1. CONCEPTUAL FRAMEWORK & ATMOSPHERIC PHYSICS

A. Air Pollutant Dispersion & Plume Behavior

When pollutants are released from an industrial stack, they do not remain concentrated at the point of emission. Wind carries the pollutants downwind while atmospheric turbulence spreads them horizontally and vertically.

The Gaussian Plume Model is a standard mathematical model used to estimate the concentration of pollutants downwind of a continuous point source such as an industrial chimney.

Gaussian Plume Model — Basic Assumptions

Continuous emission
The pollutant is released continuously at a constant emission rate, Q.
Steady wind
Wind speed, u, is assumed to remain approximately constant and acts in the downwind x-direction.
Gaussian dispersion
Pollutant concentration follows an approximately normal distribution in:
- Horizontal crosswind direction → y
- Vertical direction → z
Ground reflection
The plume is assumed to reflect from the ground surface rather than being absorbed.

Effective Stack Height

The actual chimney height is not necessarily the height at which the plume effectively disperses.

Hot flue gases have:

Thermal buoyancy due to their high temperature
Momentum due to their exit velocity

These effects cause the plume to rise above the physical stack.

Therefore:

H = hₛ + Δh

Where:

H = effective stack height
hₛ = physical stack height
Δh = plume rise

                  PLUME CENTERLINE
             . . . . . . . . . . . . . . . . .
          .´
        .´
      .´        ← Plume rise (Δh)
     /
    │
    │────────────────────────────── Effective height H
    │                              H = hₛ + Δh
    │
    │
    │  Chimney
    │
────┴──────────────────────────────────────────── Ground
          ↑
          │
       Physical
       height hₛ

Engineering Meaning

Higher effective stack height
            ↓
Plume travels higher before reaching ground
            ↓
Greater atmospheric dilution
            ↓
Generally lower ground-level concentration near the source

B. PHYSICS OF SOUND & DECIBEL MECHANICS

Sound Pressure Level — SPL

Sound pressure level is expressed using a logarithmic scale called the decibel (dB) scale.

The reference sound pressure is:

P₀ = 20 µPa = 2 × 10⁻⁵ Pa

The sound pressure level is:

Lₚ = 20 log₁₀(P / P₀)

Where:

Lₚ = sound pressure level, dB
P = RMS sound pressure, Pa
P₀ = reference sound pressure, 20 µPa

Why Are Decibels Logarithmic?

Human hearing responds to an extremely large range of sound pressures.

Therefore, a logarithmic scale is used instead of directly expressing pressure or intensity.

Important Rule

Sound levels cannot be added arithmetically.

For example:

60 dB + 60 dB ≠ 120 dB

Two identical independent 60 dB sources produce:

60 dB + 3 dB = 63 dB

Therefore:

Doubling acoustic energy → +3 dB

Equivalent Continuous Noise Level — Leq

Environmental and occupational noise often fluctuates with time.

Instead of reporting every individual sound level, the varying noise can be converted into one equivalent continuous sound level.

Leq = the constant sound level that contains the same total acoustic energy over the measurement period.

This is an energy-based logarithmic average, not an ordinary arithmetic average.

2. CORE MATHEMATICAL FORMULAS

Formula 1 — Gaussian Plume Ground-Level Concentration

For a point directly downwind of the source:

y = 0
z = 0

The simplified ground-level centerline equation is:

C = [Q / (π × u × σᵧ × σ_z)] × exp[−H² / (2σ_z²)]

Where:

C = pollutant concentration
Q = emission rate, g/s
u = wind speed, m/s
σᵧ = horizontal dispersion coefficient, m
σ_z = vertical dispersion coefficient, m
H = effective stack height, m

Exam logic

Q ↑  → C ↑
u ↑  → C ↓
H ↑  → generally C ↓ near source
σᵧ ↑ → C ↓
σ_z ↑ → C ↓

Formula 2 — Addition of Multiple Sound Sources

For n independent sound sources:

Ltotal = 10 log₁₀[Σ 10^(Lᵢ/10)]

Expanded:

Ltotal = 10 log₁₀[10^(L₁/10) + 10^(L₂/10) + ... + 10^(Lₙ/10)]

Important

Do not add dB values directly.

85 dB + 88 dB + 90 dB

NOT:

263 dB

Instead:

Convert each dB → energy term
             ↓
Add energy terms
             ↓
Take logarithm
             ↓
Combined dB

Formula 3 — Equivalent Continuous Noise Level

For varying noise levels:

Leq = 10 log₁₀[Σ fᵢ × 10^(Lᵢ/10)]

Where:

fᵢ = tᵢ / Ttotal

and:

Lᵢ = sound level during period i
tᵢ = duration of that level
Ttotal = total measurement period
fᵢ = fraction of total exposure time

The fractions must satisfy:

Σ fᵢ = 1

Formula 4 — Sound Propagation with Distance

1. Point Source

Examples:

Diesel generator
Individual machine
Isolated industrial equipment

Sound spreads approximately spherically.

