Air Pollution Control and Noise: Particulates, Gases, Vehicular Emissions and Noise
1. Particulates: measurement, settling chambers, cyclones, scrubbers, fabric filters and ESPs
Ambient particulates are measured gravimetrically: a high-volume sampler draws a known volume of air through a filter, and the concentration is the mass gained divided by the volume sampled; size-selective inlets (impactors or cyclones) make it a PM₁₀ or PM₂.₅ sampler, and beta-attenuation monitors give continuous readings. In a stack the sample must be drawn isokinetically, with the velocity into the nozzle equal to the gas velocity: sampling too slowly lets large particles, which cannot follow the diverging streamlines, enter in excess and overstate the concentration, and sampling too fast understates it. Control devices are chosen by particle size, and their performance is a fractional efficiency curve. Units in series pass the product of their penetrations: η = 1 − (1 − η₁)(1 − η₂).
A gravitational settling chamber slows the gas so that particles settle at their Stokes velocity v_t = gρ_pd²/(18μ). In laminar flow a particle is caught with efficiency η = v_tLW/Q (L and W the chamber length and width), so 100% capture needs v_t ≥ Q/(LW) and the smallest particle collected completely is d_min = √[18μQ/(gρ_pLW)] — typically tens of micrometres, so chambers are only pre-cleaners. A cyclone spins the gas and flings particles to the wall; Lapple’s model gives the cut diameter collected with 50% efficiency, d₅₀ = √[9μW/(2πN_ev_i(ρ_p − ρ_g))], with W the inlet width, N_e the number of effective turns and v_i the inlet velocity, and the efficiency for any size d is η = 1/[1 + (d₅₀/d)²]. Wet scrubbers (spray towers, venturi scrubbers) capture particles on droplets by impaction, interception and diffusion; a venturi reaches high efficiency on fine particles at the cost of pressure drop, handles hot, sticky or explosive dusts, and can absorb gases too — but it makes a wastewater. Fabric filters (baghouses) strain the gas through woven or felted bags, with the dust cake itself doing much of the filtering, and exceed 99% efficiency down to submicrometre sizes; they are sized by the air-to-cloth ratio (gas flow per unit cloth area, about 0.5–2 m/min for shaker and reverse-air cleaning, higher for pulse-jet) and limited by gas temperature and moisture. Electrostatic precipitators (ESPs) charge particles in a corona and drive them to collecting plates; the Deutsch–Anderson equation gives η = 1 − exp(−wA/Q), with w the drift velocity and A the plate area. They handle very large gas flows at low pressure drop but are sensitive to dust resistivity: too low and the dust re-entrains, too high and it causes back-corona.
| Device | Useful down to | Pressure drop | Main limitation |
|---|---|---|---|
| Settling chamber | Tens of μm | Very low | Large size; coarse particles only |
| Cyclone | About 5–10 μm | Moderate | Poor on fine particles |
| Venturi scrubber | About 1 μm and below | High | Energy use; wastewater |
| Fabric filter | Submicrometre | Moderate | Temperature, moisture, bag wear |
| ESP | Submicrometre | Low | Capital cost; dust resistivity |
2. Gaseous pollutants: measurement, absorption, adsorption, condensation, combustion and mass transfer
Standard measurement methods are chemical or instrumental: SO₂ by the West–Gaeke (pararosaniline) colorimetric method or by UV fluorescence; NO₂ by the modified Jacobs–Hochheiser method or NOₓ by chemiluminescence; CO by non-dispersive infrared; ozone by UV photometry; hydrocarbons by flame ionisation. Absorption transfers a soluble gas into a liquid in a packed or spray tower, and it is governed by interfacial mass transfer. Fick’s law gives the diffusive flux, J = −D dC/dz; in the two-film model the gas must diffuse through a gas film and then a liquid film at the interface, with equilibrium (Henry’s law) at the interface itself, so the overall resistance is the sum 1/K_G = 1/k_G + H/k_L. A very soluble gas (small H, e.g. NH₃, HCl) is gas-film controlled; a sparingly soluble one (O₂, CO₂) is liquid-film controlled, which is why aeration is designed to renew the liquid surface.
