The environmental lapse rate on a particular morning is measured as +2 °C per km (temperature rising with height). The atmospheric condition and its consequence for a stack plume are

An inversion — strongly stable, the plume is trapped and ground-level concentrations are high

Super-adiabatic — strongly unstable, the plume loops and disperses rapidly

Neutral — the plume rises in a straight cone and behaves as the design assumed

Unstable below the stack and stable above it, giving a lofting plume

Which of the following statements about noise are correct?

Doubling the distance from a point source reduces the level by about 6 dB, but from a line source by only about 3 dB

A barrier attenuates by forcing diffraction, so one that does not break the line of sight achieves very little

Control at the source is preferred over control at the receiver

Doubling the distance from a point source reduces the level by about 6 dB, but from a line source by only about 3 dB

A barrier attenuates by forcing diffraction, so one that does not break the line of sight achieves very little

Control at the source is preferred over control at the receiver

Adding a 50 dB source next to a 60 dB source raises the combined level to 110 dB

Incineration of municipal solid waste has repeatedly performed poorly in Indian cities. The most direct technical reason is that

the waste is high in organics and therefore high in moisture, so its calorific value is too low to sustain combustion

Indian municipal waste contains too much plastic, which cannot be burned at any temperature

the total quantity generated per city is too small to justify a furnace

incineration is prohibited for municipal waste under Indian rules

Air Pollution, Municipal Solid Waste and Noise

The three subjects in this chapter share a structure that is worth seeing before any of the detail: in each, the receiving medium does not dilute the way intuition expects, and every design rule follows from that. Air looks like an infinite sink and is not — a temperature inversion puts a lid on the atmosphere a few hundred metres up, and a stack that disperses perfectly on a windy afternoon traps its plume at ground level on a still winter morning, which is why air-quality standards are breached at night and in December rather than at the times a plant is busiest. Solid waste looks like a volume problem and is really a density problem — the same tonnage occupies half the space compacted, and a landfill is bought in hectares. And noise looks like it should add up and does not: two identical 60 dB sources make 63 dB, not 120, because the decibel is logarithmic, so silencing one of a pair of equal machines buys 3 dB while silencing a machine 10 dB quieter than its neighbour buys almost nothing. Each of those three facts is the reason its subject has the control technology it has, and each is examined directly rather than as background.

Air pollutants, and the atmosphere that will not always disperse them

Pollutants divide first into primary — emitted as such: particulates, SO₂, NOₓ, CO, hydrocarbons — and secondary, formed in the atmosphere from the primary ones: ozone and peroxyacetyl nitrate in photochemical smog, sulphates and nitrates in secondary particulate. The distinction matters because a secondary pollutant cannot be controlled at a stack; it is controlled only by controlling its precursors, which is why urban ozone is a NOₓ-and-hydrocarbon problem rather than an ozone problem. Particulates then divide by size, and the size is the health endpoint: PM₁₀ deposits in the upper airways, PM₂.₅ reaches the alveoli and crosses into the blood, which is why the standards for the two differ by far more than their masses suggest.

🎯 The lapse rate decides whether a stack works

A parcel of dry air lifted in the atmosphere cools at the dry adiabatic lapse rate, about 9.8 °C per km, because it expands. Whether it keeps rising depends on how the surrounding air’s temperature actually falls with height — the environmental lapse rate — and the comparison is the whole of atmospheric stability. If the environment cools faster than 9.8 °C/km (super-adiabatic), a lifted parcel finds itself warmer than its surroundings, keeps rising, and the atmosphere is unstable: plumes loop, dispersion is vigorous, ground-level concentrations are low and erratic. If the environment cools more slowly (sub-adiabatic), a lifted parcel is colder than its surroundings and sinks back: the atmosphere is stable, plumes fan out horizontally and go nowhere vertically. The extreme case is an inversion, where temperature rises with height — typically on a clear, still night when the ground has radiated its heat away — and the atmosphere is capped. This is the examinable consequence: the worst ground-level concentrations occur not when emissions are highest but when the atmosphere is most stable, which is why episodes happen on winter mornings, and why a tall stack is not a solution but a bet on the atmosphere behaving.

