Air quality and control carries 8–12 questions, and the energy and environment area (4–6) sits beside it on this page. The work runs from what a source emits, through the atmosphere, to the device that removes it.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
questions on the exam8–12
problems on this page16
areas in the specification15
Checked
What the area covers
Seven headings, from ambient air to the control device:
Ambient and indoor air quality: criteria pollutants, toxic and hazardous air pollutants.
Mass and energy balances: the STP basis, loading rates, heating values.
Emissions: emission factors and rates.
Atmospheric modeling and meteorology: stability classes, dispersion modeling, lapse rates.
Gas treatment: biofiltration, scrubbers, adsorbers, incineration, catalytic reducers.
Indoor air: air exchanges and the steady and unsteady reactor models of a room.
16 emissions and control problems
0 / 16 answered · 0 correct
Question 1 of 16 · Air Quality and Control · 8–12 items
A stack exhaust at 120 °C and 1.02 atm has a dry pollutant concentration of 25 ppmv and a volumetric flow of 2.50 m3/s (stack conditions). The pollutant's molar mass is 46.00 g/mol. Calculate the annual mass emissions in kg/yr (365 days of operation).
Given
Stack temperature = 120 °C
Stack pressure = 1.02 atm
Pollutant concentration = 25 ppmv (dry)
Stack flow = 2.50 m3/s
Molar mass M = 46.00 g/mol
Operation = 365 days/yr
Find
Annual mass emissions (kg/yr)
Solution
Convert ppmv to mg/m3 at the same conditions as the flow, then multiply by flow and time (Air quality: unit conversions).
3,780 kg/yr converts ppmv at 25 °C; the flow is at stack conditions (393 K), so the concentration must be too.
4,050 kg/yr converts ppmv at 0 °C and 1 atm (22.4 L/mol) instead of at stack conditions.
4,980 kg/yr converts ppmv at 25 °C and then scales by 393/298 instead of 298/393, correcting for the hot stack gas in the wrong direction.
9,390 kg/yr uses T = 120 instead of 393 K in the ideal gas law.
Question 2 of 16 · Energy and Environment · 4–6 items
A power plant generates 600 MW of electricity with a thermal efficiency of 35%. Cooling water from a nearby river is used to condense the steam, with a maximum allowable temperature rise of 8°C. If the specific heat of water is 4.18 kJ/kg·K, what is the required cooling water mass flow rate?
Given
Electrical output = 600 MW
Thermal efficiency = 35%
Allowed cooling water temperature rise = 8 °C
Specific heat of water = 4.18 kJ/(kg·K)
Find
Cooling water mass flow rate (kg/s)
Solution
First law on the plant, then ṁ = Q̇/(cpΔT) (Energy and environment: thermal discharge).
51,300 kg/s uses the whole 1,714 MW heat input; the 600 MW leaving as electricity is not rejected to the river.
139,000 kg/s leaves out the specific heat of 4.18 kJ/(kg·K).
Question 3 of 16 · Air Quality and Control · 8–12 items
A cyclone collector with a cut-point diameter (d50) of 5.0 μm is operating at its design flow rate of 1,000 acfm. The process flow is subsequently turned down to 250 acfm while maintaining the same gas temperature and composition. What is the new cut-point diameter?
Given
Design cut diameter d50 = 5.0 μm
Design flow = 1,000 acfm
New flow = 250 acfm
Find
New cut diameter d50 (μm)
Solution
Lapple cyclone: d50 = [9μW/(2πNeViρp)]1/2, so d50 ∝ 1/√Q at fixed geometry and gas (Air quality: cyclones).
d50,2 = 5.0 µm × √(1,000/250) = 10.0 µm
1.25 µm scales d50 directly with Q; a lower flow means lower inlet velocity and weaker separation, so the cut size grows.
2.50 µm scales with √Q; the cut diameter varies with 1/√Q.
5.00 µm assumes the cut size does not depend on flow, but it depends on the inlet velocity.
7.1 µm applies a fourth-root dependence on the flow ratio (5.0 × 41/4 = 7.07); the cut size varies with 1/√Q, a square root.
d50 ∝ 1/√Vi ∝ 1/√Q, so cutting the flow to a quarter doubles the cut size: 5.0 × √(1,000/250) = 10.0 µm.