Adsorption holds gas molecules on the surface of a porous solid — activated carbon for organic vapours, zeolites and silica gel for others — and is described by isotherms: Freundlich, q = KC^(1/n), and Langmuir, q = q_max KC/(1 + KC). A fixed bed saturates progressively from the inlet, and when the saturated front reaches the outlet the effluent concentration rises — breakthrough — after which the bed is regenerated with steam or hot gas. Condensation cools a vapour below its dew point; it recovers solvent but is economical only at high concentrations, and is often a pre-treatment. Combustion oxidises organics and CO to CO₂ and water: flares for emergency releases, thermal incinerators at high temperature with sufficient time and turbulence (the "three Ts"), and catalytic incinerators at a much lower temperature, with the catalyst liable to poisoning by lead, sulphur and phosphorus.
| Pollutant | Prevention | End-of-pipe control |
|---|---|---|
| SOₓ | Low-sulphur fuel; coal washing; fuel switching | Flue-gas desulphurisation: limestone slurry, SO₂ + CaCO₃ → CaSO₃ + CO₂, oxidised to gypsum; dry sorbent injection |
| NOₓ | Low-NOₓ burners, staged combustion, flue-gas recirculation, low excess air | SCR with NH₃ or urea over a catalyst; SNCR by injection at high temperature without one |
| CO | Complete combustion: enough air, temperature and mixing | Catalytic oxidation |
| Hydrocarbons and VOCs | Vapour recovery, sealed storage, solvent substitution | Adsorption, condensation, thermal or catalytic incineration |
3. Vehicular emission control: catalytic converters, diesel particulate filters and fuel standards
A petrol engine’s exhaust carries CO, unburnt hydrocarbons and NOₓ. The three-way catalytic converter — platinum, palladium and rhodium on a ceramic honeycomb — oxidises CO and HC and reduces NOₓ at the same time, but only in a narrow window around the stoichiometric air–fuel ratio (about 14.7 : 1 by mass for petrol), which an oxygen (lambda) sensor holds by feedback. Lean running would leave NOₓ unreduced; rich running would leave CO and HC unoxidised. Leaded fuel poisons the catalyst, which is one reason unleaded petrol had to come first. A diesel engine runs lean, so a three-way catalyst cannot reduce its NOₓ: it uses a diesel oxidation catalyst for CO and HC, a diesel particulate filter (DPF) — a wall-flow ceramic monolith that traps soot and is periodically regenerated by burning it off — and selective catalytic reduction with urea solution for NOₓ, often with exhaust gas recirculation to lower combustion temperature. Fuel standards make these devices possible: sulphur poisons catalysts and forms sulphate particles, so low-sulphur fuel is paired with each emission stage. India’s Bharat Stage norms moved nationwide from BS-IV directly to BS-VI on 1 April 2020, skipping BS-V; BS-VI fuel carries at most 10 ppm sulphur, and the stage effectively requires DPFs and SCR on diesel vehicles.
4. Noise: sources, effects, standards, SPL, L_eq, L_dn, frequency analysis and control
Noise is unwanted sound, from road, rail and air traffic, industry and construction, loudspeakers and generators. The ear responds to pressure over a range of about a million to one, so sound is measured on a logarithmic scale. The sound pressure level is L_p = 20 log₁₀(p/p₀) with p₀ = 20 μPa, the threshold of hearing; the intensity level is L_I = 10 log₁₀(I/I₀) with I₀ = 10⁻¹² W/m². Levels therefore add logarithmically: L_total = 10 log₁₀ Σ10^(L_i/10), so two equal sources are 3 dB louder than one, and a source 10 dB quieter than another adds only 0.4 dB. In the open, a point source falls by 20 log(r₂/r₁) — 6 dB per doubling of distance — and a line source such as a busy road by 10 log(r₂/r₁), 3 dB per doubling.
A fluctuating level is summarised by the equivalent continuous level, the steady level with the same energy: L_eq = 10 log₁₀[(1/T)Σt_i 10^(L_i/10)]. The day–night level L_dn adds a 10 dB penalty to the nine night hours (22:00–07:00), L_dn = 10 log₁₀{(1/24)[15 × 10^(L_d/10) + 9 × 10^((L_n + 10)/10)]}. Statistical levels L₁₀, L₅₀ and L₉₀ are the levels exceeded 10, 50 and 90% of the time — L₉₀ is the background, L₁₀ the intrusive peaks. Measurements are made with a sound level meter through the A-weighting filter, which follows the ear’s reduced sensitivity at low frequencies, giving dB(A). Frequency analysis splits the sound into octave bands (each band’s upper limit twice its lower, with centre frequencies 31.5, 63, 125, 250, 500, 1000, 2000, 4000 and 8000 Hz) or one-third-octave bands, to find which frequencies dominate and choose the control. Health effects run from annoyance, speech interference and sleep disturbance, through raised blood pressure and stress, to temporary threshold shift and permanent noise-induced hearing loss, which typically appears first as a dip in hearing near 4 kHz. India’s Noise Pollution (Regulation and Control) Rules, 2000 set ambient limits in dB(A) L_eq of 75 (day) and 70 (night) for industrial areas, 65 and 55 for commercial, 55 and 45 for residential, and 50 and 40 for silence zones.