L₂ = L₁ − 20 log₁₀(r₂/r₁)

For every doubling of distance:

Sound level decreases by approximately 6 dB

2. Line Source

Examples:

Continuous highway traffic
Long railway/traffic corridor

Sound spreads approximately cylindrically.

L₂ = L₁ − 10 log₁₀(r₂/r₁)

For every doubling of distance:

Sound level decreases by approximately 3 dB

3. SOLVED NUMERICAL EXAMPLES

Problem 1 — Gaussian Plume Dispersion

Question

A coal-fired power plant emits SO₂ at:

Q = 80 g/s

Effective stack height:

H = 60 m

Wind speed:

u = 4.0 m/s

At 1 km downwind:

σᵧ = 70 m

σ_z = 50 m

Calculate ground-level SO₂ concentration on the plume centerline.

Step 1 — Write the formula

C = [Q / (π × u × σᵧ × σ_z)] × exp[−H² / (2σ_z²)]

Step 2 — Calculate the denominator

π × 4 × 70 × 50

= 28,000π

≈ 87,964.59

Therefore:

Q / denominator = 80 / 87,964.59

= 9.0946 × 10⁻⁴ g/m³

Convert to µg/m³:

9.0946 × 10⁻⁴ g/m³ × 10⁶ µg/g

= 909.46 µg/m³

Step 3 — Calculate the exponential term

−H² / (2σ_z²)

= −60² / [2 × 50²]

= −3600 / 5000

= −0.72

Therefore:

exp(−0.72) ≈ 0.48675

Step 4 — Calculate concentration

C = 909.46 × 0.48675

C ≈ 442.68 µg/m³

Answer

Ground-level SO₂ concentration ≈ 442.68 µg/m³

Problem 2 — Addition of Multiple Industrial Noise Sources

Question

Three machines produce:

Machine 1 = 85 dB
Machine 2 = 88 dB
Machine 3 = 90 dB

Calculate the combined noise level.

Step 1 — Formula

Ltotal = 10 log₁₀[10^(85/10) + 10^(88/10) + 10^(90/10)]

Step 2 — Calculate the energy terms

10^(85/10) = 10⁸·⁵ = 316,227,766

10^(88/10) = 10⁸·⁸ = 630,957,344

10^(90/10) = 10⁹ = 1,000,000,000

Step 3 — Add

Total = 316,227,766 + 630,957,344 + 1,000,000,000

= 1,947,185,110

Step 4 — Convert back to dB

Ltotal = 10 log₁₀(1,947,185,110)

≈ 92.89 dB

Answer

Combined noise level ≈ 92.9 dB

Important observation

The highest source is only 90 dB, but the combined level is 92.9 dB, not 263 dB.

Problem 3 — Equivalent Continuous Noise Level

Question

An industrial worker experiences:

Noise Level	Duration
90 dB	3 hours
95 dB	2 hours
70 dB	3 hours

Total duration = 8 hours

Calculate the 8-hour Leq.

Step 1 — Calculate time fractions

For 90 dB:

f₁ = 3/8 = 0.375

For 95 dB:

f₂ = 2/8 = 0.250

For 70 dB:

f₃ = 3/8 = 0.375

Step 2 — Apply the Leq formula

**Leq = 10 log₁₀[
0.375 × 10⁹

0.250 × 10⁹·⁵
0.375 × 10⁷
]**

Step 3 — Calculate weighted contributions

90 dB:

0.375 × 1,000,000,000

= 375,000,000

95 dB:

0.250 × 3,162,277,660

= 790,569,415

70 dB:

0.375 × 10,000,000

= 3,750,000

Step 4 — Add

Weighted sum = 375,000,000 + 790,569,415 + 3,750,000

= 1,169,319,415

Step 5 — Convert to dB

Leq = 10 log₁₀(1,169,319,415)

≈ 90.68 dB

Answer

8-hour Leq ≈ 90.7 dB

Important observation

The 95 dB exposure lasts only 2 hours, but because the decibel scale is logarithmic, it contributes strongly to the total acoustic energy.

Problem 4 — Sound Attenuation with Distance

Part 1 — Point Source

Question

A diesel generator produces:

L₁ = 95 dB at r₁ = 10 m

Find the level at:

r₂ = 50 m

Step 1 — Formula

L₂ = L₁ − 20 log₁₀(r₂/r₁)

Step 2 — Substitute

L₂ = 95 − 20 log₁₀(50/10)

= 95 − 20 log₁₀(5)

Since:

log₁₀(5) ≈ 0.6990

Therefore:

L₂ = 95 − 13.98

= 81.02 dB

Answer

Point-source noise at 50 m = 81.02 dB

Part 2 — Line Source

Question

A continuous highway produces:

L₁ = 80 dB at 10 m

Find the level at:

r₂ = 50 m

Step 1 — Formula

L₂ = L₁ − 10 log₁₀(r₂/r₁)

