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10 FE practice problems: BOD and dissolved oxygen, with solutions

By Dr. Mohamed Elansary, PhD Environmental EngineeringUpdated

These ten original problems practice BOD and dissolved oxygen, a topic from the FE Environmental exam specification, using the Environmental Engineering, Streeter Phelps 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, BOD and dissolved oxygen

Just below a wastewater outfall, a mixed stream has an ultimate BOD of 11.6 mg/L and a dissolved-oxygen deficit of 0.7 mg/L. The deoxygenation rate constant is 0.31/day and the reaeration rate constant is 0.87/day (base e); the saturation DO is 8.6 mg/L. Find the dissolved oxygen 3.9 days downstream (travel time), in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, Streeter Phelps, FE Reference Handbook 10.6

Show the worked solution
Answer
6.88 mg/L
Given
L0 = 11.6 mg/L, Da = 0.7 mg/L, kd = 0.31/day, kr = 0.87/day, DOsat = 8.6 mg/L, t = 3.9 days
Find
DO at 3.9 days (mg/L)
Handbook
Environmental Engineering, Stream Modeling, Streeter Phelps, page 327
Equation
D = [kd·Lo/(kr - kd)][exp(-kd t) - exp(-kr t)] + Da exp(-kr t); DO = DOsat - D
Substitute
  1. D = [kd·Lo/(kr - kd)](e^(-kd t) - e^(-kr t)) + Da·e^(-kr t)
  2. = [0.31(11.6)/(0.87 - 0.31)](e^(-1.209) - e^(-3.393)) + 0.7·e^(-3.393) = 1.724 mg/L
  3. DO = DOsat - D = 8.6 - 1.724 = 6.88 mg/L
Result
6.88 mg/L, 3 significant figures
Check
the critical (lowest-DO) point is at tc = 1.64 days of travel; DO cannot exceed saturation.
Common wrong answers
  • 1.72 mg/L: is the oxygen deficit, not the dissolved oxygen
  • 3.62 mg/L: swapped the deoxygenation and reaeration constants

Problem 2 · FE Environmental, BOD and dissolved oxygen

Just below a wastewater outfall, a mixed stream has an ultimate BOD of 12.5 mg/L and a dissolved-oxygen deficit of 1.4 mg/L. The deoxygenation rate constant is 0.12/day and the reaeration rate constant is 0.70/day (base e); the saturation DO is 9.1 mg/L. Find the dissolved oxygen 0.9 days downstream (travel time), in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, Streeter Phelps, FE Reference Handbook 10.6

Show the worked solution
Answer
7.41 mg/L
Given
L0 = 12.5 mg/L, Da = 1.4 mg/L, kd = 0.12/day, kr = 0.70/day, DOsat = 9.1 mg/L, t = 0.9 days
Find
DO at 0.9 days (mg/L)
Handbook
Environmental Engineering, Stream Modeling, Streeter Phelps, page 327
Equation
D = [kd·Lo/(kr - kd)][exp(-kd t) - exp(-kr t)] + Da exp(-kr t); DO = DOsat - D
Substitute
  1. D = [kd·Lo/(kr - kd)](e^(-kd t) - e^(-kr t)) + Da·e^(-kr t)
  2. = [0.12(12.5)/(0.70 - 0.12)](e^(-0.108) - e^(-0.630)) + 1.4·e^(-0.630) = 1.690 mg/L
  3. DO = DOsat - D = 9.1 - 1.690 = 7.41 mg/L
Result
7.41 mg/L, 3 significant figures
Check
the critical (lowest-DO) point is at tc = 1.70 days of travel; DO cannot exceed saturation.
Common wrong answers
  • 2.34 mg/L: swapped the deoxygenation and reaeration constants
  • 8.16 mg/L: left out the initial deficit term Da·e^(-kr t)
  • 1.69 mg/L: is the oxygen deficit, not the dissolved oxygen

Problem 3 · FE Environmental, BOD and dissolved oxygen

Just below a wastewater outfall, a mixed stream has an ultimate BOD of 9.0 mg/L and a dissolved-oxygen deficit of 2.8 mg/L. The deoxygenation rate constant is 0.22/day and the reaeration rate constant is 0.34/day (base e); the saturation DO is 8.2 mg/L. Find the dissolved oxygen 1.2 days downstream (travel time), in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, Streeter Phelps, FE Reference Handbook 10.6