Control follows the chain source–path–receiver. At the source: quieter design and maintenance, balancing and vibration isolation, mufflers and silencers on intakes and exhausts, and replacing impact processes. On the path: distance, barriers and earth berms that block the line of sight (most effective at high frequencies, whose short wavelengths diffract less over the top), enclosures and acoustic absorbing linings that cut reverberation; the transmission loss of a single wall rises about 6 dB for each doubling of its mass per unit area or of frequency (the mass law). At the receiver: ear plugs and muffs, acoustic cabins, and limiting the time of exposure. Land-use planning — keeping housing away from highways, airports and industry — is the cheapest mitigation of all.
Key takeaways
- Stack sampling must be isokinetic; units in series pass the product of penetrations. Settling chamber η = v_tLW/Q; Lapple d₅₀ = √[9μW/(2πN_ev_iρ_p)] with η = 1/[1 + (d₅₀/d)²]; ESP η = 1 − exp(−wA/Q); baghouses are sized by air-to-cloth ratio.
- Fick J = −D dC/dz; two-film 1/K_G = 1/k_G + H/k_L, soluble gases gas-film controlled; Freundlich q = KC^(1/n); incinerators need time, temperature and turbulence.
- SO₂ by limestone FGD (CaCO₃/SO₂ = 100/64 by mass at stoichiometry); NOₓ by combustion modification, SCR and SNCR; CO and HC by complete combustion and oxidation catalysts.
- A three-way catalyst works only near the stoichiometric air–fuel ratio; diesels need an oxidation catalyst, a DPF and SCR; India moved to BS-VI with 10 ppm sulphur fuel on 1 April 2020.
- L_p = 20 log(p/20 μPa); levels add as 10 log Σ10^(L/10); point source −6 dB per doubling of distance; L_eq is an energy average; L_dn adds 10 dB to the night; residential limits are 55/45 dB(A).
Practice questions (20)
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.
A cyclone with 80% efficiency is followed by a fabric filter with 98% efficiency. What is the overall collection efficiency of the pair, in per cent? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 99.6
The penetrations multiply: (1 − 0.80)(1 − 0.98) = 0.2 × 0.02 = 0.004, so η = 99.6%. Adding the efficiencies (178%) or averaging them (89%) is meaningless.A settling chamber 10 m long and 2 m wide treats 5 m³/s of air (μ = 1.8 × 10⁻⁵ Pa·s) carrying dust of density 2500 kg/m³. Assuming laminar flow and Stokes’ law, what is the smallest particle collected with 100% efficiency, in μm? Give the answer to one decimal place. (g = 9.81 m/s²)
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Answer: 57.5
Complete capture needs v_t = Q/(LW) = 5/20 = 0.25 m/s. From Stokes, d = √(18μv_t/(gρ_p)) = √(18 × 1.8 × 10⁻⁵ × 0.25/(9.81 × 2500)) = √(3.303 × 10⁻⁹) = 5.75 × 10⁻⁵ m = 57.5 μm. The chamber height cancels, just as the depth does in a water settling tank.In the chamber of the previous question (L = 10 m, W = 2 m, Q = 5 m³/s, ρ_p = 2500 kg/m³, μ = 1.8 × 10⁻⁵ Pa·s), what is the collection efficiency for 30 μm particles, in per cent? Give the answer to the nearest whole number. (g = 9.81 m/s²)
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Answer: 27
v_t = gρ_pd²/(18μ) = 9.81 × 2500 × (3 × 10⁻⁵)²/(18 × 1.8 × 10⁻⁵) = 0.0681 m/s. η = v_tLW/Q = 0.0681 × 20/5 = 0.272, i.e. 27%. The shortcut (30/57.5)² = 0.272 gives the same, since efficiency scales with d².A cyclone has an inlet width of 0.25 m, six effective turns and an inlet velocity of 15 m/s, and treats air (μ = 1.8 × 10⁻⁵ Pa·s) carrying particles of density 2000 kg/m³. Neglecting the gas density, what is Lapple’s cut diameter, in μm? Give the answer to two decimal places.
