10 FE practice problems: landfill capacity and incineration, with solutions
These ten original problems practice landfill capacity and incineration, a topic from the FE Environmental exam specification, using the Environmental Engineering, Landfill 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, Landfill capacity and incineration
A city of 186,000 people generates 2.5 kg of solid waste per person per day. The waste is compacted in the landfill to 900 kg/m³, and daily cover uses 1 m³ of soil for every 5 m³ of compacted refuse. The landfill has 4,200,000 m³ of capacity. With no growth, the landfill life is most nearly:
Handbook: Environmental Engineering, Landfill, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (18.6 years)
- Given
- pop = 186,000, g = 2.5 kg, rho = 900 kg/m³, ratio = 5, V = 4,200,000 m³, one = 1
- Find
- landfill life (years)
- Handbook
- Environmental Engineering, Landfill (densities of compacted waste), page 332
- Equation
Life = capacity/[(people × generation × 365/compacted density) × (1 + cover/refuse)]- Substitute
Refuse per year = 186,000 × 2.5 kg/day × 365 days ÷ 900 kg/m³ = 188,600 m³With cover: 188,600 × (1 + 1/5) = 226,300 m³ per yearLife = 4,200,000/226,300 = 18.6 years
- Result
- 18.6 years, 3 significant figures
- Check
- cover soil uses part of the airspace, so the life is shorter than the refuse-only figure.
- Why the others are wrong
- A: used 5 volumes of cover per volume of refuse (inverted ratio)
- C: took cover as 1/6 of the refuse instead of 1/5
- D: left out the daily cover soil
Problem 2 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 82.5 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.999%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (8.25 × 10⁻⁴ kg/h)
- Given
- Win = 82.5 kg/h, DRE = 99.999%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 82.5 × (1 - 0.99999) = 8.25 × 10⁻⁴ kg/h
- Result
- 8.25 × 10⁻⁴ kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- A: used 99.9999% instead of 99.999%
- C: used 99.99% instead of 99.999%
- D: converted the result to lb/h but kept the kg/h label
Problem 3 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 116.5 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.9%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (0.116 kg/h)
- Given
- Win = 116.5 kg/h, DRE = 99.9%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 116.5 × (1 - 0.999) = 0.116 kg/h
- Result
- 0.116 kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- B: used 99.99% instead of 99.9%
- C: converted the result to lb/h but kept the kg/h label
- D: used 99.0% instead of 99.9%
Problem 4 · FE Environmental, Landfill capacity and incineration
A city of 429,000 people generates 1.4 kg of solid waste per person per day. The waste is compacted in the landfill to 650 kg/m³, and daily cover uses 1 m³ of soil for every 3 m³ of compacted refuse. The landfill has 13,500,000 m³ of capacity. With no growth, the landfill life is most nearly:
Handbook: Environmental Engineering, Landfill, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (30.0 years)
- Given
- pop = 429,000, g = 1.4 kg, rho = 650 kg/m³, ratio = 3, V = 13,500,000 m³, one = 1
- Find
- landfill life (years)
- Handbook
- Environmental Engineering, Landfill (densities of compacted waste), page 332
- Equation
Life = capacity/[(people × generation × 365/compacted density) × (1 + cover/refuse)]- Substitute
Refuse per year = 429,000 × 1.4 kg/day × 365 days ÷ 650 kg/m³ = 337,300 m³With cover: 337,300 × (1 + 1/3) = 449,700 m³ per yearLife = 13,500,000/449,700 = 30.0 years
- Result
- 30.0 years, 3 significant figures
- Check
- cover soil uses part of the airspace, so the life is shorter than the refuse-only figure.