Step 2 — Substitute

L₂ = 80 − 10 log₁₀(50/10)

= 80 − 10 log₁₀(5)

= 80 − 6.99

= 73.01 dB

Answer

Line-source noise at 50 m = 73.01 dB

4. POINT SOURCE vs LINE SOURCE

Feature	Point Source	Line Source
Typical example	Generator, isolated machine	Highway traffic
Sound spreading	Spherical	Cylindrical
Equation	L₂ = L₁ − 20 log₁₀(r₂/r₁)	L₂ = L₁ − 10 log₁₀(r₂/r₁)
Distance doubling	−6 dB	−3 dB
Reason	Energy spreads over increasing spherical area	Energy spreads over a cylindrical geometry

Memory Trick

POINT → 20 → 6

LINE → 10 → 3

5. FORMULA SUMMARY — QUICK REFERENCE

Parameter / Model	Standard Formula	Important Variables
Gaussian plume	C = [Q/(πuσᵧσ_z)] × exp[−H²/(2σ_z²)]	Q = g/s; u = m/s; H, σᵧ, σ_z = m
Effective stack height	H = hₛ + Δh	Physical height + plume rise
Sound pressure level	Lₚ = 20 log₁₀(P/P₀)	P₀ = 20 µPa
Decibel addition	Ltotal = 10 log₁₀[Σ10^(Lᵢ/10)]	Multiple independent sources
Equivalent noise level	Leq = 10 log₁₀[Σfᵢ10^(Lᵢ/10)]	fᵢ = tᵢ/T
Point-source attenuation	L₂ = L₁ − 20 log₁₀(r₂/r₁)	−6 dB per doubling
Line-source attenuation	L₂ = L₁ − 10 log₁₀(r₂/r₁)	−3 dB per doubling

6. RAPID EXAM RECALL

AIR POLLUTION
────────────────────────────────────

Gaussian plume
        ↓
Q = emission rate
u = wind speed
σy = horizontal spread
σz = vertical spread
H = effective stack height

H = hs + Δh

Ground-level centerline:
C = Q/(πuσyσz) × exp[−H²/(2σz²)]


NOISE
────────────────────────────────────

dB is logarithmic

Two identical sources:
60 dB + 60 dB = 63 dB

NOT 120 dB


MULTIPLE SOURCES
────────────────────────────────────

Ltotal = 10 log₁₀[Σ10^(Li/10)]


FLUCTUATING NOISE
────────────────────────────────────

Leq = 10 log₁₀[Σfi10^(Li/10)]


DISTANCE
────────────────────────────────────

POINT SOURCE
20 log
→ −6 dB per doubling

LINE SOURCE
10 log
→ −3 dB per doubling

Core memory chain

PLUME → STACK HEIGHT → GROUND CONCENTRATION

NOISE → LOG SCALE → ENERGY ADDITION → Leq → DISTANCE ATTENUATION

16.6: Air & Noise Pollution Calculations

1. CONCEPTUAL FRAMEWORK & ATMOSPHERIC PHYSICS

A. Air Pollutant Dispersion & Plume Behavior

Gaussian Plume Model — Basic Assumptions

Effective Stack Height

Engineering Meaning

B. PHYSICS OF SOUND & DECIBEL MECHANICS

Sound Pressure Level — SPL

Why Are Decibels Logarithmic?

Important Rule

Equivalent Continuous Noise Level — Leq

2. CORE MATHEMATICAL FORMULAS

Formula 1 — Gaussian Plume Ground-Level Concentration

Exam logic

Formula 2 — Addition of Multiple Sound Sources

Important

Formula 3 — Equivalent Continuous Noise Level

Formula 4 — Sound Propagation with Distance

1. Point Source

2. Line Source

3. SOLVED NUMERICAL EXAMPLES

Problem 1 — Gaussian Plume Dispersion

Question

Step 1 — Write the formula

Step 2 — Calculate the denominator

Step 3 — Calculate the exponential term

Step 4 — Calculate concentration

Answer

Problem 2 — Addition of Multiple Industrial Noise Sources

Question

Step 1 — Formula

Step 2 — Calculate the energy terms

Step 3 — Add

Step 4 — Convert back to dB

Answer

Important observation

Problem 3 — Equivalent Continuous Noise Level

Question

Step 1 — Calculate time fractions

Step 2 — Apply the Leq formula

Step 3 — Calculate weighted contributions

Step 4 — Add

Step 5 — Convert to dB

Answer

Important observation

Problem 4 — Sound Attenuation with Distance

Part 1 — Point Source

Question

Step 1 — Formula

Step 2 — Substitute

Answer

Part 2 — Line Source

Question

Step 1 — Formula

Step 2 — Substitute

Answer

4. POINT SOURCE vs LINE SOURCE

Memory Trick

5. FORMULA SUMMARY — QUICK REFERENCE

6. RAPID EXAM RECALL

Core memory chain

Related Notes