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BOD5 for Mixed Lagoons in Series

Environmental Engineering · FE Reference Handbook section

Environmental Engineering
4 formulas
10 exam-style examples
~53 min
All Environmental Engineering lectures

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
First-order BOD decay — solve for remaining BOD — BOD5 for Mixed Lagoons in Series

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 210.0 mg/L; rate constant (k) = 0.1900 1/day; time (t) = 9.5000 day, determine the remaining BOD (Lt) in mg/L.

Given

  • ultimateBOD(L0)=210.0mg/Lultimate BOD (L_{0}) = 210.0 mg/L
  • rateconstant(k)=0.19001/dayrate constant (k) = 0.1900 1/day
  • time(t)=9.5000daytime (t) = 9.5000 day

Find

remaining BOD (Lt), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except Lt is given, so isolate Lt symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that Lt stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=210.0mg/L,rateconstant(k)=0.19001/day,time(t)=9.5000dayList the givens: ultimate BOD (L_{0}) = 210.0 mg/L, rate constant (k) = 0.1900 1/day, time (t) = 9.5000 day
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    Lt=34.5396 mg/LLt = 34.5396\ \text{mg/L}
  6. Step 6 — Check: returning Lt = 34.5396 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Lt=34.5396 mg/LLt = 34.5396\ \text{mg/L}

Why the other options are there

  • 69.0793 — kept a factor of two that cancels in the correct rearrangement.
  • 17.2698 — dropped that same factor in the other direction.
  • 37.9936 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 2
First-order BOD decay — solve for ultimate BOD — BOD5 for Mixed Lagoons in Series (2)

A environmental engineering problem uses First-order BOD decay. Given rate constant (k) = 0.0700 1/day; time (t) = 3.5000 day; remaining BOD (Lt) = 303.9 mg/L, determine the ultimate BOD (L0) in mg/L.

Given

  • rateconstant(k)=0.07001/dayrate constant (k) = 0.0700 1/day
  • time(t)=3.5000daytime (t) = 3.5000 day
  • remainingBOD(Lt)=303.9mg/Lremaining BOD (Lt) = 303.9 mg/L

Find

ultimate BOD (L0), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except L0 is given, so isolate L0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that L0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:rateconstant(k)=0.07001/day,time(t)=3.5000day,remainingBOD(Lt)=303.9mg/LList the givens: rate constant (k) = 0.0700 1/day, time (t) = 3.5000 day, remaining BOD (Lt) = 303.9 mg/L
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    L0=388.3 mg/LL_{0} = 388.3\ \text{mg/L}
  6. Step 6 — Check: returning L0 = 388.3 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
L0=388.3 mg/LL_{0} = 388.3\ \text{mg/L}

Why the other options are there

  • 776.5 — kept a factor of two that cancels in the correct rearrangement.
  • 194.1 — dropped that same factor in the other direction.
  • 427.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 3
First-order BOD decay — solve for rate constant — BOD5 for Mixed Lagoons in Series (3)

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 375.0 mg/L; time (t) = 7.5000 day; remaining BOD (Lt) = 371.3 mg/L, determine the rate constant (k) in 1/day.

Given

  • ultimateBOD(L0)=375.0mg/Lultimate BOD (L_{0}) = 375.0 mg/L
  • time(t)=7.5000daytime (t) = 7.5000 day
  • remainingBOD(Lt)=371.3mg/Lremaining BOD (Lt) = 371.3 mg/L

Find

rate constant (k), in 1/day

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that k stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=375.0mg/L,time(t)=7.5000day,remainingBOD(Lt)=371.3mg/LList the givens: ultimate BOD (L_{0}) = 375.0 mg/L, time (t) = 7.5000 day, remaining BOD (Lt) = 371.3 mg/L
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    k=0.0013 1/dayk = 0.0013\ \text{1/day}
  6. Step 6 — Check: returning k = 0.0013 1/day to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
k=0.0013 1/dayk = 0.0013\ \text{1/day}

Why the other options are there

  • 0.0026 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0007 — dropped that same factor in the other direction.
  • 0.0015 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 4
First-order BOD decay — solve for remaining BOD (case 2) — BOD5 for Mixed Lagoons in Series (4)

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 260.0 mg/L; rate constant (k) = 0.0700 1/day; time (t) = 4.5000 day, determine the remaining BOD (Lt) in mg/L.