Question 4 of 16 · Air Quality and Control · 8–12 items
A fabric filter baghouse operates with a filtration velocity (V) of 2.0 m/min and an inlet dust loading (C) of 1.0 g m−3. The specific cake resistance coefficient (K) is 10 Pa · min · m/g, and the clean filter pressure drop is 200 Pa. What is the total pressure drop after 10 minutes of operation?
Given
Filtration velocity V = 2.0 m/min
Inlet dust loading C = 1.0 g/m3
Specific cake resistance K = 10 Pa·min·m/g
Clean filter pressure drop ΔPclean = 200 Pa
Operating time t = 10 min
Find
Total pressure drop (Pa)
Solution
Filter drag model: clean-cloth drop plus a cake drop that grows with time (Air quality: fabric filters).
Question 5 of 16 · Air Quality and Control · 8–12 items
An electrostatic precipitator (ESP) achieves a collection efficiency of 99.0% at a gas flow rate of 10,000 acfm. If the gas flow rate is increased to 20,000 acfm while all other parameters remain constant, what is the new predicted efficiency?
Given
Initial collection efficiency η1 = 99.0%
Initial gas flow Q1 = 10,000 acfm
New gas flow Q2 = 20,000 acfm
Find
New collection efficiency (%)
Solution
Deutsch–Anderson: η = 1 − e−Aw/Q (Air quality: electrostatic precipitators).
49.5% halves the efficiency; doubling the flow halves the exponent Aw/Q, not η.
−ln(1 − 0.99) = 4.605 at 10,000 acfm; doubling Q halves Aw/Q to 2.303, so η = 1 − e−2.303 = 90.0%.
98.0% doubles the penetration (1% → 2%); penetration goes from e−4.6 to e−2.3, a tenfold rise.
99.0% assumes efficiency does not depend on flow, but a higher Q lowers Aw/Q.
99.5% moves in the wrong direction: more gas through the same plate area lowers efficiency.
Question 6 of 16 · Air Quality and Control · 8–12 items
A treatment system uses four identical baghouse modules in parallel to treat a gas stream with an inlet PM concentration of 20 g m−3. During an upset, one module is fully bypassed (0% removal) while the other three operate at 99% efficiency. Assuming flow is split equally, what is the combined outlet concentration?
Given
Modules in parallel = 4
Inlet PM concentration = 20 g/m3
Bypassed modules = 1 (0 % removal)
Efficiency of the other three = 99 %
Find
Combined outlet concentration (g/m3)
Solution
Flow-weighted average of the parallel outlets (Air quality: control-device performance).
0.200 g/m3 assumes all four modules still remove 99%; the bypassed quarter passes untreated.
5.00 g/m3 counts only the bypassed stream and forgets the 0.2 g/m3 still leaving the three working modules.
Three quarters of the flow leaves at 0.2 g/m3 and one quarter at 20 g/m3: 0.75 × 0.2 + 0.25 × 20 = 5.15 g/m3.
15.1 g/m3 swaps the fractions, as if three modules were bypassed and one worked.
20.0 g/m3 is the inlet concentration, as if no module were working.
Question 7 of 16 · Air Quality and Control · 8–12 items
During a particulate matter stack test, the sampling nozzle velocity is maintained at 10 m/s while the stack gas velocity is 20 m/s. How does this sub-isokinetic sampling condition bias the reported particulate mass concentration?
Solution
Sub-isokinetic sampling (Vnozzle < Vstack) over-collects inertial particles, biasing mass concentration high (Air quality: source sampling).
Inertial particles are over-collected when sampling sub-isokinetically, so the sample is enriched in coarse, not fine, particles.
Large particles are the ones most affected, because their inertia carries them across the diverging streamlines into the nozzle.
Underestimation comes from super-isokinetic sampling, when the nozzle is faster than the stack gas.
With the nozzle slower than the gas, excess gas flows around it, but heavier particles cannot follow the diverging streamlines and enter anyway, inflating the collected mass.
Only isokinetic sampling (nozzle velocity equal to gas velocity) gives an unbiased particle sample.
Question 8 of 16 · Air Quality and Control · 8–12 items
A facility has three emission sources with the operating parameters listed in the table below. Calculate the total hourly particulate matter (PM) emissions from the facility in lb/hr.