Show the worked solution
Answer
4.64 mg/L
Given
L0 = 9.0 mg/L, Da = 2.8 mg/L, kd = 0.22/day, kr = 0.34/day, DOsat = 8.2 mg/L, t = 1.2 days
Find
DO at 1.2 days (mg/L)
Handbook
Environmental Engineering, Stream Modeling, Streeter Phelps, page 327
Equation
D = [kd·Lo/(kr - kd)][exp(-kd t) - exp(-kr t)] + Da exp(-kr t); DO = DOsat - D
Substitute
  1. D = [kd·Lo/(kr - kd)](e^(-kd t) - e^(-kr t)) + Da·e^(-kr t)
  2. = [0.22(9.0)/(0.34 - 0.22)](e^(-0.264) - e^(-0.408)) + 2.8·e^(-0.408) = 3.561 mg/L
  3. DO = DOsat - D = 8.2 - 3.561 = 4.64 mg/L
Result
4.64 mg/L, 3 significant figures
Check
the critical (lowest-DO) point is at tc = 2.08 days of travel; DO cannot exceed saturation.
Common wrong answers
  • 3.56 mg/L: is the oxygen deficit, not the dissolved oxygen
  • 6.50 mg/L: left out the initial deficit term Da·e^(-kr t)
  • 3.42 mg/L: swapped the deoxygenation and reaeration constants

Problem 4 · FE Environmental, BOD and dissolved oxygen

A 5-day BOD test gives BOD5 = 252 mg/L. The BOD rate constant is 0.10/day (base e). Find the ultimate carbonaceous BOD (Lo), in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, BOD Exertion, FE Reference Handbook 10.6

Show the worked solution
Answer
640 mg/L
Given
BOD5 = 252 mg/L, k = 0.10/day, t5 = 5-day
Find
ultimate BOD Lo (mg/L)
Handbook
Environmental Engineering, BOD Exertion, page 327
Equation
BODt = Lo(1 - e^(-kt)) with t = 5 days
Substitute
  1. BOD5 = Lo(1 - e^(-5k)), so Lo = BOD5/(1 - e^(-5k)) = 252/(1 - e^(-0.50)) = 252/0.3935 = 640 mg/L
Result
640 mg/L, 3 significant figures
Check
Lo is larger than BOD5, since only 39.3% of the ultimate BOD is exerted in 5 days.
Common wrong answers
  • 415 mg/L: divided by e^(-5k), the fraction remaining, instead of the fraction exerted
  • 99.2 mg/L: multiplied BOD5 by (1 - e^(-5k)) instead of dividing
  • 369 mg/L: used base 10 although k is base e

Problem 5 · FE Environmental, BOD and dissolved oxygen

A wastewater has an ultimate carbonaceous BOD (Lo) of 50 mg/L and a BOD rate constant of 0.18/day (base e). Find the BOD exerted after 1 days, in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, BOD Exertion, FE Reference Handbook 10.6

Show the worked solution
Answer
8.24 mg/L
Given
L0 = 50 mg/L, k = 0.18/day, t = 1 days
Find
BOD exerted after 1 days (mg/L)
Handbook
Environmental Engineering, BOD Exertion, page 327
Equation
BODt = Lo(1 - e^(-kt))
Substitute
  1. BODt = Lo(1 - e^(-kt)) = 50(1 - e^(-0.18 × 1)) = 50(1 - 0.8353) = 8.24 mg/L
Result
8.24 mg/L, 3 significant figures
Check
BODt is below Lo = 50 mg/L and approaches it as t grows.
Common wrong answers
  • 17.0 mg/L: used base 10 although the handbook's k is base e
  • 41.8 mg/L: is the BOD still remaining, not the BOD exerted
  • 9.00 mg/L: used Lo·k·t (a straight line)

Problem 6 · FE Environmental, BOD and dissolved oxygen

A wastewater has an ultimate carbonaceous BOD (Lo) of 215 mg/L and a BOD rate constant of 0.30/day (base e). Find the BOD exerted after 19 days, in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, BOD Exertion, FE Reference Handbook 10.6

Show the worked solution
Answer
214 mg/L
Given
L0 = 215 mg/L, k = 0.30/day, t = 19 days
Find
BOD exerted after 19 days (mg/L)
Handbook
Environmental Engineering, BOD Exertion, page 327
Equation
BODt = Lo(1 - e^(-kt))
Substitute
  1. BODt = Lo(1 - e^(-kt)) = 215(1 - e^(-0.30 × 19)) = 215(1 - 0.003346) = 214 mg/L
Result
214 mg/L, 3 significant figures
Check
BODt is below Lo = 215 mg/L and approaches it as t grows.

Problem 7 · FE Environmental, BOD and dissolved oxygen

A wastewater has an ultimate carbonaceous BOD (Lo) of 70 mg/L and a BOD rate constant of 0.30/day (base e). Find the BOD exerted after 6 days, in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, BOD Exertion, FE Reference Handbook 10.6

Show the worked solution
Answer
58.4 mg/L
Given
L0 = 70 mg/L, k = 0.30/day, t = 6 days
Find
BOD exerted after 6 days (mg/L)
Handbook
Environmental Engineering, BOD Exertion, page 327
Equation
BODt = Lo(1 - e^(-kt))
Substitute
  1. BODt = Lo(1 - e^(-kt)) = 70(1 - e^(-0.30 × 6)) = 70(1 - 0.1653) = 58.4 mg/L
Result
58.4 mg/L, 3 significant figures
Check
BODt is below Lo = 70 mg/L and approaches it as t grows.
Common wrong answers
  • 68.9 mg/L: used base 10 although the handbook's k is base e
  • 11.6 mg/L: is the BOD still remaining, not the BOD exerted