Given

  • ultimateBOD(L0)=260.0mg/Lultimate BOD (L_{0}) = 260.0 mg/L
  • rateconstant(k)=0.07001/dayrate constant (k) = 0.0700 1/day
  • time(t)=4.5000daytime (t) = 4.5000 day

Find

remaining BOD (Lt), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except Lt is given, so isolate Lt symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that Lt stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=260.0mg/L,rateconstant(k)=0.07001/day,time(t)=4.5000dayList the givens: ultimate BOD (L_{0}) = 260.0 mg/L, rate constant (k) = 0.0700 1/day, time (t) = 4.5000 day
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    Lt=189.7 mg/LLt = 189.7\ \text{mg/L}
  6. Step 6 — Check: returning Lt = 189.7 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Lt=189.7 mg/LLt = 189.7\ \text{mg/L}

Why the other options are there

  • 379.5 — kept a factor of two that cancels in the correct rearrangement.
  • 94.8726 — dropped that same factor in the other direction.
  • 208.7 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 5
First-order BOD decay — solve for ultimate BOD (case 2) — BOD5 for Mixed Lagoons in Series (5)

A environmental engineering problem uses First-order BOD decay. Given rate constant (k) = 0.1000 1/day; time (t) = 9.0000 day; remaining BOD (Lt) = 245.6 mg/L, determine the ultimate BOD (L0) in mg/L.

Given

  • rateconstant(k)=0.10001/dayrate constant (k) = 0.1000 1/day
  • time(t)=9.0000daytime (t) = 9.0000 day
  • remainingBOD(Lt)=245.6mg/Lremaining BOD (Lt) = 245.6 mg/L

Find

ultimate BOD (L0), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except L0 is given, so isolate L0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that L0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:rateconstant(k)=0.10001/day,time(t)=9.0000day,remainingBOD(Lt)=245.6mg/LList the givens: rate constant (k) = 0.1000 1/day, time (t) = 9.0000 day, remaining BOD (Lt) = 245.6 mg/L
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    L0=604.1 mg/LL_{0} = 604.1\ \text{mg/L}
  6. Step 6 — Check: returning L0 = 604.1 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
L0=604.1 mg/LL_{0} = 604.1\ \text{mg/L}

Why the other options are there

  • 1,208 — kept a factor of two that cancels in the correct rearrangement.
  • 302.0 — dropped that same factor in the other direction.
  • 664.5 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 6
First-order BOD decay — solve for rate constant (case 2) — BOD5 for Mixed Lagoons in Series (6)

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 220.0 mg/L; time (t) = 3.0000 day; remaining BOD (Lt) = 188.3 mg/L, determine the rate constant (k) in 1/day.

Given

  • ultimateBOD(L0)=220.0mg/Lultimate BOD (L_{0}) = 220.0 mg/L
  • time(t)=3.0000daytime (t) = 3.0000 day
  • remainingBOD(Lt)=188.3mg/Lremaining BOD (Lt) = 188.3 mg/L

Find

rate constant (k), in 1/day

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that k stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=220.0mg/L,time(t)=3.0000day,remainingBOD(Lt)=188.3mg/LList the givens: ultimate BOD (L_{0}) = 220.0 mg/L, time (t) = 3.0000 day, remaining BOD (Lt) = 188.3 mg/L
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    k=0.0519 1/dayk = 0.0519\ \text{1/day}
  6. Step 6 — Check: returning k = 0.0519 1/day to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
k=0.0519 1/dayk = 0.0519\ \text{1/day}

Why the other options are there

  • 0.1037 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0259 — dropped that same factor in the other direction.
  • 0.0571 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 7
First-order BOD decay — solve for remaining BOD (case 3) — BOD5 for Mixed Lagoons in Series (7)

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 355.0 mg/L; rate constant (k) = 0.0800 1/day; time (t) = 3.0000 day, determine the remaining BOD (Lt) in mg/L.

Given

  • ultimateBOD(L0)=355.0mg/Lultimate BOD (L_{0}) = 355.0 mg/L
  • rateconstant(k)=0.08001/dayrate constant (k) = 0.0800 1/day
  • time(t)=3.0000daytime (t) = 3.0000 day

Find

remaining BOD (Lt), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except Lt is given, so isolate Lt symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that Lt stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=355.0mg/L,rateconstant(k)=0.08001/day,time(t)=3.0000dayList the givens: ultimate BOD (L_{0}) = 355.0 mg/L, rate constant (k) = 0.0800 1/day, time (t) = 3.0000 day
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    Lt=279.3 mg/LLt = 279.3\ \text{mg/L}
  6. Step 6 — Check: returning Lt = 279.3 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Lt=279.3 mg/LLt = 279.3\ \text{mg/L}

Why the other options are there

  • 558.5 — kept a factor of two that cancels in the correct rearrangement.
  • 139.6 — dropped that same factor in the other direction.
  • 307.2 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 8
First-order BOD decay — solve for ultimate BOD (case 3) — BOD5 for Mixed Lagoons in Series (8)

A environmental engineering problem uses First-order BOD decay. Given rate constant (k) = 0.2100 1/day; time (t) = 2.0000 day; remaining BOD (Lt) = 256.5 mg/L, determine the ultimate BOD (L0) in mg/L.

