Air Quality and Control · Study · FE Environmental · FE → PE Prep
Air Quality and Control
8% of exam
Ambient and indoor air quality, mass and energy balances, emission factors and rates, atmospheric dispersion and stability classes, and gas and particulate control technologies.
5 concepts
A. Ambient and indoor air quality
Ambient Air Quality and Criteria Pollutants
Know the six criteria pollutants and the NAAQS, compute the Air Quality Index, and trace photochemical smog, ozone formation, acid deposition, and primary versus secondary pollutants.
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C. Emissions
Emission Factors and Mass Balances
Estimate pollutant emission rates from AP-42 emission factors, close steady-state mass balances on processes and control devices, combine efficiencies in series, and convert ppm to mg/m³.
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Predict downwind ground-level concentrations from an elevated source using the Gaussian dispersion model, Pasquill stability classes, effective stack height, and the lapse rate.
Almost every air-dispersion question on the FE Environmental exam is a single equation wearing different costumes: the Gaussian plume. A continuous source — a stack, a flare, a vent — releases pollutant at rate Q, the wind carries it downwind at speed u, and turbulence spreads it into an ever-widening cone whose cross-section is a two-dimensional bell curve. Your job is almost always the same: find the concentration that a receptor on the ground actually breathes, some distance downwind. Points are lost not in the exponential but in three bookkeeping decisions — using the effective stack height instead of the physical one, reading the right dispersion coefficients σy and σz for the stability class, and keeping Q in μg/s so the answer lands in μg/m3. This material lives in the FE Reference Handbook under Environmental Engineering, "Atmospheric Dispersion Modeling (Gaussian)."
The full Gaussian plume equation
The model treats the time-averaged plume as a Gaussian distribution in the crosswind (y) and vertical (z) directions superimposed on steady downwind advection. The centerline rides at the effective stack height H, and the ground at z=0 acts as a mirror that reflects the plume back upward — that reflection is the second exponential term. Here C
The cases you actually compute: ground level and centerline
Two simplifications cover the vast majority of exam problems. First, the receptor is almost always at ground level, z=0, where the two reflection terms become identical and add to give a factor of two. Second, the worst-case receptor sits on the plume centerline, y=0, where the crosswind exponential is one. Imposing both at once collapses the full model into the compact ground-level, centerline concentration below — memorise this form, because it is the one you will substitute into most often.
C(x,0,0)=πuσyσzQexp(−2σz2H2)
Effective stack height and plume rise
The plume does not disperse from the top of the stack — buoyant, hot exhaust and exit momentum carry it higher before it levels off and begins to spread. The effective stack height H is the physical height h plus this plume rise Δh. Because H enters the ground-level equation inside a squared exponential, modest changes have an outsized effect: a taller effective stack pushes the centerline up and the ground-level concentration plummets. The handbook supplies H=h+Δh
Pasquill-Gifford stability classes and dispersion coefficients
Turbulence — and therefore σy and σz — is governed by atmospheric stability, sorted into six Pasquill classes from A (extremely unstable, strong daytime sun, light wind) through D (neutral, overcast or windy) to F (moderately stable, clear calm night). You select the class from a table keyed on surface wind speed, solar insolation by day, and cloud cover by night, then read σy
Maximum ground-level concentration
As you walk downwind, the ground-level centerline concentration first rises (the plume is bending down toward the ground) then falls (dilution wins). The peak occurs where the vertical spread reaches σz=H/2. Substituting that condition into the ground-level equation makes the exponential equal
Lapse rate, mixing height, and plume behaviour
Stability ultimately comes from comparing the actual (environmental) lapse rate Γ — how fast ambient temperature falls with height — to the dry adiabatic lapse rate ΓAD=0.98∘C per 100m
Exam strategy
Work every plume problem in the same order. (1) Convert Q to μg/s immediately (1g/s=106μg/s) so the answer is in μg/m3
Problem. A power-plant stack emits SO2 at Q=80g/s. Wind speed at stack height is u=5m/s
Common pitfalls
•Using the physical stack height h instead of the effective height H=h+Δh. Plume rise can easily double the effective height and cut ground concentration by an order of magnitude — always add Δh first.
•Leaving Q
References
NCEES FE Reference Handbook — Environmental Engineering
Turner, Workbook of Atmospheric Dispersion Estimates — Pasquill-Gifford curves and stability classification
Davis & Cornwell, Introduction to Environmental Engineering — air dispersion chapter
F. Particle treatment technologies
Particulate Control: Cyclones, ESPs, and Baghouses
Size and rate the three workhorse particulate collectors — cyclone cut diameter, the ESP Deutsch-Anderson equation, and baghouse air-to-cloth ratio — and match each device to particle size.
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G. Indoor air quality modeling and controls
Indoor Air Quality Modeling and Ventilation
Model indoor pollutant concentrations with the well-mixed mass balance, find the steady state, track transient build-up and decay through the air-exchange rate, and size ventilation.
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is the steady-state concentration at point
(x,y,z)
in
μg/m3
,
Q
the emission rate in
μg/s
,
u
the wind speed at stack height in
m/s
, and
σy,σz
the horizontal and vertical dispersion parameters in metres, both functions of downwind distance
itself (from buoyancy and momentum flux, e.g. Holland's or Briggs' formulas) is generally given to you or stated in the problem.