Particulate control, in the order of the sizes they can reach
Device	Mechanism	Practical lower size	What limits it
Gravity settling chamber	Stokes settling — the same `v ≥ Q/(LW)` as a sedimentation tank	~50–60 μm	`v_s ∝ d²`, so anything fine settles far too slowly
Cyclone	Centrifugal force replaces gravity, many times larger	~10 μm	Pressure drop rises steeply as the body is narrowed
Fabric filter (baghouse)	Sieving and impaction on the deposited cake	< 1 μm	Temperature and moisture — the fabric has limits
Electrostatic precipitator	Charging the particle, then collecting it on a plate	< 1 μm	Resistivity of the dust — too high or too low both fail
Wet scrubber	Impaction on droplets; also absorbs gases	~1 μm	Creates a wastewater stream — one problem for another

Municipal solid waste: a density problem wearing a volume problem’s clothes

Generation is a per-capita rate, typically 0.3 to 0.6 kg per person per day in Indian cities and rising with income. A city of 100,000 at 0.5 kg/capita/day makes 50 tonnes a day — a number that means nothing until it is turned into a volume, and the volume depends entirely on the density. Loose, at about 300 kg/m³, that tonnage is 166.7 m³/day; compacted in the vehicle at 600 kg/m³, 83.3 m³/day. The whole economics of collection is in that halving: a compactor truck costs more and carries twice as much, so the choice turns on haul distance, and the transfer station exists precisely to make long hauls carry compacted rather than loose waste.

Characteristics — the composition (organics, paper, plastics, metals, glass, inerts), the moisture content, and the calorific value. These three are not independent: Indian MSW is high in organics and therefore high in moisture, and high moisture is what makes its calorific value too low to burn without support fuel. That single chain is why incineration has repeatedly failed here and composting and bio-methanation have not.
Segregation at source is the step everything else depends on. Mixed waste cannot be composted (glass and metal contaminate it), cannot be recycled economically (sorting mixed waste costs more than the material is worth) and burns badly. Nothing downstream recovers what mixing destroys.
Engineered systems for solid waste management run through reuse and recycling, energy recovery by incineration or bio-methanation, treatment by composting, and only then disposal. Engineered disposal is a sanitary landfill, not a dump: a low-permeability liner below, leachate collection above it, daily cover, and gas wells for the methane that anaerobic decomposition produces. The two things a landfill must contain are exactly the two things it produces, and both are managed rather than prevented.

⚠️ A landfill is bought in hectares, so the design variable is depth

Take the city above. Twenty years of 83.3 m³/day compacted is 83.3 × 365 × 20 = 608,333 m³, and at a working depth of 10 m that is 6.08 hectares — before cover material, before access roads, before the buffer. Two consequences follow that candidates miss. First, the cover is not free: daily cover typically consumes 20 to 25 per cent of the airspace, so the real land requirement is a quarter larger than the waste volume suggests, and a question that gives you a cover ratio expects you to apply it. Second, going deeper is far cheaper than going wider, since area scales with the reciprocal of depth — doubling the depth to 20 m halves the land to 3.04 ha — which is why landfills are excavated as well as raised, and why the binding constraint is usually groundwater, not land: you can only go as deep as the water table and the liner allow. And note what the compaction did. Had this city landfilled its waste loose at 300 kg/m³, the same twenty years would need 12.17 hectares. Compaction did not reduce the waste at all; it halved the land.