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Answer: 5.98
d₅₀ = √[9μW/(2πN_ev_iρ_p)] = √[9 × 1.8 × 10⁻⁵ × 0.25/(2π × 6 × 15 × 2000)] = √(4.05 × 10⁻⁵/1.131 × 10⁶) = √(3.581 × 10⁻¹¹) = 5.98 × 10⁻⁶ m = 5.98 μm. Particles of this size are caught half the time; a 10 μm particle would be caught with 1/[1 + (0.598)²] = 74%.A cyclone has a cut diameter of 5 μm. By Lapple’s efficiency curve, what fraction of 10 μm particles does it collect, in per cent?
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Answer: 80
η = 1/[1 + (d₅₀/d)²] = 1/[1 + (5/10)²] = 1/1.25 = 0.8, i.e. 80%. Using d/d₅₀ in the bracket gives 1/(1 + 4) = 20%, the efficiency the curve would give a particle half the cut size.An electrostatic precipitator has 5000 m² of collecting plate and treats 200 m³/s of flue gas; the particle drift velocity is 0.1 m/s. What is its collection efficiency by the Deutsch–Anderson equation, in per cent? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 91.8
η = 1 − exp(−wA/Q) = 1 − exp(−0.1 × 5000/200) = 1 − e^(−2.5) = 1 − 0.0821 = 0.918, i.e. 91.8%. The linear form wA/Q = 2.5 would suggest more than 100%, which is why the exponential law is needed.An ESP must remove 99% of the dust from 100 m³/s of gas, with a drift velocity of 0.08 m/s. What plate area is required, in m²? Give the answer to the nearest whole number.
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Answer: 5756
A = −(Q/w) ln(1 − η) = −(100/0.08) ln(0.01) = 1250 × 4.605 = 5756 m². Raising the efficiency from 99% to 99.9% would need 1250 × 6.908 = 8635 m² — each extra "nine" costs the same additional area.A baghouse treats 600 m³/min of gas at an air-to-cloth ratio of 1.5 m/min, using bags 0.2 m in diameter and 5 m long. How many bags are required?
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Answer: 128
Cloth area = Q/(A/C) = 600/1.5 = 400 m². Each bag offers πdL = π × 0.2 × 5 = 3.1416 m², so 400/3.1416 = 127.3, rounded up to 128 bags. Rounding down to 127 would run the filter above its design air-to-cloth ratio.In stack sampling, drawing the sample at a velocity lower than the stack-gas velocity (sub-isokinetic sampling) will:
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Answer: A — overestimate the particulate concentration, because excess large particles enter the nozzle
When the nozzle draws more slowly than the gas, streamlines diverge around it; the gas goes round but large, inertial particles keep going straight in. The sample is enriched in large particles and the measured concentration is too high. Super-isokinetic sampling does the opposite. Gases follow the streamlines, so they are unaffected.An activated carbon follows the Freundlich isotherm q = KC^(1/n) with K = 20 and 1/n = 0.5 (q in mg/g, C in mg/m³). What is its equilibrium loading at C = 16 mg/m³, in mg/g?
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Answer: 80
q = 20 × 16^0.5 = 20 × 4 = 80 mg/g. Because 1/n < 1, loading rises less than proportionally with concentration: quadrupling C only doubles q.A limestone scrubber is operated at a Ca/S stoichiometric ratio of 1.1. How much CaCO₃ is needed per kg of SO₂ removed, in kg? Give the answer to two decimal places. (CaCO₃ = 100, SO₂ = 64)
Numerical answer — type the value.
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Answer: 1.72
SO₂ + CaCO₃ → CaSO₃ + CO₂: one mole each, so 100/64 = 1.5625 kg CaCO₃ per kg SO₂ at stoichiometry; × 1.1 = 1.72 kg. Using the ratio of sulphur (32) gives twice as much and confuses S with SO₂.A gas with diffusivity 2 × 10⁻⁵ m²/s diffuses across a stagnant film 0.01 m thick, with a concentration difference of 0.5 mol/m³ across it. What is the steady diffusive flux by Fick’s law, in mol/(m²·s)?