H=h+Δh
and
σz
off the Pasquill-Gifford curves at your downwind distance
x
. The unstable classes spread a plume rapidly (large
σ
, low concentrations close in); the stable classes keep it tight and ribbon-like. The single most common error is mixing up classes or reading the curves at the wrong
x
. Remember: overcast forces Class D regardless of wind speed.
e−1
, giving the clean maximum below. You evaluate
σy
at the same downwind distance where
σz=H/2
. This tells a designer the highest concentration any ground receptor will ever see and roughly where it sits.
Cmax=πeuσyσzQσz=H/2
(
5.4∘F
per
1,000ft
). If
Γ>ΓAD
the air is unstable (a rising parcel stays warmer and keeps rising — looping plume); if
Γ=ΓAD
it is neutral (coning); if
Γ<ΓAD
it is stable (a rising parcel cools below its surroundings and sinks — fanning, poor dispersion). A temperature inversion (
Γ<0
, temperature rising with height) is the extreme stable case and caps vertical mixing. The mixing height is the altitude of that cap; plumes trapped beneath it cannot vent upward, and a plume above an inversion that mixes down to the ground produces fumigation, the worst-case ground exposure.
Inversion (Γ<0) is the strongly stable limit that caps mixing and traps the plume.
ppb / concentration conversionppb=PMW(μg/m3)RT
Convert mass concentration to volume mixing ratio. R=0.0821L⋅atm/(mol⋅K), T in K, P in atm, MW in g/mol.
and the effective stack height is
H=50m
. Under neutral (Class D) conditions, at
x=1km
downwind the Pasquill-Gifford curves give
σy=68m
and
σz=33m
. Find the ground-level concentration on the plume centerline.
Solution. Convert the source: Q=80g/s=8.0×107μg/s.
Use the ground-level centerline form (z=0,y=0):
C=πuσyσzQexp(−2σz2H2).
Exponent: 2σz2H2=2(33)2502=21782500=1.148, so e−1.148=0.317.
Prefactor: πuσyσz=π(5)(68)(33)=35,250m3/s.
C=35,2508.0×107(0.317)=(2270)(0.317)=720μg/m3.
**Answer: C≈720μg/m3.** Units check: (μg/s)/(m3/s)=μg/m3. The exponent is order one, confirming the plume has reached the ground at this distance.
C=π(5)(68)(33)8.0×107e−1.148≈720μg/m3
Maximum ground-level concentration
Problem. For the same stack (Q=80g/s, u=5m/s, H=50m, Class D), find the maximum ground-level concentration. At the downwind distance where σz=H/2 (just beyond 1km for Class D), the curves give σy≈70m.
Solution. The maximum occurs where the vertical spread satisfies σz=H/2=50/1.414=35.4m
Cmax=πe(5)(70)(35.4)8.0×107≈757μg/m3
Stability class from the lapse rate
Problem. On a given morning the air temperature is 22.0∘C at the 2m reference height and 20.0∘C at 302m. Determine the environmental lapse rate, classify the stability, and describe the expected plume behaviour.
Solution. Take both differences in the same direction, top minus bottom: ΔT=Ttop−Tbottom=20.0−22.0=−2.0∘C
Γ=−ΔzΔT=−300m−2.0∘C=0.667∘C/100m<ΓAD
in g/s. The handbook Gaussian equation expects
Q
in
μg/s
to give
C
in
μg/m3
; multiply g/s by
106
before substituting.
•Forgetting the factor of two at the ground. The general equation has two reflection terms; at z=0 they are equal and add, which is why the ground-level form has π (not 2π) in the denominator.
•Reading σy,σz at the wrong distance or wrong stability class. They depend on both x and class; an overcast sky forces Class D regardless of wind speed.
•Confusing mass concentration (μg/m3) with volume mixing ratio (ppm/ppb). Convert with ppb=(μg/m3)RT/(P⋅MW) — never equate them directly.
•Sign error in the lapse rate: a stable atmosphere has Γ<ΓAD (small or negative slope), an unstable one Γ>ΓAD. An inversion (T rising with height) is strongly stable, not unstable.
•Assuming the maximum concentration is at the receptor you were given. The peak is at σz=H/2; you must read σy at that location, not at an arbitrary downwind distance.
in the previous example — as it must, since the peak sits slightly farther downwind, where
σz
has grown from
33
to
35.4m
while
σy
is still near
70m
. A computed maximum that fell below an already-known receptor value would flag an inconsistent
σy
. The procedure — set
σz=H/2
, then read
σy
at that same distance — is the exam method.
over
Δz=302−2=300m
(temperature decreases with height).
Γ=−ΔzΔT=−300m−2.0∘C=0.667∘C/100m.
Compare to the dry adiabatic rate
ΓAD=0.98∘C/100m
. Since
Γ=0.667<0.98=ΓAD
, the atmosphere is **stable**.
**Answer:
Γ=0.667∘C/100m
, stable.** A stable layer suppresses vertical mixing, so the plume disperses laterally but barely vertically — a flat, ribbon-like ·fanning· plume with poor ground-level dispersion near the source. Sanity check: temperature still decreases with height (no inversion), so this is stable but not the trapped inversion limit.