Noise, and the arithmetic that refuses to add

Sound pressure level is defined as L = 20 log₁₀(p/p₀) with p₀ = 20 μPa, and its consequences run through every question in this section. Because the scale is logarithmic, levels do not add: two sources must be combined through their intensities, L = 10 log₁₀(Σ 10^(Lᵢ/10)). Two equal 60 dB sources give 63.01 dB. Ten equal 60 dB sources give exactly 70 dB. And a 50 dB source beside a 60 dB one gives 60.41 dB — the quieter machine, ten decibels down, contributes four-tenths of a decibel. That last figure is the single most practically important number in noise control, and it is why the first question about any noise problem is not "how loud is everything" but "what is the loudest thing", because until the dominant source is dealt with nothing else matters.

The three log rules that carry the marks
Situation	Rule	Worked value
Two equal sources	Add `10 log₁₀ 2`	+3.01 dB — 60 + 60 = 63.01
Doubling distance from a point source	Lose `20 log₁₀ 2`	−6.02 dB
Doubling distance from a line source (a road)	Lose `10 log₁₀ 2`	−3.01 dB — half as much relief

🧠 L_eq is not an average of the levels

The equivalent continuous level L_eq = 10 log₁₀[(1/T) Σ tᵢ · 10^(Lᵢ/10)] is the steady level carrying the same energy as the varying one, and because it averages energies rather than decibels it is always pulled towards the loudest period. Take a day of 8 hours at 80 dB, 8 at 70 and 8 at 60: the arithmetic mean of the three numbers is 70 dB, and the true L_eq is 75.68 dB — nearly six decibels higher, and the arithmetic mean is not an approximation of it but a different quantity altogether. Read the terms and the reason is plain: 10^8 against 10^7 against 10^6 means the loudest eight hours contribute about 90 per cent of the day’s energy, and the quietest eight contribute under one per cent. The practical corollary is the one worth carrying into an exam and out of it: a short, very loud period dominates a daily exposure completely, so noise control is about peaks and their duration, not about the average condition — and a mitigation that shaves 2 dB off the quiet hours changes L_eq by almost nothing.

Control of noise pollution follows the same order everywhere and the order is not negotiable: at the source first (a quieter machine, a resilient mounting, a silencer), then along the path (distance, a barrier, an enclosure), then at the receiver last (ear protection, a sealed window), because protection at the receiver protects only whoever is wearing it. Barriers deserve one caution: they work by forcing the sound to diffract over an edge, so their attenuation depends on the path-length difference rather than on their mass, and a barrier that does not break the line of sight does essentially nothing. Indian ambient standards are set by area category — industrial, commercial, residential and silence zone — with a lower limit at night than by day in every category, which is the regulation acknowledging that the same level is more intrusive when the background is quieter.

Air quality standards and the index; the impacts, limits and measurement of noise

Air quality standards in India are the National Ambient Air Quality Standards, and they are set per pollutant with different averaging periods — 24 hours for particulates and sulphur dioxide, 8 hours for carbon monoxide and ozone, annual for the chronic exposures. That structure matters more than the numbers: a pollutant can meet its annual standard and breach its 24-hour standard on fifty days of the year, so compliance is a statement about a period and never about a moment. The air quality index then collapses several pollutants into one figure for the public, and the way it does so is the examinable part.

⚠️ The AQI is the worst sub-index, not the average of them

Each pollutant is first converted to a sub-index by linear interpolation between the breakpoints of its band. Indian PM₂.₅ breakpoints put 61–90 μg/m³ on the 101–200 band, so a reading of 85 gives 101 + (200 − 101)/(90 − 61) × (85 − 61) = 182.93. Do the same for the others and suppose PM₁₀ returns 95, NO₂ 60 and SO₂ 40. The AQI is then the maximum of the sub-indices — 182.93, reported as the PM₂.₅ band — and not their mean, which would be 94.48 and would describe the air as satisfactory when one pollutant alone is nearly at the "poor" threshold. The factor of two between those two numbers is the entire design intent of the index: a person breathes every pollutant at once, so being safe on three of four is not being safe. Two consequences follow. The AQI names its responsible pollutant alongside its value, because the number without it does not say what to do. And an index built on a maximum cannot be averaged over space or time to get a regional or annual figure — an "average AQI" is a quantity with no meaning, and reporting one is the commonest misuse of the scale. Air pollution control then acts on whichever pollutant the index named, which is why the control-device table earlier in this chapter is organised by particle size rather than by industry.

