Air Pollution: The Atmosphere, Pollutants and Standards, Meteorology and Dispersion

The first of two chapters for Section 6 of the GATE Environmental Science and Engineering (ES) paper, Air and Noise Pollution. The section’s five sub-headings split cleanly into what air pollution is and how it moves, and how it is removed and measured — plus noise, a different physical problem. This chapter takes the first and the fourth sub-headings: the structure and composition of the atmosphere; natural and anthropogenic sources; the atmospheric processes of emission, transformation, transport and removal; indoor air pollution; the effects on health and the environment; gases and particulate matter, primary and secondary and criteria pollutants; ambient and emission standards, air quality indices, visibility and radiative effects; and air quality management — point, line and area sources, the emission inventory, meteorology and dispersion, stability and inversion, mixing height, wind roses, Gaussian plume modelling and the influence of topography. The next chapter takes the control of particulates and gases, vehicular emissions and noise.

1. Structure and composition of the atmosphere; sources and atmospheric processes

The atmosphere is layered by its temperature profile. In the troposphere, from the ground to about 8 km at the poles and 16–18 km at the equator, temperature falls with height at about 6.5 °C/km on average; it holds most of the mass and all the weather, and nearly all air pollution lives in its lowest kilometre or two. Above the tropopause the stratosphere warms with height to about 50 km because its ozone absorbs solar ultraviolet; the mesosphere cools again to about 85 km, and the thermosphere warms above it. Dry air is about 78% N₂, 21% O₂ and 0.93% Ar by volume, with CO₂ now above 400 ppm, and trace amounts of Ne, He, CH₄, N₂O, H₂ and O₃; water vapour varies from nearly zero to about 4%.

Natural sources include volcanoes (SO₂, ash), wind-blown dust, sea salt, forest fires, pollen and the biogenic VOCs (isoprene, terpenes) of vegetation. Anthropogenic sources are dominated by combustion — power plants and industry (SO₂, NOₓ, particulates), vehicles (CO, NOₓ, hydrocarbons, fine particles), domestic solid fuels and open burning of crop residue and waste — together with industrial processes, solvents and agriculture, which emits most of the ammonia. Once emitted, pollutants go through four atmospheric processes: emission; transformation, chiefly oxidation by the OH radical (SO₂ to sulphate, NO₂ to nitric acid and nitrate, VOCs to secondary organic aerosol, and the NOₓ–VOC photochemistry that makes ozone); transport by the mean wind and dilution by turbulence; and removal by dry deposition (settling and uptake at surfaces) and wet deposition (in-cloud rainout and below-cloud washout).

ℹ️ Converting ppm to μg/m³
At 25 °C and 1 atm one mole of gas occupies 24.45 L, so C (μg/m³) = ppm × M × 1000/24.45. For SO₂ (M = 64), 0.1 ppm = 0.1 × 64 × 1000/24.45 = 262 μg/m³. At 0 °C the molar volume is 22.4 L and the same ppm gives a larger mass concentration.

2. Gases and particulates, primary and secondary pollutants, indoor air, and effects

A primary pollutant is emitted as such (SO₂, NO, CO, soot, lead); a secondary pollutant forms in the air from precursors (ozone, sulphate and nitrate aerosol, peroxyacetyl nitrate, most PM₂.₅ mass in many cities). Particulate matter is classed by aerodynamic diameter: PM₁₀ (inhalable, reaching the airways) and PM₂.₅ (fine, reaching the alveoli), with coarse particles mostly mechanical in origin and fine particles mostly from combustion and secondary formation. The criteria pollutants — those regulated by ambient standards on the basis of health criteria — are classically six: particulate matter, ozone, carbon monoxide, sulphur dioxide, nitrogen dioxide and lead; India’s National Ambient Air Quality Standards (2009) cover twelve, adding PM₂.₅ separately, ammonia, benzene, benzo(a)pyrene, arsenic and nickel.

Indoor air pollution can exceed outdoor levels by far. Cooking and heating with solid fuels in poorly ventilated homes produce high PM₂.₅ and CO; building materials and furnishings release formaldehyde and other VOCs; soil gas brings radon, a radioactive decay product of uranium and a cause of lung cancer; tobacco smoke, combustion appliances (NO₂), moulds and bioaerosols add to it; and "sick building syndrome" describes complaints tied to poor ventilation. The effects on health follow from where a pollutant acts: fine particles penetrate to the alveoli and are linked to cardiovascular and respiratory disease; CO binds haemoglobin some 200 times more strongly than oxygen, forming carboxyhaemoglobin and starving tissues of oxygen; SO₂ and NO₂ irritate the airways; ozone inflames the lungs; lead is a neurotoxin, especially for children. The effects on the environment include acid deposition on soils, lakes and buildings, ozone injury to crops and forests, fluoride damage to vegetation, soiling and corrosion of materials, and haze.