Source ID
Activity Rate
Emission Factor (EF)
Control Efficiency
Boiler A
50 MMBtu/hr
0.1 lb/MMBtu
80%
Process B
500 gal/hr
20 lb/1,000 gal
0%
Heater C
20 MMBtu/hr
0.1 lb/MMBtu
50%
Given
Boiler A activity rate = 50 MMBtu/hr
Boiler A emission factor EF = 0.1 lb/MMBtu
Boiler A control efficiency = 80%
Process B activity rate = 500 gal/hr
Process B emission factor EF = 20 lb/1,000 gal
Process B control efficiency = 0%
Heater C activity rate = 20 MMBtu/hr
Heater C emission factor EF = 0.1 lb/MMBtu
Heater C control efficiency = 50%
Find
Total controlled PM emissions (lb/hr)
Solution
Emissions = activity × emission factor × (1 − control efficiency), summed over sources (Air quality: emission inventories).
2.0 lb/hr adds the two controlled combustion sources and leaves out Process B.
6.0 lb/hr applies an average control efficiency (65%) to the whole facility, including the uncontrolled process.
Boiler A 5 × 0.20 = 1.0, Process B 500 × 20/1,000 = 10.0, Heater C 2 × 0.50 = 1.0; total = 12.0 lb/hr.
15.0 lb/hr multiplies by the control efficiency instead of (1 − efficiency), so Boiler A keeps 80% of its emissions.
17.0 lb/hr is the uncontrolled total; it ignores the 80% and 50% controls.
Question 9 of 16 · Energy and Environment · 4–6 items
An environmental audit of a manufacturing facility produced the greenhouse gas (GHG) inventory data shown below. Based on this data, what are the total Scope 1 emissions in kg CO2e?
Source
Classification
GHG
Mass (kg)
GWP
Natural Gas Boiler
On-site combustion
CO2
1,000
1
Purchased Electricity
Grid mix
CO2
5,000
1
Fleet Vehicles
On-site transport
CH4
10
25
Refrigeration Leak
Fugitive emissions
HFC-134a
1
1,430
Given
Natural gas boiler (on-site combustion): CO2 = 1,000 kg, GWP = 1
Scope 1 = direct emissions (on-site combustion, owned vehicles, fugitive releases) × GWP; purchased electricity is Scope 2 (Energy and environment: GHG accounting).
1,000 × 1 + 10 × 25 + 1 × 1,430 = 2,680 kg CO2e
1,251 weights the methane by its GWP of 25 but counts the 1 kg of HFC-134a at raw mass, skipping its GWP of 1,430.
2,440 counts the 10 kg of methane without its GWP of 25.
Scope 1 covers direct sources: boiler 1,000 + fleet 10 × 25 = 250 + refrigerant leak 1 × 1,430 = 1,430, total 2,680 kg CO2e.
6,011 adds raw masses of all gases, including purchased electricity, with no GWP weighting.
7,680 includes purchased electricity, which is an indirect (Scope 2) emission.
Question 10 of 16 · Air Quality and Control · 8–12 items
A gas stream is treated by two control devices in series: Device 1 with efficiency η1 followed by Device 2 with efficiency η2. A fraction B of the inlet flow bypasses Device 1 completely due to a duct leak, recombining with the treated stream before entering Device 2. Which expression represents the overall removal efficiency of the system?
Solution
Collectors in series multiply penetrations; a bypass around Device 1 lowers its effective efficiency to (1 − B)η1 (Air quality: control-device performance).
ηoverall = 1 − [1 − (1 − B)η1](1 − η2)
This is the series formula with no bypass; the leak lets fraction B skip Device 1.
Only (1 − B) of the flow sees Device 1, so its effective efficiency is (1 − B)η1; the recombined stream then passes Device 2.
This treats the bypass as skipping both devices, but the bypassed gas still goes through Device 2.
This puts the bypass on Device 2, but the leak is around Device 1.
Subtracting B directly mixes a flow fraction with an efficiency; the bypass reduces the first stage's efficiency multiplicatively.
Question 11 of 16 · Air Quality and Control · 8–12 items
A 100 m3 room has a ventilation rate of 200 m3/h and a first-order contaminant deposition rate constant k = 1.0 h−1. A continuous source starts emitting at t=0. Calculate the time required for the concentration to reach approximately 95% of its steady-state value.
Given
Room volume V = 100 m3
Ventilation rate Q = 200 m3/h
Deposition rate constant k = 1.0 h−1
Target = 95% of steady-state concentration
Find
Time to reach 95% of steady state (min)
Solution
One-zone box model with ventilation and first-order loss (Air quality: indoor air).