The impacts of noise run in the same order as its control hierarchy, from the reversible to the permanent. Short exposure to high levels causes a temporary threshold shift — the dulled hearing after a loud concert, which recovers in hours. Repeated exposure makes it a permanent threshold shift, which does not recover, because the loss is of hair cells that are not replaced. Below the levels that damage hearing there are effects that are no less real for being harder to measure: sleep disturbance, raised blood pressure, impaired concentration and task performance, and simple annoyance, which is the impact that actually generates complaints and is therefore what regulation mostly addresses.

Indian permissible limits of noise pollution, dB(A) — CPCB ambient standards by area
Area category	Day (6 am – 10 pm)	Night (10 pm – 6 am)	Night is lower by
Industrial	75	70	5 dB
Commercial	65	55	10 dB
Residential	55	45	10 dB
Silence zone	50	40	10 dB

🎯 The measurement of noise is weighted, and that is why the limits are dB(A)

Every limit in the table is in dB(A), not dB, and the letter is doing real work. The ear is far less sensitive at low frequencies than in the speech range, so a sound-level meter applies the A-weighting filter — cutting the bass heavily, leaving 1–4 kHz almost untouched — to make the reading correspond to what a person actually hears. A machine rumbling at 40 Hz can be loud in unweighted dB and modest in dB(A), which is the honest reason the two must never be compared. The meter itself then measures a fast or slow exponential average for a fluctuating source, and for anything that varies over a period the quantity reported is L_eq, computed as the energy average worked earlier in this chapter. Two features of the table are the examinable ones. Night limits are lower — by 5 dB in an industrial area and by 10 everywhere else — which is the regulation acknowledging that the same level is more intrusive when the background is quieter and people are asleep. And a silence zone, defined as within 100 m of a hospital, educational institution or court, carries the tightest limit of all, at a level a busy conversation would breach. Read against the arithmetic earlier in this chapter, the 10 dB gap between day and night is a demand for a tenfold reduction in sound energy, not a small tightening — which is why night-time construction needs a genuinely different method and not merely more care.

Key takeaways

Secondary pollutants form in the atmosphere and cannot be controlled at a stack — only their precursors can.
Compare the environmental lapse rate with the dry adiabatic 9.8 °C/km: faster is unstable, slower is stable, rising is an inversion.
A gravity settling chamber uses the same v ≥ Q/(LW) as a sedimentation tank; here it gives a cut size of about 64 μm.
Compaction from 300 to 600 kg/m³ does not reduce the waste — it halves the land the landfill needs.
Two equal 60 dB sources give 63.01 dB; a source 10 dB quieter adds only 0.41 dB.
L_eq averages energy, so it is pulled to the loudest period: 8h at 80/70/60 dB gives 75.68, not the mean 70.

Practice questions (8)

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.

