10 FE practice problems: air quality: plume and control efficiency, with solutions
These ten original problems practice air quality: plume and control efficiency, a topic from the FE Environmental exam specification, using the Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian) part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.
How to use these problems
Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.
The problems
Problem 1 · FE Environmental, Air quality: plume and control efficiency
A stack emits 31.9 g/s of SO2 with an effective stack height of 35 m. The wind speed is 5.4 m/s. At the receptor distance, the stability-class charts give σy = 179.5 m and σz = 64 m. The ground-level concentration directly downwind of the stack is most nearly:
Handbook: Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (141 µg/m³)
- Given
- Qs = 31.9 g/s, H = 35 m, u = 5.4 m/s, sy = 179.5 m, sz = 64 m
- Find
- ground-level centerline concentration (µg/m³)
- Handbook
- Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), page 318
- Equation
C = [Q/(2π u σy σz)] exp(-y²/2σy²) [exp(-(z - H)²/2σz²) + exp(-(z + H)²/2σz²)], at y = z = 0- Substitute
At ground level (z = 0) on the centerline (y = 0), the Gaussian equation reduces to C = Q/(π u σy σz) exp[-H²/(2σz²)]C = 31.9/(π × 5.4 × 179.5 × 64) × exp[-35²/(2 × 64²)] g/m³ = 141 µg/m³
- Result
- 141 µg/m³, 3 significant figures
- Check
- the exponential factor 0.8611 shows how much the stack height lowers the ground value.
- Why the others are wrong
- B: left out the exponential term for the effective stack height
- C: used 1/(2π) without the ground-reflection term
- D: did not square H/σz in the exponent
Problem 2 · FE Environmental, Air quality: plume and control efficiency
An electrostatic precipitator with 6,600 m² of collection plates treats 232 m³/s of flue gas carrying 860 kg/h of fly ash. The drift velocity is 0.08 m/s. The fly ash emission rate leaving the precipitator is most nearly:
Handbook: Environmental Engineering, Electrostatic Precipitator Efficiency, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (88.3 kg/h)
- Given
- E_in = 860 kg/h, A = 6,600 m², Q = 232 m³/s, W = 0.08 m/s
- Find
- outlet emission rate (kg/h)
- Handbook
- Environmental Engineering, Electrostatic Precipitator Efficiency (Deutsch-Anderson equation), page 325
- Equation
η = 1 - e^(-WA/Q) (Deutsch-Anderson); emitted = inlet × e^(-WA/Q)- Substitute
η = 1 - e^(-WA/Q) = 1 - e^(-0.08 × 6,600/232) = 1 - e^(-2.276) = 89.73%Emitted = 860 × (1 - η) = 860 × e^(-2.276) = 88.3 kg/h
- Result
- 88.3 kg/h, 3 significant figures
- Check
- the emitted fraction is the penetration e^(-WA/Q); collected plus emitted equals the inlet rate.
- Why the others are wrong
- A: swapped the gas flow and the plate area
- B: used 10^(-WA/Q) instead of e^(-WA/Q)
- D: treated the gas flow as m³/min
Problem 3 · FE Environmental, Air quality: plume and control efficiency
An electrostatic precipitator with 2,600 m² of collection plates treats 192 m³/s of flue gas carrying 300 kg/h of fly ash. The drift velocity is 0.16 m/s. The fly ash emission rate leaving the precipitator is most nearly:
Handbook: Environmental Engineering, Electrostatic Precipitator Efficiency, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (34.4 kg/h)
- Given
- E_in = 300 kg/h, A = 2,600 m², Q = 192 m³/s, W = 0.16 m/s
- Find
- outlet emission rate (kg/h)
- Handbook
- Environmental Engineering, Electrostatic Precipitator Efficiency (Deutsch-Anderson equation), page 325
- Equation
η = 1 - e^(-WA/Q) (Deutsch-Anderson); emitted = inlet × e^(-WA/Q)- Substitute
η = 1 - e^(-WA/Q) = 1 - e^(-0.16 × 2,600/192) = 1 - e^(-2.167) = 88.54%Emitted = 300 × (1 - η) = 300 × e^(-2.167) = 34.4 kg/h
- Result
- 34.4 kg/h, 3 significant figures
- Check
- the emitted fraction is the penetration e^(-WA/Q); collected plus emitted equals the inlet rate.