180 min uses deposition alone (τ = 1/k = 1 h) and ignores ventilation.
20 min is one time constant, V/(Q + kV); one τ reaches only 63% of steady state.
90 min uses τ = V/Q = 0.5 h and ignores the deposition loss.
30 min is V/Q, the time constant without deposition, and it is only one τ.
Question 12 of 16 · Air Quality and Control · 8–12 items
A building ventilation system supplies 1,000 m3/h of outdoor air and recirculates 3,000 m3/h through a filter with 50% efficiency (η= 0.5). The room volume is 500 m3 and a source emits contaminant at 500 mg/h. Calculate the steady-state concentration assuming perfect mixing.
Given
Outdoor air supply QOA = 1,000 m3/h
Recirculated flow Qrec = 3,000 m3/h
Filter efficiency η = 0.5
Room volume V = 500 m3
Source emission rate S = 500 mg/h
Find
Steady-state concentration (mg/m3)
Solution
Steady state: source = removal by outdoor air + filtered recirculation (Air quality: indoor air).
Clean-air delivery = 1,000 + 0.5 × 3,000 = 2,500 m3/h, so C = 500/2,500 = 0.200 mg/m3.
0.250 mg/m3 applies the 50% filter to the total 4,000 m3/h; only the recirculated air passes the filter.
0.333 mg/m3 credits only the filtered recirculation (0.5 × 3,000 m3/h) and forgets the 1,000 m3/h of outdoor air.
0.500 mg/m3 counts only outdoor air and gives no credit to the filter.
1.00 mg/m3 divides the source by the room volume; at steady state the room volume drops out.
Question 13 of 16 · Air Quality and Control · 8–12 items
A contaminant source (S = 200 mg/h) is activated in a clean room (V = 100 m3, Q = 200 m3/h). Calculate the integrated exposure (dose) in mg · h/m3 experienced by an occupant during the first 30 minutes (T=0.5 h).
Given
Source emission rate S = 200 mg/h
Room volume V = 100 m3
Ventilation rate Q = 200 m3/h
Exposure time T = 0.5 h
Find
Integrated exposure (mg·h/m3)
Solution
Integrate the one-zone build-up curve (Air quality: indoor air).
Question 14 of 16 · Air Quality and Control · 8–12 items
In a classroom with 25 occupants, the steady-state CO2 concentration is 1400 ppm and outdoor CO2 is 400 ppm. Assume a CO2 generation rate of 0.018 m3/h per person. Using this ventilation rate, estimate the infection probability for a 3-hour exposure to one infector (q=60 quanta/h, p = 0.5 m3/h).
Given
Occupants N = 25
Indoor CO2 = 1,400 ppm
Outdoor CO2 = 400 ppm
CO2 generation = 0.018 m3/h per person
Exposure time t = 3 h
Quanta generation q = 60 quanta/h
Breathing rate p = 0.5 m3/h
Infectors I = 1
Find
Infection probability (%)
Solution
CO2 balance gives the ventilation rate; Wells–Riley gives the infection probability (Air quality: indoor air).
22% computes e0.2 − 1, flipping the sign of the exponent.
24% uses the indoor 1,400 ppm instead of the 1,000 ppm rise above outdoor air, which understates ventilation.
33% leaves out the 0.5 m3/h breathing rate, doubling the inhaled dose.
99% uses the CO2 of one person instead of 25, so the ventilation comes out 25 times too low.
Question 15 of 16 · Energy and Environment · 4–6 items
A 500 MW natural gas power plant has a capital cost of $1,200/kW, a capital recovery factor (CRF) of 0.12, fixed O&M of $25/kW-yr, variable O&M of $4.00/MWh, a heat rate of 9,500 Btu/kWh, a fuel cost of $3.50/MMBtu, and a capacity factor of 0.70. Calculate the levelized cost of energy (LCOE) in $/MWh.
Given
Plant capacity = 500 MW
Capital cost = $1,200/kW
Capital recovery factor CRF = 0.12
Fixed O&M = $25/kW-yr
Variable O&M = $4.00/MWh
Heat rate = 9,500 Btu/kWh
Fuel cost = $3.50/MMBtu
Capacity factor = 0.70
Find
LCOE ($/MWh)
Solution
LCOE = annualized fixed costs ÷ annual energy + variable costs (Energy and environment: cost of energy).