⚠️ Ozone is not emitted
Ground-level ozone is the textbook secondary pollutant: it forms from NOₓ and VOCs in sunlight, so it peaks in the afternoon and often downwind of a city rather than at its centre, where fresh NO titrates it (NO + O₃ → NO₂ + O₂). Cutting NOₓ at the centre can therefore raise ozone there before it lowers it regionally.

3. Ambient and emission standards, air quality indices, visibility and radiative effects

An ambient standard limits the concentration in the air people breathe, for a stated averaging time; an emission standard limits what a source may release, as a concentration in the stack gas, a mass rate or a mass per unit of product. Ambient standards set the goal; emission standards, stack-height rules and fuel standards are the tools that meet it. Each pollutant is set against averaging periods that match its harm — 24 hours and a year for particulates, 8 hours for CO and ozone, 1 hour for peaks — and compliance is judged over those periods, not at an instant.

Selected National Ambient Air Quality Standards, India (2009), industrial, residential, rural and other areas
PollutantAnnual24-hour (or shorter)
PM₁₀ (μg/m³)60100
PM₂.₅ (μg/m³)4060
SO₂ (μg/m³)5080
NO₂ (μg/m³)4080
O₃ (μg/m³)—100 (8-hour); 180 (1-hour)
CO (mg/m³)—2 (8-hour); 4 (1-hour)
Lead (μg/m³)0.51.0

An air quality index turns several pollutants into one number for the public. Each pollutant’s concentration C_p is converted to a sub-index by linear interpolation within its band, I_p = I_lo + (I_hi − I_lo)(C_p − BP_lo)/(BP_hi − BP_lo), and the AQI is the maximum sub-index, reported with the responsible pollutant. India’s National AQI uses eight pollutants and six categories from Good (0–50) to Severe (401–500). Visibility is reduced by particles that scatter and absorb light; the visual range follows Koschmieder’s relation L_v = 3.912/b_ext, with b_ext the extinction coefficient, and fine particles near the wavelength of light (0.1–1 μm) are the most effective. The radiative effects of aerosols are direct — sulphate scatters sunlight and cools, black carbon absorbs and warms — and indirect, as particles act as cloud condensation nuclei and make clouds brighter and longer-lived.

⚠️ The AQI is a maximum, not a mean
If PM₂.₅ is 105 μg/m³ and its band is 91–120 μg/m³ for index values 201–300, its sub-index is 201 + (99/29) × 14 = 248.8. If every other pollutant scores below 100, the AQI is still 249, "Poor", driven by PM₂.₅. Averaging the sub-indices would hide exactly the pollutant that needs action.

4. Air quality management: sources, inventory, stability, inversion, mixing height and wind roses

For management, sources are grouped by geometry: point sources (stacks), line sources (highways, railways), and area sources (a city’s domestic fuel use, a landfill, a field being burnt). An emission inventory estimates each source’s release as activity × emission factor × (1 − control efficiency) — tonnes of coal burnt times kg SO₂ per tonne, vehicle-kilometres times g/km — and adds them by pollutant, sector and grid cell. It is the starting point for dispersion modelling, for apportioning ambient pollution among sources, and for choosing where control buys the most improvement.

Dispersion is governed by atmospheric stability, found by comparing the environmental lapse rate (ELR) with the dry adiabatic lapse rate Γ_d = g/c_p ≈ 9.8 °C/km, at which a rising parcel of dry air cools. If the environment cools faster than Γ_d (superadiabatic), a displaced parcel keeps rising and the air is unstable; if it cools at Γ_d it is neutral; if slower (subadiabatic), isothermal or warming with height (inversion), it is stable and vertical mixing is suppressed. Inversions form by radiation from the ground on clear, calm nights, by subsidence of air in high-pressure systems, at fronts, and by advection of warm air over a cold surface. The mixing height is the depth through which pollutants mix: the height at which the dry adiabat drawn from the afternoon maximum surface temperature meets the environmental profile. With the mean wind through that layer it gives the ventilation coefficient, mixing height × wind speed (m²/s), a single measure of how well a region disperses what it emits. Wind speed increases with height, often represented by the power law u = u₁(z/z₁)^p. A wind rose shows, for a site, how often the wind blows from each direction and at what speed — used to place industries downwind of towns and to site monitors.