Given

  • rateconstant(k)=0.21001/dayrate constant (k) = 0.2100 1/day
  • time(t)=2.0000daytime (t) = 2.0000 day
  • remainingBOD(Lt)=256.5mg/Lremaining BOD (Lt) = 256.5 mg/L

Find

ultimate BOD (L0), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except L0 is given, so isolate L0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that L0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:rateconstant(k)=0.21001/day,time(t)=2.0000day,remainingBOD(Lt)=256.5mg/LList the givens: rate constant (k) = 0.2100 1/day, time (t) = 2.0000 day, remaining BOD (Lt) = 256.5 mg/L
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    L0=390.4 mg/LL_{0} = 390.4\ \text{mg/L}
  6. Step 6 — Check: returning L0 = 390.4 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
L0=390.4 mg/LL_{0} = 390.4\ \text{mg/L}

Why the other options are there

  • 780.8 — kept a factor of two that cancels in the correct rearrangement.
  • 195.2 — dropped that same factor in the other direction.
  • 429.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 9
First-order BOD decay — solve for rate constant (case 3) — BOD5 for Mixed Lagoons in Series (9)

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 320.0 mg/L; time (t) = 7.5000 day; remaining BOD (Lt) = 168.8 mg/L, determine the rate constant (k) in 1/day.

Given

  • ultimateBOD(L0)=320.0mg/Lultimate BOD (L_{0}) = 320.0 mg/L
  • time(t)=7.5000daytime (t) = 7.5000 day
  • remainingBOD(Lt)=168.8mg/Lremaining BOD (Lt) = 168.8 mg/L

Find

rate constant (k), in 1/day

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except k is given, so isolate k symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that k stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=320.0mg/L,time(t)=7.5000day,remainingBOD(Lt)=168.8mg/LList the givens: ultimate BOD (L_{0}) = 320.0 mg/L, time (t) = 7.5000 day, remaining BOD (Lt) = 168.8 mg/L
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    k=0.0853 1/dayk = 0.0853\ \text{1/day}
  6. Step 6 — Check: returning k = 0.0853 1/day to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
k=0.0853 1/dayk = 0.0853\ \text{1/day}

Why the other options are there

  • 0.1706 — kept a factor of two that cancels in the correct rearrangement.
  • 0.0426 — dropped that same factor in the other direction.
  • 0.0938 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

Example 10
First-order BOD decay — solve for remaining BOD (case 4) — BOD5 for Mixed Lagoons in Series (10)

A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 385.0 mg/L; rate constant (k) = 0.1800 1/day; time (t) = 8.0000 day, determine the remaining BOD (Lt) in mg/L.

Given

  • ultimateBOD(L0)=385.0mg/Lultimate BOD (L_{0}) = 385.0 mg/L
  • rateconstant(k)=0.18001/dayrate constant (k) = 0.1800 1/day
  • time(t)=8.0000daytime (t) = 8.0000 day

Find

remaining BOD (Lt), in mg/L

Start with the thinking

  • The governing relation printed in this handbook section is First-order BOD decay.
  • Everything except Lt is given, so isolate Lt symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    Lt=L0e−ktL_t = L_0 e^{-k t}
  2. Step 2 — Rearrange the relation so that Lt stands alone on the left-hand side.

  3. Step 3

    Listthegivens:ultimateBOD(L0)=385.0mg/L,rateconstant(k)=0.18001/day,time(t)=8.0000dayList the givens: ultimate BOD (L_{0}) = 385.0 mg/L, rate constant (k) = 0.1800 1/day, time (t) = 8.0000 day
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    Lt=91.2172 mg/LLt = 91.2172\ \text{mg/L}
  6. Step 6 — Check: returning Lt = 91.2172 mg/L to

    Lt=L0e−ktL_t = L_0 e^{-k t}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Lt=91.2172 mg/LLt = 91.2172\ \text{mg/L}

Why the other options are there

  • 182.4 — kept a factor of two that cancels in the correct rearrangement.
  • 45.6086 — dropped that same factor in the other direction.
  • 100.3 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → BOD5 for Mixed Lagoons in Series

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