- Why the others are wrong
- B: swapped the gas flow and the plate area
- C: used 10^(-WA/Q) instead of e^(-WA/Q)
- D: treated the gas flow as m³/min
Problem 4 · FE Environmental, Air quality: plume and control efficiency
An electrostatic precipitator with 4,400 m² of collection plates treats 279 m³/s of flue gas carrying 770 kg/h of fly ash. The drift velocity is 0.15 m/s. The fly ash emission rate leaving the precipitator is most nearly:
Handbook: Environmental Engineering, Electrostatic Precipitator Efficiency, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (72.3 kg/h)
- Given
- E_in = 770 kg/h, A = 4,400 m², Q = 279 m³/s, W = 0.15 m/s
- Find
- outlet emission rate (kg/h)
- Handbook
- Environmental Engineering, Electrostatic Precipitator Efficiency (Deutsch-Anderson equation), page 325
- Equation
η = 1 - e^(-WA/Q) (Deutsch-Anderson); emitted = inlet × e^(-WA/Q)- Substitute
η = 1 - e^(-WA/Q) = 1 - e^(-0.15 × 4,400/279) = 1 - e^(-2.366) = 90.61%Emitted = 770 × (1 - η) = 770 × e^(-2.366) = 72.3 kg/h
- Result
- 72.3 kg/h, 3 significant figures
- Check
- the emitted fraction is the penetration e^(-WA/Q); collected plus emitted equals the inlet rate.
- Why the others are wrong
- A: treated the gas flow as m³/min
- C: swapped the gas flow and the plate area
- D: is the rate collected on the plates, not the rate emitted
Problem 5 · FE Environmental, Air quality: plume and control efficiency
An electrostatic precipitator treats 287 m³/s of flue gas. The particle drift velocity is 0.16 m/s. The collection plate area needed for 98% efficiency is most nearly:
Handbook: Environmental Engineering, Electrostatic Precipitator Efficiency, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (7,020 m²)
- Given
- Q = 287 m³/s, W = 0.16 m/s, eta = 98%
- Find
- collection area A (m²)
- Handbook
- Environmental Engineering, Electrostatic Precipitator Efficiency (Deutsch-Anderson equation), page 325
- Equation
η = 1 - e^(-WA/Q) (Deutsch-Anderson)- Substitute
η = 1 - e^(-WA/Q), so A = -(Q/W) ln(1 - η)A = -(287/0.16) ln(1 - 0.98) = 1,794 × 3.912 = 7,020 m²
- Result
- 7,020 m², 3 significant figures
- Check
- each added 'nine' of efficiency costs another equal slice of plate area, because A grows with -ln(1 - η).
- Why the others are wrong
- A: used log10 instead of the natural log
- B: multiplied Q by W instead of dividing
- D: used A = ηQ/W (a straight-line efficiency)
Problem 6 · FE Environmental, Air quality: plume and control efficiency
An electrostatic precipitator with 1,500 m² of collection plates treats 66 m³/s of flue gas carrying 970 kg/h of fly ash. The drift velocity is 0.07 m/s. The fly ash emission rate leaving the precipitator is most nearly:
Handbook: Environmental Engineering, Electrostatic Precipitator Efficiency, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (198 kg/h)
- Given
- E_in = 970 kg/h, A = 1,500 m², Q = 66 m³/s, W = 0.07 m/s
- Find
- outlet emission rate (kg/h)
- Handbook
- Environmental Engineering, Electrostatic Precipitator Efficiency (Deutsch-Anderson equation), page 325
- Equation
η = 1 - e^(-WA/Q) (Deutsch-Anderson); emitted = inlet × e^(-WA/Q)- Substitute
η = 1 - e^(-WA/Q) = 1 - e^(-0.07 × 1,500/66) = 1 - e^(-1.591) = 79.63%Emitted = 970 × (1 - η) = 970 × e^(-1.591) = 198 kg/h
- Result
- 198 kg/h, 3 significant figures
- Check
- the emitted fraction is the penetration e^(-WA/Q); collected plus emitted equals the inlet rate.
- Why the others are wrong
- A: used 10^(-WA/Q) instead of e^(-WA/Q)
- B: treated the gas flow as m³/min
- C: is the rate collected on the plates, not the rate emitted
Problem 7 · FE Environmental, Air quality: plume and control efficiency
An electrostatic precipitator treats 42 m³/s of flue gas. The particle drift velocity is 0.15 m/s. The collection plate area needed for 90% efficiency is most nearly:
Handbook: Environmental Engineering, Electrostatic Precipitator Efficiency, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (645 m²)
- Given
- Q = 42 m³/s, W = 0.15 m/s, eta = 90%
- Find
- collection area A (m²)
- Handbook
- Environmental Engineering, Electrostatic Precipitator Efficiency (Deutsch-Anderson equation), page 325
- Equation
η = 1 - e^(-WA/Q) (Deutsch-Anderson)- Substitute
η = 1 - e^(-WA/Q), so A = -(Q/W) ln(1 - η)A = -(42/0.15) ln(1 - 0.9) = 280.0 × 2.303 = 645 m²
- Result
- 645 m², 3 significant figures
- Check
- each added 'nine' of efficiency costs another equal slice of plate area, because A grows with -ln(1 - η).