Stability and plume shape
ConditionPlumeGround-level impact
Superadiabatic (unstable), sunny dayLooping — large eddies carry it up and downHigh, intermittent, close to the stack
Neutral, overcast or windyConing — a symmetrical coneModerate, further downwind
Inversion throughout (stable night)Fanning — spreads sideways, little verticallyLow while it lasts
Unstable below, inversion above the stackFumigation — mixed down to the groundThe worst case: high concentrations
Inversion below, unstable aboveLofting — dispersed upwardThe best case: little reaches the ground

5. Gaussian plume dispersion, plume rise and topographic influences

The Gaussian plume model treats a continuous point source of strength Q (g/s) in a steady wind u (m/s) at the stack height, with the plume spreading as normal distributions across wind (σ_y) and vertically (σ_z), both growing with distance downwind at rates set by the Pasquill stability class (A, very unstable, to F, stable). With the ground treated as a reflecting surface by an image source, C(x, y, z) = [Q/(2πuσ_yσ_z)] exp(−y²/2σ_y²) {exp[−(z − H)²/2σ_z²] + exp[−(z + H)²/2σ_z²]}. At ground level on the plume centreline (y = 0, z = 0) this becomes C = [Q/(πuσ_yσ_z)] exp(−H²/2σ_z²), and for a ground-level source (H = 0) simply Q/(πuσ_yσ_z). The ground-level maximum occurs roughly where σ_z = H/√2, so raising H both lowers the peak and pushes it further away.

H is the effective stack height, the physical height plus the plume rise Δh from the exit momentum and buoyancy of the gas. Holland’s equation is a common estimate: Δh = (v_s d/u)[1.5 + 2.68 × 10⁻³ p d (T_s − T_a)/T_s], with v_s the exit velocity (m/s), d the stack diameter (m), p the pressure (mbar) and T_s, T_a the stack-gas and air temperatures (K); Briggs’ formulas are the more modern alternative. In India, stack height for SO₂ dispersion is set by the CPCB relation H = 14Q^0.3, with Q the SO₂ emission in kg/h. Topographic influences can defeat all of this: valleys trap cold air and inversions at night and channel the wind along their axis; slopes generate drainage (downslope) winds at night; coasts have sea breezes by day that can carry a shoreline plume inland and fumigate it; tall buildings create a turbulent wake that drags a short plume down (downwash), which is why a stack is made at least about 2.5 times the height of nearby buildings; and cities form a heat island that alters the local circulation.

⚠️ π, not 2π, at ground level with reflection
The full equation has 2π in the denominator and two exponential terms. On the ground (z = 0) the two terms are equal and add, so the factor becomes Q/(πuσ_yσ_z). For Q = 100 g/s, u = 5 m/s, σ_y = 60 m, σ_z = 30 m and H = 50 m: C = 100/(π × 5 × 60 × 30) × e^(−2500/1800) = 3.537 × 10⁻³ × 0.2494 = 8.82 × 10⁻⁴ g/m³ = 882 μg/m³. Using 2π halves the answer.

Key takeaways

  • Nearly all pollution sits in the lowest part of the troposphere; dry air is 78% N₂, 21% O₂, 0.93% Ar; pollutants are emitted, transformed (mostly by OH), transported and removed by dry and wet deposition.
  • Ozone, sulphate and nitrate are secondary; the classic criteria pollutants are PM, O₃, CO, SO₂, NO₂ and lead; India’s 2009 NAAQS cover twelve, e.g. PM₂.₅ 40 annual and 60 over 24 h (μg/m³).
  • Sub-index I = I_lo + (I_hi − I_lo)(C − BP_lo)/(BP_hi − BP_lo), and the AQI is the largest sub-index; visual range L_v = 3.912/b_ext.
  • ELR > 9.8 °C/km is unstable, < 9.8 stable, inversion strongly stable; fumigation is the worst plume; mixing height from the afternoon adiabat; ventilation coefficient = mixing height × wind speed.
  • Ground-level centreline C = [Q/(πuσ_yσ_z)] exp(−H²/2σ_z²) with H = stack height + plume rise (Holland); CPCB stack height H = 14Q^0.3 for Q in kg/h of SO₂.