- Why the others are wrong
- A: used 3 volumes of cover per volume of refuse (inverted ratio)
- C: took cover as 1/4 of the refuse instead of 1/3
- D: left out the daily cover soil
Problem 5 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 41 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.9999%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (4.10 × 10⁻⁵ kg/h)
- Given
- Win = 41 kg/h, DRE = 99.9999%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 41 × (1 - 0.999999) = 4.10 × 10⁻⁵ kg/h
- Result
- 4.10 × 10⁻⁵ kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- B: used 99.99999% instead of 99.9999%
- C: converted the result to lb/h but kept the kg/h label
- D: used 99.999% instead of 99.9999%
Problem 6 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 187.5 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.9%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (0.187 kg/h)
- Given
- Win = 187.5 kg/h, DRE = 99.9%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 187.5 × (1 - 0.999) = 0.187 kg/h
- Result
- 0.187 kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- B: used 99.99% instead of 99.9%
- C: converted the result to lb/h but kept the kg/h label
- D: divided the feed by the DRE value
Problem 7 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 61 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.9999%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (6.10 × 10⁻⁵ kg/h)
- Given
- Win = 61 kg/h, DRE = 99.9999%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 61 × (1 - 0.999999) = 6.10 × 10⁻⁵ kg/h
- Result
- 6.10 × 10⁻⁵ kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- A: converted the result to lb/h but kept the kg/h label
- B: used 99.99999% instead of 99.9999%
- C: used 99.999% instead of 99.9999%
Problem 8 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 69.5 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.999%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (6.95 × 10⁻⁴ kg/h)
- Given
- Win = 69.5 kg/h, DRE = 99.999%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 69.5 × (1 - 0.99999) = 6.95 × 10⁻⁴ kg/h
- Result
- 6.95 × 10⁻⁴ kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- B: used 99.99% instead of 99.999%
- C: converted the result to lb/h but kept the kg/h label
- D: used 99.9999% instead of 99.999%
Problem 9 · FE Environmental, Landfill capacity and incineration
A city of 310,000 people generates 2.0 kg of solid waste per person per day. The waste is compacted in the landfill to 900 kg/m³, and daily cover uses 1 m³ of soil for every 5 m³ of compacted refuse. The landfill has 9,100,000 m³ of capacity. With no growth, the landfill life is most nearly:
Handbook: Environmental Engineering, Landfill, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (30.2 years)
- Given
- pop = 310,000, g = 2.0 kg, rho = 900 kg/m³, ratio = 5, V = 9,100,000 m³, one = 1
- Find
- landfill life (years)
- Handbook
- Environmental Engineering, Landfill (densities of compacted waste), page 332
- Equation
Life = capacity/[(people × generation × 365/compacted density) × (1 + cover/refuse)]- Substitute
Refuse per year = 310,000 × 2.0 kg/day × 365 days ÷ 900 kg/m³ = 251,400 m³With cover: 251,400 × (1 + 1/5) = 301,700 m³ per yearLife = 9,100,000/301,700 = 30.2 years
- Result
- 30.2 years, 3 significant figures
- Check
- cover soil uses part of the airspace, so the life is shorter than the refuse-only figure.
- Why the others are wrong
- A: used 5 volumes of cover per volume of refuse (inverted ratio)
- C: took cover as 1/6 of the refuse instead of 1/5
- D: used 1,000 kg/m³ (water) instead of the compacted density
Problem 10 · FE Environmental, Landfill capacity and incineration
A hazardous waste incinerator is fed 6 kg/h of a principal organic hazardous constituent (POHC). The permit requires a destruction and removal efficiency (DRE) of 99.9999%. The largest stack emission rate of the POHC that meets the permit is most nearly:
Handbook: Environmental Engineering, Incineration, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (6.00 × 10⁻⁶ kg/h)
- Given
- Win = 6 kg/h, DRE = 99.9999%
- Find
- maximum POHC emission rate (kg/h)
- Handbook
- Environmental Engineering, Incineration (destruction and removal efficiency), page 325
- Equation
DRE = (Win - Wout)/Win × 100%- Substitute
DRE = (Win - Wout)/Win × 100%, so Wout = Win(1 - DRE/100)Wout = 6 × (1 - 0.999999) = 6.00 × 10⁻⁶ kg/h
- Result
- 6.00 × 10⁻⁶ kg/h, 3 significant figures
- Check
- each extra 'nine' in the DRE cuts the allowed emission by a factor of 10.
- Why the others are wrong
- A: converted the result to lb/h but kept the kg/h label
- B: used 99.999% instead of 99.9999%
- D: used 99.99999% instead of 99.9999%
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is landfill capacity and incineration in the FE Reference Handbook?
Look in the Environmental Engineering, Landfill 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.