Two machines each produce a sound pressure level of 60 dB at a point. Operating together, the level at that point in dB is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 63.01
63.01 dB. Decibels are logarithmic, so they cannot be added directly — the intensities add, and the level is recovered from their sum: L = 10 log₁₀(10^(60/10) + 10^(60/10)) = 10 log₁₀(2 × 10⁶) = 60 + 10 log₁₀ 2 = 60 + 3.01 = 63.01 dB. The general rule falls straight out: doubling the number of equal sources adds 3.01 dB, whatever the level. Ten equal 60 dB sources give exactly 70 dB, and a hundred give 80. Now use the same formula on unequal sources, where it produces the result that actually governs noise control. A 50 dB source beside a 60 dB one gives 10 log₁₀(10⁶ + 10⁵) = 60.41 dB — the quieter machine, only ten decibels down, contributes 0.41 dB, which no instrument in a real environment would resolve. So silencing it entirely achieves nothing measurable. That is why noise surveys look for the dominant source rather than cataloguing everything: until the loudest source is dealt with, work on any other is wasted, and after it is dealt with the ranking has changed and the question must be asked again. Candidates lose this mark two ways — by answering 120 (adding the levels) or 60 (assuming the second source cannot matter).
A worker is exposed to 80 dB for 8 hours, 70 dB for 8 hours and 60 dB for 8 hours in a day. The 24-hour equivalent continuous sound level L_eq in dB is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 75.68
75.68 dB. L_eq = 10 log₁₀[(1/T) Σ tᵢ · 10^(Lᵢ/10)] = 10 log₁₀[(8 × 10⁸ + 8 × 10⁷ + 8 × 10⁶)/24]. The bracket is (8 × 10⁸ + 0.8 × 10⁸ + 0.08 × 10⁸)/24 = 8.88 × 10⁸/24 = 3.70 × 10⁷, and 10 log₁₀(3.70 × 10⁷) = 75.68 dB. Set that against the arithmetic mean of 80, 70 and 60, which is 70 dB — nearly six decibels lower and simply the wrong quantity. L_eq averages energy, not level, and energy is exponential in the level, so a period ten decibels louder carries ten times the weight per unit time. Look at the terms: the 80 dB block contributes 8 × 10⁸ out of a total of 8.88 × 10⁸, which is 90 per cent of the day’s energy from one third of the day, and the 60 dB block contributes under one per cent. The practical reading is what makes this examinable rather than arithmetic. A control measure that removes the quietest eight hours entirely changes L_eq by 0.04 dB — nothing. A measure that takes 3 dB off the loudest eight hours takes about 2.5 dB off the whole day. Exposure is governed by peaks and how long they last, which is exactly why occupational limits are written as a level and a duration, halving the permitted time for every few decibels rather than setting one number.
The environmental lapse rate on a particular morning is measured as +2 °C per km (temperature rising with height). The atmospheric condition and its consequence for a stack plume are
1. An inversion — strongly stable, the plume is trapped and ground-level concentrations are high
2. Super-adiabatic — strongly unstable, the plume loops and disperses rapidly
3. Neutral — the plume rises in a straight cone and behaves as the design assumed
4. Unstable below the stack and stable above it, giving a lofting plume
Show answer
Answer: A — An inversion — strongly stable, the plume is trapped and ground-level concentrations are high
An inversion, and the worst case for ground-level concentration. Compare the environmental lapse rate with the dry adiabatic rate of −9.8 °C/km. Here the environment is at +2 °C/km: not merely slower than adiabatic but the opposite sign — the air is warmer higher up. A parcel of plume rising from the stack cools adiabatically at 9.8 °C/km while its surroundings warm, so within a very short rise it is far colder and denser than the air around it and sinks straight back. Vertical motion is suppressed completely; the atmosphere has a lid on it. The plume can only fan out horizontally at its release height, and if the inversion sits above the stack the pollutants accumulate beneath it — the classic fumigation case, where the morning sun then heats the ground, breaks the inversion from below, and delivers the whole night’s accumulated plume to ground level at once. This is why severe air pollution episodes occur on still, clear winter mornings rather than at times of peak emission: the emissions have not changed, the atmosphere’s capacity to accept them has collapsed. It is also the honest limit of the tall-stack strategy — a stack disperses only as well as the atmosphere lets it, and buys nothing at all under an inversion that caps it. Option (b) has the sign backwards. Option (c) describes a lapse rate equal to the adiabatic. Option (d) describes lofting, which needs the inversion below the stack rather than above it, and is the one favourable case rather than this one.