Practice questions (15)

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. Temperature increases with altitude in the stratosphere chiefly because:

    1. ozone absorbs solar ultraviolet radiation
    2. it is closer to the Sun
    3. water vapour condenses and releases latent heat
    4. the greenhouse effect is strongest there
    Show answer

    Answer: A — ozone absorbs solar ultraviolet radiation

    Stratospheric ozone absorbs UV between about 200 and 320 nm and converts it to heat, so the layer warms upward to the stratopause near 50 km. The few tens of kilometres make no difference to the distance from the Sun, and the stratosphere is too dry for latent heating.
  2. A power plant burns 500 t of coal per day containing 0.8% sulphur, and its flue-gas desulphurisation unit removes 85% of the SO₂. What mass of SO₂ is emitted per day, in tonnes? (S = 32, O = 16)

    Numerical answer — type the value.

    Show answer

    Answer: 1.2

    Sulphur burnt = 500 × 0.008 = 4 t/d; each tonne of S gives 64/32 = 2 t of SO₂, so 8 t/d is formed; 15% escapes: 8 × 0.15 = 1.2 t/d. Forgetting the factor 2 gives 0.6 t/d, reporting sulphur rather than SO₂.
  3. What is the concentration of 0.1 ppm of SO₂ at 25 °C and 1 atm, in μg/m³? Give the answer to the nearest whole number. (Molar volume 24.45 L/mol; SO₂ = 64 g/mol)

    Numerical answer — type the value.

    Show answer

    Answer: 262

    C = ppm × M × 1000/24.45 = 0.1 × 64 × 1000/24.45 = 261.8 ≈ 262 μg/m³. Using 22.4 L (0 °C) gives 286 μg/m³; the temperature of the conversion must match the stated conditions.
  4. Which of the following are secondary pollutants?

    1. Ground-level ozone
    2. Peroxyacetyl nitrate (PAN)
    3. Carbon monoxide from a vehicle exhaust
    4. Sulphate aerosol formed from SO₂
    Show answer

    Answer: A — Ground-level ozone; B — Peroxyacetyl nitrate (PAN); D — Sulphate aerosol formed from SO₂

    Ozone and PAN form photochemically from NOₓ and VOCs, and sulphate forms by oxidation of SO₂ in the air or in cloud droplets — all secondary. Carbon monoxide from an exhaust is emitted directly, so it is primary.
  5. Carbon monoxide is toxic at low concentrations mainly because it:

    1. binds haemoglobin far more strongly than oxygen, reducing oxygen transport
    2. dissolves in the airways to form carbonic acid
    3. is a strong oxidant that damages lung tissue
    4. deposits as fine particles in the alveoli
    Show answer

    Answer: A — binds haemoglobin far more strongly than oxygen, reducing oxygen transport

    CO forms carboxyhaemoglobin with an affinity roughly 200 times that of O₂, so even small concentrations occupy a large share of the haemoglobin and starve the heart and brain of oxygen. It is not an acid gas, not an oxidant — it is a reducing gas — and not a particle.
  6. In an air quality index, PM₂.₅ breakpoints of 91 and 120 μg/m³ correspond to index values of 201 and 300. What is the PM₂.₅ sub-index for a concentration of 105 μg/m³? Give the answer to the nearest whole number.

    Numerical answer — type the value.

    Show answer

    Answer: 249

    I = I_lo + (I_hi − I_lo)(C − BP_lo)/(BP_hi − BP_lo) = 201 + (300 − 201)(105 − 91)/(120 − 91) = 201 + 99 × 14/29 = 201 + 47.8 = 248.8 ≈ 249. Interpolating from zero instead of from the band’s lower breakpoint is the usual mistake.
  7. Haze gives an extinction coefficient of 0.3 km⁻¹. Using Koschmieder’s relation, what is the visual range, in km? Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 13.04

    L_v = 3.912/b_ext = 3.912/0.3 = 13.04 km. The constant 3.912 is −ln 0.02, the natural log of the 2% contrast threshold at which an object is just visible.
  8. Which of the following statements about air quality standards and indices are correct?