- Why the others are wrong
- A: used A = ηQ/W (a straight-line efficiency)
- B: used ln(η) instead of ln(1 - η)
- D: used log10 instead of the natural log
Problem 8 · FE Environmental, Air quality: plume and control efficiency
A stack emits 4.2 g/s of SO2 with an effective stack height of 120 m. The wind speed is 6.4 m/s. At the receptor distance, the stability-class charts give σy = 135 m and σz = 90.5 m. The ground-level concentration directly downwind of the stack is most nearly:
Handbook: Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (7.10 µg/m³)
- Given
- Qs = 4.2 g/s, H = 120 m, u = 6.4 m/s, sy = 135 m, sz = 90.5 m
- Find
- ground-level centerline concentration (µg/m³)
- Handbook
- Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), page 318
- Equation
C = [Q/(2π u σy σz)] exp(-y²/2σy²) [exp(-(z - H)²/2σz²) + exp(-(z + H)²/2σz²)], at y = z = 0- Substitute
At ground level (z = 0) on the centerline (y = 0), the Gaussian equation reduces to C = Q/(π u σy σz) exp[-H²/(2σz²)]C = 4.2/(π × 6.4 × 135 × 90.5) × exp[-120²/(2 × 90.5²)] g/m³ = 7.10 µg/m³
- Result
- 7.10 µg/m³, 3 significant figures
- Check
- the exponential factor 0.4152 shows how much the stack height lowers the ground value.
- Why the others are wrong
- A: used 1/(2π) without the ground-reflection term
- C: did not square H/σz in the exponent
- D: left out the exponential term for the effective stack height
Problem 9 · FE Environmental, Air quality: plume and control efficiency
A stack emits 33.4 g/s of SO2 with an effective stack height of 45 m. The wind speed is 5.4 m/s. At the receptor distance, the stability-class charts give σy = 83.5 m and σz = 77 m. The ground-level concentration directly downwind of the stack is most nearly:
Handbook: Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (258 µg/m³)
- Given
- Qs = 33.4 g/s, H = 45 m, u = 5.4 m/s, sy = 83.5 m, sz = 77 m
- Find
- ground-level centerline concentration (µg/m³)
- Handbook
- Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), page 318
- Equation
C = [Q/(2π u σy σz)] exp(-y²/2σy²) [exp(-(z - H)²/2σz²) + exp(-(z + H)²/2σz²)], at y = z = 0- Substitute
At ground level (z = 0) on the centerline (y = 0), the Gaussian equation reduces to C = Q/(π u σy σz) exp[-H²/(2σz²)]C = 33.4/(π × 5.4 × 83.5 × 77) × exp[-45²/(2 × 77²)] g/m³ = 258 µg/m³
- Result
- 258 µg/m³, 3 significant figures
- Check
- the exponential factor 0.8430 shows how much the stack height lowers the ground value.
- Why the others are wrong
- B: used 1/(2π) without the ground-reflection term
- C: left out the 2 in 2σz²
- D: did not square H/σz in the exponent
Problem 10 · FE Environmental, Air quality: plume and control efficiency
A stack emits 3.1 g/s of SO2 with an effective stack height of 55 m. The wind speed is 4.8 m/s. At the receptor distance, the stability-class charts give σy = 38.5 m and σz = 61 m. The ground-level concentration directly downwind of the stack is most nearly:
Handbook: Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (58.3 µg/m³)
- Given
- Qs = 3.1 g/s, H = 55 m, u = 4.8 m/s, sy = 38.5 m, sz = 61 m
- Find
- ground-level centerline concentration (µg/m³)
- Handbook
- Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian), page 318
- Equation
C = [Q/(2π u σy σz)] exp(-y²/2σy²) [exp(-(z - H)²/2σz²) + exp(-(z + H)²/2σz²)], at y = z = 0- Substitute
At ground level (z = 0) on the centerline (y = 0), the Gaussian equation reduces to C = Q/(π u σy σz) exp[-H²/(2σz²)]C = 3.1/(π × 4.8 × 38.5 × 61) × exp[-55²/(2 × 61²)] g/m³ = 58.3 µg/m³
- Result
- 58.3 µg/m³, 3 significant figures
- Check
- the exponential factor 0.6660 shows how much the stack height lowers the ground value.
- Why the others are wrong
- A: left out the 2 in 2σz²
- B: used 1/(2π) without the ground-reflection term
- C: left out the exponential term for the effective stack height
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is air quality: plume and control efficiency in the FE Reference Handbook?
Look in the Environmental Engineering, Atmospheric Dispersion Modeling (Gaussian) part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.
Are these real FE exam questions?
No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.
How do I check my answer before opening the solution?
Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.
Sources
- NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.