Numerical answer — type the value.
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Answer: 0.001
J = D ΔC/Δz = 2 × 10⁻⁵ × 0.5/0.01 = 1 × 10⁻³ mol/(m²·s). In the two-film model this is one of the two resistances in series; halving the film thickness by stronger turbulence doubles the flux. Enter it as 0.001.Which of the following statements about gaseous pollutant control are correct?
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Answer: A — Absorption of a highly soluble gas such as NH₃ is gas-film controlled; B — Selective catalytic reduction of NOₓ uses ammonia or urea as the reductant; D — Breakthrough in an adsorber occurs when the saturated zone reaches the bed outlet
(a) with a small Henry constant the liquid-side resistance H/k_L is negligible. (b) SCR reduces NOₓ to N₂ with NH₃ over a catalyst. (d) defines breakthrough. (c) is false: the catalyst exists precisely to let oxidation proceed at a much lower temperature, saving fuel.A three-way catalytic converter cannot be used to control NOₓ from a diesel engine mainly because:
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Answer: A — the diesel exhaust contains excess oxygen, so NOₓ is not reduced
A diesel always runs lean. In a three-way catalyst NOₓ is reduced by the CO and hydrocarbons in the exhaust, and with excess oxygen present those reductants are oxidised first, so NOₓ passes through. Diesels therefore use SCR with urea for NOₓ, an oxidation catalyst for CO and HC, and a DPF for soot.Three machines produce 85, 88 and 90 dB at a point when run separately. What is the combined sound pressure level when all three run together, in dB? Give the answer to one decimal place.
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Answer: 92.9
L = 10 log(10^8.5 + 10^8.8 + 10^9.0) = 10 log(3.162 × 10⁸ + 6.310 × 10⁸ + 10 × 10⁸) = 10 log(1.947 × 10⁹) = 92.9 dB. Adding the numbers (263 dB) or averaging them (87.7 dB) treats a logarithmic scale as linear.A worker is exposed to 80 dB(A) for 4 hours and 70 dB(A) for 4 hours of an 8-hour shift. What is the equivalent continuous level L_eq over the shift, in dB(A)? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 77.4
L_eq = 10 log[(1/8)(4 × 10⁸ + 4 × 10⁷)] = 10 log(5.5 × 10⁷) = 77.4 dB(A). The arithmetic mean, 75, underweights the louder half, which carries ten times the energy.A residential site has L_eq = 60 dB(A) over the 15 daytime hours and 50 dB(A) over the 9 night hours (22:00–07:00). What is its day–night level L_dn, in dB(A)?
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Answer: 60
The night level is penalised by 10 dB, to 60. L_dn = 10 log{(1/24)[15 × 10⁶ + 9 × 10^((50 + 10)/10)]} = 10 log[(15 × 10⁶ + 9 × 10⁶)/24] = 10 log(10⁶) = 60 dB(A). Without the penalty the 24-hour L_eq would be 10 log[(15 × 10⁶ + 9 × 10⁵)/24] = 58.2 dB(A).A small machine, radiating as a point source in the open, gives 90 dB at 1 m. What is the level at 8 m, in dB? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 71.9
L₂ = L₁ − 20 log(r₂/r₁) = 90 − 20 log 8 = 90 − 18.06 = 71.9 dB — 6 dB for each of the three doublings of distance. The line-source rule, 10 log 8 = 9 dB, would give 81 dB.What is the sound pressure level of a sound with an rms pressure of 2 Pa, in dB? (Reference pressure 20 μPa)
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Answer: 100
L_p = 20 log(p/p₀) = 20 log(2/(2 × 10⁻⁵)) = 20 log 10⁵ = 100 dB. Using 10 log for a pressure ratio gives 50 dB; the factor 20 arises because intensity goes as pressure squared.Which of the following statements about noise are correct?
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Answer: A — A-weighting reduces the contribution of low frequencies to match the ear’s sensitivity; B — Noise-induced hearing loss typically shows first as a dip near 4 kHz; D — In India the ambient limit for a silence zone is lower than for a residential area
(a) A-weighting follows the ear’s reduced low-frequency sensitivity. (b) the 4 kHz notch is the classic audiogram sign. (d) silence zones are 50/40 dB(A) against 55/45 for residential areas. (c) is false: long low-frequency waves diffract over a barrier easily, so barriers work best on high frequencies.