A gravity settling chamber is 6 m long and 2 m wide and treats 3 m³/s of gas. Particle density is 2000 kg/m³ and gas viscosity is 1.8 × 10⁻⁵ Pa·s. Taking g = 9.81 m/s², the smallest particle diameter completely removed, in micrometres, is ______ (round to one decimal place)
Numerical answer — type the value.
Show answer
Answer: 64.3
64.3 μm. The first move is the one that makes this question easy, and it is to recognise that you have already done it — a settling chamber is a sedimentation tank with gas instead of water, and the same algebra applies. The particle must fall the full chamber height in the time the gas takes to cross the length, and when you write that out the height cancels: the required settling velocity is v = Q/(LW) = 3/(6 × 2) = 0.25 m/s, the surface overflow rate again, with no depth in it. Then invert Stokes for the diameter: v = gρ_p d²/18μ, so d = √(18μv/(gρ_p)) = √(18 × 1.8×10⁻⁵ × 0.25/(9.81 × 2000)) = √(4.128 × 10⁻⁹) = 6.425 × 10⁻⁵ m = 64.3 μm. That number is the whole reason the device exists in the position it does. A 50 μm particle settles at only 0.151 m/s, well below the 0.25 required, so it is not completely collected; anything in the respirable range is untouched, since v_s ∝ d² means a 10 μm particle settles 41 times more slowly than a 64 μm one. A settling chamber is therefore always a pre-cleaner — it takes the coarse load off a cyclone, baghouse or precipitator downstream and protects them from abrasion — and never the final device. The design lever is the same one as in water treatment: since height does not appear, the way to improve a chamber is to insert horizontal trays that multiply the collecting area within the same box.
A city of 100,000 people generates municipal solid waste at 0.5 kg per capita per day. Compacted to a density of 600 kg/m³, the volume required for 20 years of disposal at a working depth of 10 m, in hectares, is ______ (round to two decimal places, ignoring cover material)
Numerical answer — type the value.
Show answer
Answer: 6.08
6.08 hectares. Chain it through. Generation = 100,000 × 0.5 = 50,000 kg/day = 50 tonnes/day. Compacted volume = 50,000/600 = 83.33 m³/day. Over twenty years, 83.33 × 365 × 20 = 608,333 m³. At 10 m depth the plan area is 60,833 m², and 60,833/10,000 = 6.08 hectares. Three things in that chain carry marks in their own right. The density is doing more work than the tonnage: at a loose 300 kg/m³ the same waste would need 12.17 hectares, so compaction did not reduce the waste at all — it halved the land. The depth is the cheap lever, because area varies as its reciprocal: 20 m instead of 10 m halves the requirement to 3.04 ha, which is why landfills are excavated as well as raised, and why the binding constraint is usually the water table rather than the land price. And the question told you to ignore cover, which real design cannot: daily and intermediate cover typically consume 20 to 25 per cent of the airspace, so a real 20-year cell for this city needs closer to 7.6 hectares, before access roads, the buffer zone and the leachate pond. A landfill is an engineered structure with a liner beneath it and gas wells through it, and it is sized on airspace rather than on waste — which is why the number that matters to a municipality is cubic metres of void, not tonnes of waste.
Which of the following statements about noise are correct?
1. Doubling the distance from a point source reduces the level by about 6 dB, but from a line source by only about 3 dB
2. A barrier attenuates by forcing diffraction, so one that does not break the line of sight achieves very little
3. Control at the source is preferred over control at the receiver
4. Adding a 50 dB source next to a 60 dB source raises the combined level to 110 dB
Show answer
Answer: A — Doubling the distance from a point source reduces the level by about 6 dB, but from a line source by only about 3 dB; B — A barrier attenuates by forcing diffraction, so one that does not break the line of sight achieves very little; C — Control at the source is preferred over control at the receiver
