emissions from owned boilers and vehicles are Scope 1, and the factory is responsible for Scope 2 through its consumption even though it has no stack for them.

  7. Which of the following pairs of control technology and the pollutant it is mainly designed to remove is correct?

    1. an electrostatic precipitator, mainly for removing gaseous SO₂ from flue gas
    2. a fabric filter, mainly for removing gaseous NOx from flue gas
    3. a cyclone separator, mainly for removing gaseous CO from flue gas
    4. selective catalytic reduction with ammonia, for NOx
    Show answer

    Answer: D — selective catalytic reduction with ammonia, for NOx

    SCR reduces NOx to N₂ and water over a catalyst using ammonia. Precipitators, fabric filters and cyclones collect particles and do nothing for gases such as SO₂, NOx or CO.

  8. A plant manager notes that a published CO₂ emission factor for a fuel can be applied to any boiler, but a published NOx factor gives a poor estimate for a particular boiler. The reason is that

    1. CO₂ is not formed in boilers at all, so any factor at all will do for CO₂ and NOx alike
    2. NOx is formed only from nitrogen in the fuel and never from the nitrogen in the air
    3. CO₂ follows from the fuel carbon, while NOx depends on burner design, load, excess air and flame temperature
    4. CO₂ is measured in ppm and NOx in percent, so the two emission factors cannot be compared
    Show answer

    Answer: C — CO₂ follows from the fuel carbon, while NOx depends on burner design, load, excess air and flame temperature

    Complete combustion fixes CO₂ per unit of fuel carbon; NOx is set by combustion conditions, so a generic factor can be far off. NOx forms from both fuel nitrogen and air nitrogen, and CO₂ is formed in all hydrocarbon combustion.

  9. Which of the following statements about particulate matter are correct?

    1. PM₂.₅ refers to particles of aerodynamic diameter up to 2.5 µm
    2. Fine particles penetrate deeper into the lungs than coarse particles
    3. Particulates come only from mineral ash and never from incomplete combustion
    4. A cyclone separator is less effective on fine particles than on coarse ones
    Show answer

    Answer: A — PM₂.₅ refers to particles of aerodynamic diameter up to 2.5 µm; B — Fine particles penetrate deeper into the lungs than coarse particles; D — A cyclone separator is less effective on fine particles than on coarse ones

    PM sizes are defined by aerodynamic diameter, finer particles go deeper, and cyclones rely on inertia, which is weak for small particles. Soot from incomplete combustion of oil, gas and biomass is also particulate matter.

  10. Which of the following statements about greenhouse-gas sources are correct?

    1. Methane is emitted from landfills, rice paddies and livestock
    2. Nitrous oxide is emitted from fertilised soils
    3. Carbon dioxide arises only from the burning of fuels
    4. Cement manufacture releases CO₂ from the limestone as well as from the fuel
    Show answer

    Answer: A — Methane is emitted from landfills, rice paddies and livestock; B — Nitrous oxide is emitted from fertilised soils; D — Cement manufacture releases CO₂ from the limestone as well as from the fuel

    Landfills, paddies and livestock are standard methane sources, fertilised soils give N₂O, and calcination of limestone adds process CO₂ in cement. CO₂ also arises from such process sources, so the claim that it comes only from fuel is false.

  11. Which of the following statements about combustion emissions are correct?

    1. Lowering the peak flame temperature by staging or flue-gas recirculation reduces thermal NOx
    2. Adding more excess air always reduces both NOx and CO in every boiler
    3. Burning a fuel of lower sulphur content reduces SO₂ emission in proportion
    4. For complete combustion the CO₂ released per unit mass of fuel carbon is fixed by stoichiometry
    Show answer

    Answer: A — Lowering the peak flame temperature by staging or flue-gas recirculation reduces thermal NOx; C — Burning a fuel of lower sulphur content reduces SO₂ emission in proportion; D — For complete combustion the CO₂ released per unit mass of fuel carbon is fixed by stoichiometry

    Thermal NOx falls with flame temperature, SO₂ is proportional to fuel sulphur, and CO₂ is 44/12 per unit carbon in complete combustion. More air tends to reduce CO but can raise NOx where the added oxygen outweighs the cooling, so the claim of always reducing both is false.

  12. A factory in a year buys 2.0 × 10⁶ kWh of grid electricity and burns 10 000 GJ of natural gas. Take the grid emission factor as 0.82 kg CO₂ per kWh and the natural-gas factor as 56 kg CO₂ per GJ. The total annual CO₂ emission counted in its inventory is ____ tonnes (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 2200

    Electricity: 2.0 × 10⁶ × 0.82 = 1.64 × 10⁶ kg = 1640 t. Gas: 10 000 × 56 = 5.6 × 10⁵ kg = 560 t. Total = 2200 t CO₂. Forgetting the division by 1000 to convert kg to tonnes is the usual slip.

  13. A coal-fired boiler burns 500 t of coal per hour containing 1.2 % sulphur by mass. All sulphur is oxidised to SO₂ (S = 32, O = 16), and a flue-gas desulphurisation unit removes 90 % of the SO₂. The SO₂ released to the atmosphere is ____ t/h (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 1.2

    Sulphur = 500 × 0.012 = 6 t/h. SO₂ formed = 6 × 64/32 = 12 t/h. Released = 12 × (1 − 0.90) = 1.2 t/h. Stopping at 12 t/h omits the scrubber, and using 6 t/h forgets the oxygen.

  14. Methane burns completely with 20 % excess air: CH₄ + 2(1.2)(O₂ + 3.76N₂) → products. Treat air as 21 % O₂ and 79 % N₂ by volume. The volume percentage of CO₂ in the dry flue gas is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 9.6

    Air supplied: 2.4 kmol O₂ with 9.024 kmol N₂. Dry products: CO₂ 1; O₂ left 2.4 − 2 = 0.4; N₂ 9.024; total 10.424 kmol. CO₂ = 1/10.424 = 9.59 %, so 9.6. The theoretical-air value is 11.7 %, which this excess dilutes.

  15. A site emits 1000 kg of CH₄, 10 kg of N₂O and 5000 kg of CO₂ in a year. Taking GWP values over the chosen horizon as 28 for CH₄ and 265 for N₂O (and 1 for CO₂), the total emission is ____ kg CO₂-equivalent (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 35650

    CO₂e = 5000 × 1 + 1000 × 28 + 10 × 265 = 5000 + 28 000 + 2650 = 35 650 kg. Adding the raw masses gives only 6010 kg, which ignores the weighting.

  16. An electrostatic precipitator treats a gas flow of 60 m³/s. It has a total collecting plate area of 3000 m² and an effective migration velocity of 0.08 m/s. Using the Deutsch–Anderson equation η = 1 − exp(−wA/Q), the collection efficiency is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 98.2

    wA/Q = 0.08 × 3000/60 = 4.0. η = 1 − e⁻⁴ = 1 − 0.018316 = 0.98168 = 98.2 %. Using wA/Q directly as a fraction is not possible here, since it exceeds 1.