$56.54/MWh spreads the fixed costs over 8.76 MWh per kW per year, as if the plant ran all year; at a 0.70 capacity factor it makes 6.13 MWh.
$41.33/MWh leaves out the annualized capital cost of $23.48/MWh.
$31.56/MWh leaves out the fuel cost (9.5 MMBtu/MWh × $3.50 = $33.25/MWh).
$237.02/MWh divides the full $1,200/kW capital cost by one year's output instead of annualizing it with the CRF.
Per kW: 8.76 × 0.70 = 6.13 MWh/yr; capital 144/6.13 = 23.48, fixed O&M 25/6.13 = 4.08, variable O&M 4.00, fuel 9.5 × 3.50 = 33.25; total $64.81/MWh.
Question 16 of 16 · Energy and Environment · 4–6 items
A utility portfolio consists of two plants. Plant A has 600 MW of capacity, a 0.80 capacity factor and an intensity of 900 kg CO2/MWh; Plant B has 400 MW of capacity, a 0.25 capacity factor and zero emissions. What is the weighted average CO2 emissions intensity for the combined generation?
Given
Plant A: 600 MW, capacity factor 0.80, 900 kg CO2/MWh
Plant B: 400 MW, capacity factor 0.25, 0 kg CO2/MWh
Find
Weighted average CO2 intensity (kg/MWh)
Solution
Average intensity = total emissions ÷ total energy (Energy and environment: emissions accounting).
432 kg/MWh divides Plant A's emissions (600 × 0.80 × 900) by the total installed capacity (1,000 MW) instead of by the energy generated.
450 kg/MWh is the simple average of 900 and 0, ignoring how much each plant generates.
540 kg/MWh weights by installed capacity (600 of 1,000 MW); intensity must be weighted by energy generated.
686 kg/MWh weights by capacity factor alone (0.80 of 1.05) and ignores the plant sizes.
Energy weights: A 600 × 0.80 = 480 MW average, B 400 × 0.25 = 100 MW; intensity = 480 × 900/580 = 745 kg/MWh.
Answer every question to see your score
Common traps
ppm in air needs a temperature and a pressure. It is a volume (mole) fraction. With MW in g/mol and the molar volume Vm in L/mol:
C (µg/m3) = ppm × MW × 1,000 / Vm
Vm is 22.4 L/mol at 1 atm and 0 °C, and 24.45 L/mol at 1 atm and 25 °C; stack gas is at neither, so compute it from RT/P.
Stability from the lapse rate. Dry air cools about 9.8 °C per km as it rises. An atmosphere cooling faster than that is unstable; one cooling slower, or warming with height in an inversion, is stable. Pasquill classes run from A (very unstable) to F (stable), read from wind speed and sunlight or cloud cover.
Effective stack height, not the physical one. A Gaussian plume is centered at H = h + Δh, where Δh is the plume rise. At ground level on the centerline, with reflection from the ground:
C = Q / (π u σy σz) × exp(−H2 / 2σz2)
Dry basis and oxygen correction. Limits are often stated dry at a reference oxygen level. Correct a measured value with Cref = Cmeas (20.9 − O2,ref) / (20.9 − O2,meas), oxygen in percent by volume.
Energy and Environment
The specification's energy area (4–6 questions) covers conventional and alternative energy sources and their impacts: greenhouse gases, carbon footprint, heat rejected to water and the water a plant needs. Three habits prevent most slips:
Power is not energy. MW is a rate; annual energy = capacity × capacity factor × 8,760 h.
Carbon is not CO2. One unit mass of carbon burns to 44/12 units of CO2.
CO2-equivalent multiplies each gas by its global warming potential. Use the values the problem gives, since published sets differ between assessment reports.
Cycle efficiencies and gas-law work belong to FE Mechanical thermodynamics; heating values and combustion of fuels and wastes stay with air.
Air law in one line
Clean Air Act (CAA): EPA sets National Ambient Air Quality Standards for six criteria pollutants (carbon monoxide, lead, nitrogen dioxide, ozone, particulate matter and sulfur dioxide) and regulates hazardous air pollutants separately (EPA).
Does the FE Environmental exam test air dispersion modeling?
Yes. Atmospheric modeling and meteorology, including stability classes, dispersion modeling and lapse rates, is one of the seven subtopic groups of this area.
Where is energy and environment covered?
On this page, in its own section. The specification gives it 4–6 questions.
Air quality drills on your phone
FE Environmental questions, answers and worked solutions in a study app.