(a), (b) and (c). (a) is correct and the difference between the two geometries is the reason traffic noise is such a difficult problem. A point source radiates over a sphere, so intensity falls as 1/r² and the level by 20 log₁₀ 2 = 6.02 dB per doubling. A road behaves as a line source: the wavefront is cylindrical, intensity falls as 1/r, and the level by only 10 log₁₀ 2 = 3.01 dB. Moving a house from 25 m to 50 m from a factory buys 6 dB; moving it the same distance from a highway buys 3. (b) is correct and is the most common design mistake in barrier work: a noise barrier does not stop sound, it makes the sound take a longer path over its top edge, and the attenuation depends on the path-length difference rather than on the barrier’s mass or thickness. A barrier with a gap, or one too low to break the line of sight from source to receiver, does almost nothing however solid it is. (c) is correct and is the standard hierarchy — source, then path, then receiver — for the plain reason that protection at the receiver protects only the person wearing it, and stops working the moment they take it off. (d) is wrong and is the arithmetic the whole section is built to prevent: levels do not add. 10 log₁₀(10⁶ + 10⁵) = 60.41 dB, so the 50 dB source raises the total by 0.41 dB — undetectable in any real environment, and a complete waste of money to silence while the 60 dB source is still running.
Incineration of municipal solid waste has repeatedly performed poorly in Indian cities. The most direct technical reason is that
1. the waste is high in organics and therefore high in moisture, so its calorific value is too low to sustain combustion
2. Indian municipal waste contains too much plastic, which cannot be burned at any temperature
3. the total quantity generated per city is too small to justify a furnace
4. incineration is prohibited for municipal waste under Indian rules
Show answer
Answer: A — the waste is high in organics and therefore high in moisture, so its calorific value is too low to sustain combustion
High organics means high moisture means low calorific value. The three characteristics the syllabus names — composition, moisture content and calorific value — are not independent measurements of a waste; they are a chain, and this is the chain in action. Indian municipal waste is typically 40 to 60 per cent food and garden waste by mass, and that fraction is mostly water. Water in the feed does not merely fail to burn; it consumes heat as it evaporates, so it subtracts from the energy released twice over. The net calorific value falls below the threshold at which combustion is self-sustaining, and a furnace that is not self-sustaining must be fed auxiliary fuel — at which point the plant is buying diesel to destroy waste, and the economics that justified it have inverted. Several Indian waste-to-energy plants have been built and shut for exactly this reason. What this argues for is not that thermal treatment is impossible but that it must follow segregation: separate the wet organic fraction and it becomes an excellent feed for composting or bio-methanation, where its moisture is a help rather than a hindrance, while the dry rejects left behind have a calorific value that will actually burn. Option (b) is wrong on the chemistry — plastics burn very well and are among the highest-calorific components; their real problem is emissions control, not combustion. Option (c) is wrong: quantity is not the constraint in a city of any size. Option (d) is simply false; the rules regulate incineration rather than prohibit it.
Ten identical machines each produce 60 dB at a receiver. Operating all ten together, the sound level at the receiver in dB is ______
Numerical answer — type the value.
Show answer
Answer: 70
Exactly 70 dB. L = 10 log₁₀(10 × 10^(60/10)) = 60 + 10 log₁₀ 10 = 60 + 10 = 70 dB. The result is worth holding as a rule rather than a calculation: multiplying the number of equal sources by ten adds exactly 10 dB, just as doubling them adds 3.01 dB. Ten machines are ten times the acoustic power and only ten decibels louder, which is what the logarithm buys — and it cuts both ways, which is the part that matters for control. To get 10 dB back from this workshop you would have to switch off nine of the ten machines; switching off five leaves 10 log₁₀(5 × 10⁶) = 67 dB, a reduction of just 3 dB for halving the plant. That is the arithmetic behind an uncomfortable engineering fact: where many equal sources contribute, partial control achieves very little, and the only effective strategies are to treat every source (quieter machines, enclosures on all of them) or to treat the path (a barrier, distance) or the receiver. It is the mirror image of the dominant-source case, where one loud machine among quiet ones means fixing that one solves the problem. Diagnosing which of the two situations you are in — one dominant source, or many equal ones — is the first step in any noise survey, because the right action is completely different in each.