Problem 8 · FE Environmental, BOD and dissolved oxygen

Just below a wastewater outfall, a mixed stream has an ultimate BOD of 16.6 mg/L and a dissolved-oxygen deficit of 2.9 mg/L. The deoxygenation rate constant is 0.35/day and the reaeration rate constant is 0.50/day (base e); the saturation DO is 7.6 mg/L. Find the dissolved oxygen 3.9 days downstream (travel time), in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, Streeter Phelps, FE Reference Handbook 10.6

Show the worked solution
Answer
2.81 mg/L
Given
L0 = 16.6 mg/L, Da = 2.9 mg/L, kd = 0.35/day, kr = 0.50/day, DOsat = 7.6 mg/L, t = 3.9 days
Find
DO at 3.9 days (mg/L)
Handbook
Environmental Engineering, Stream Modeling, Streeter Phelps, page 327
Equation
D = [kd·Lo/(kr - kd)][exp(-kd t) - exp(-kr t)] + Da exp(-kr t); DO = DOsat - D
Substitute
  1. D = [kd·Lo/(kr - kd)](e^(-kd t) - e^(-kr t)) + Da·e^(-kr t)
  2. = [0.35(16.6)/(0.50 - 0.35)](e^(-1.365) - e^(-1.950)) + 2.9·e^(-1.950) = 4.794 mg/L
  3. DO = DOsat - D = 7.6 - 4.794 = 2.81 mg/L
Result
2.81 mg/L, 3 significant figures
Check
the critical (lowest-DO) point is at tc = 1.86 days of travel; DO cannot exceed saturation.
Common wrong answers
  • 0.601 mg/L: swapped the deoxygenation and reaeration constants
  • 3.22 mg/L: left out the initial deficit term Da·e^(-kr t)
  • 4.79 mg/L: is the oxygen deficit, not the dissolved oxygen

Problem 9 · FE Environmental, BOD and dissolved oxygen

Just below a wastewater outfall, a mixed stream has an ultimate BOD of 9.6 mg/L and a dissolved-oxygen deficit of 1.1 mg/L. The deoxygenation rate constant is 0.36/day and the reaeration rate constant is 0.75/day (base e); the saturation DO is 9.1 mg/L. Find the dissolved oxygen 1.9 days downstream (travel time), in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, Streeter Phelps, FE Reference Handbook 10.6

Show the worked solution
Answer
6.50 mg/L
Given
L0 = 9.6 mg/L, Da = 1.1 mg/L, kd = 0.36/day, kr = 0.75/day, DOsat = 9.1 mg/L, t = 1.9 days
Find
DO at 1.9 days (mg/L)
Handbook
Environmental Engineering, Stream Modeling, Streeter Phelps, page 327
Equation
D = [kd·Lo/(kr - kd)][exp(-kd t) - exp(-kr t)] + Da exp(-kr t); DO = DOsat - D
Substitute
  1. D = [kd·Lo/(kr - kd)](e^(-kd t) - e^(-kr t)) + Da·e^(-kr t)
  2. = [0.36(9.6)/(0.75 - 0.36)](e^(-0.684) - e^(-1.425)) + 1.1·e^(-1.425) = 2.605 mg/L
  3. DO = DOsat - D = 9.1 - 2.605 = 6.50 mg/L
Result
6.50 mg/L, 3 significant figures
Check
the critical (lowest-DO) point is at tc = 1.54 days of travel; DO cannot exceed saturation.
Common wrong answers
  • 2.60 mg/L: is the oxygen deficit, not the dissolved oxygen
  • 6.76 mg/L: left out the initial deficit term Da·e^(-kr t)
  • 3.67 mg/L: swapped the deoxygenation and reaeration constants

Problem 10 · FE Environmental, BOD and dissolved oxygen

A wastewater has an ultimate carbonaceous BOD (Lo) of 140 mg/L and a BOD rate constant of 0.36/day (base e). Find the BOD exerted after 16 days, in mg/L.

Enter your answer in mg/L, 3 significant figures.

Handbook: Environmental Engineering, BOD Exertion, FE Reference Handbook 10.6

Show the worked solution
Answer
140 mg/L
Given
L0 = 140 mg/L, k = 0.36/day, t = 16 days
Find
BOD exerted after 16 days (mg/L)
Handbook
Environmental Engineering, BOD Exertion, page 327
Equation
BODt = Lo(1 - e^(-kt))
Substitute
  1. BODt = Lo(1 - e^(-kt)) = 140(1 - e^(-0.36 × 16)) = 140(1 - 0.003151) = 140 mg/L
Result
140 mg/L, 3 significant figures
Check
BODt is below Lo = 140 mg/L and approaches it as t grows.

A new problem posts every day inside the lab, with the full worked solution the same evening.

Frequently asked questions

Where is BOD and dissolved oxygen in the FE Reference Handbook?

Look in the Environmental Engineering, Streeter Phelps 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

  1. NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
  2. NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.