    1. An ambient standard limits the concentration in the air, for a stated averaging time
    2. An emission standard limits what a source may release
    3. The AQI is the average of the pollutant sub-indices
    4. Black carbon aerosol absorbs sunlight and has a warming effect
    Show answer

    Answer: A — An ambient standard limits the concentration in the air, for a stated averaging time; B — An emission standard limits what a source may release; D — Black carbon aerosol absorbs sunlight and has a warming effect

    (a) and (b) define the two kinds of standard. (d) is the direct radiative effect of absorbing aerosol, the opposite of scattering sulphate. (c) is false: the AQI is the maximum sub-index, reported with the pollutant responsible.
  9. A plume that is carried down to the ground in the morning, when the ground-heated unstable layer grows up to an inversion lying just above the stack, is described as:

    1. fumigation
    2. lofting
    3. fanning
    4. coning
    Show answer

    Answer: A — fumigation

    Fumigation is unstable air below and an inversion lid above: the plume cannot escape upward and is mixed rapidly down, giving the highest ground-level concentrations. Lofting is the reverse (inversion below, unstable above), fanning is a stable layer throughout, and coning is neutral.
  10. The morning environmental temperature profile near a city starts at 20 °C at the ground and falls at 5 °C/km. The afternoon maximum surface temperature is 30 °C. Taking the dry adiabatic lapse rate as 9.8 °C/km, what is the maximum mixing height, in m? Give the answer to the nearest whole number.

    Numerical answer — type the value.

    Show answer

    Answer: 2083

    The dry adiabat from 30 °C is T = 30 − 9.8z and the environment is T = 20 − 5z (z in km). They meet where 30 − 9.8z = 20 − 5z, so 4.8z = 10 and z = 2.083 km = 2083 m. Dividing the 10 °C difference by 9.8 alone (1020 m) ignores that the environment is also cooling with height.
  11. The wind speed at 10 m is 4 m/s and the wind profile follows a power law with exponent 0.25. What is the wind speed at 160 m, in m/s?

    Numerical answer — type the value.

    Show answer

    Answer: 8

    u = u₁(z/z₁)^p = 4 × (160/10)^0.25 = 4 × 16^0.25 = 4 × 2 = 8 m/s. A mixing height of 1200 m with a mean wind of 3 m/s would give a ventilation coefficient of 3600 m²/s — the product, not the ratio, of the two.
  12. A stack emits SO₂ at 100 g/s with an effective height of 50 m into a 5 m/s wind. At a point downwind where σ_y = 60 m and σ_z = 30 m, what is the ground-level concentration on the plume centreline, in μg/m³, with total reflection at the ground? Give the answer to the nearest whole number.

    Numerical answer — type the value.

    Show answer

    Answer: 882

    C = [Q/(πuσ_yσ_z)] exp(−H²/2σ_z²) = [100/(π × 5 × 60 × 30)] × exp(−2500/1800) = 3.537 × 10⁻³ × 0.2494 = 8.82 × 10⁻⁴ g/m³ = 882 μg/m³. With 2π in the denominator (no reflection) it would be 441 μg/m³.
  13. Gas leaves a stack of 2 m diameter at 15 m/s and 400 K into air at 300 K and 1000 mbar, with a wind of 5 m/s. Using Holland’s equation, Δh = (v_s d/u)[1.5 + 2.68 × 10⁻³ p d (T_s − T_a)/T_s], what is the plume rise, in m? Give the answer to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 17.04

    v_s d/u = 15 × 2/5 = 6 m. The bracket is 1.5 + 2.68 × 10⁻³ × 1000 × 2 × 100/400 = 1.5 + 1.34 = 2.84. Δh = 6 × 2.84 = 17.04 m. The momentum term alone (6 × 1.5 = 9 m) ignores the buoyancy of the hot gas, which here contributes almost half the rise.
  14. Using the CPCB relation H = 14Q^0.3, with Q the SO₂ emission in kg/h, what stack height is required for an emission of 100 kg/h of SO₂, in m? Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 55.7

    H = 14 × 100^0.3 = 14 × 10^0.6 = 14 × 3.981 = 55.7 m. Because of the small exponent, doubling the emission raises the required height by only 2^0.3 = 1.23 times.
  15. Which of the following statements about dispersion are correct?

    1. Raising the effective stack height lowers the maximum ground-level concentration and moves it further downwind
    2. For a given source, the ground-level concentration is inversely proportional to the wind speed at the stack height
    3. Building downwash is avoided by making a stack much shorter than nearby buildings
    4. Valleys can trap night-time inversions and pollutants
    Show answer

    Answer: A — Raising the effective stack height lowers the maximum ground-level concentration and moves it further downwind; B — For a given source, the ground-level concentration is inversely proportional to the wind speed at the stack height; D — Valleys can trap night-time inversions and pollutants

    (a) follows from the exp(−H²/2σ_z²) term and the maximum at σ_z ≈ H/√2. (b) u appears in the denominator, because a faster wind stretches the plume over more air (ignoring its effect on plume rise). (d) cold air drains into valleys at night. (c) is false: downwash is avoided by making the stack taller — about 2.5 times the building height — so the plume clears the wake.