A wastewater has BOD_u = 320 mg/L with k = 0.23 /day (base e). What is BOD₅ and the remaining oxygen demand at day 5?

Start with the thinking

• BOD₅ is the amount exerted, not what remains.
• Exponential decay uses base e with this k.

Step-by-step solution

1. Decay factor

   $$e^{-kt} = e^{-0.23(5)} = e^{-1.15} = 0.3166$$

2. Exerted demand

   $$BOD_{5} = BOD_u(1 - e^{-kt}) = 320(1 - 0.3166)$$

3. Evaluate

   $$BOD_{5} = 320(0.6834) = 219 mg/L$$

4. Remaining

   $$320 - 219 = 101 mg/L$$

Why the other options are there

• 101 mg/L (remaining reported as BOD₅)
• 320 mg/L (ultimate reported)

Reference: FE Reference Handbook — Environmental — BOD kinetics

water treatment plant sedimentation basin surface loading rate design Given plan area (A) = 3,304 m^2; surface loading (overflow) rate (v) = 19.0000 m/day, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Water Treatment.
• Everything except Q is given, so isolate Q 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.
• Water treatment plant sedimentation basins are sized using surface loading rate relating flow, area, and overflow velocity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = A \times v$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = A\,v$$

3. Step 3 — List the givens: plan area (A) = 3,304 m^2, surface loading (overflow) rate (v) = 19.0000 m/day.

4. Step 4 — Substitute the given values:

   $$Q = 3304\,19.0000$$

5. Step 5 — Evaluate:

   Q = 62776\ \text{m^3/day}

6. Step 6 — Check: returning Q = 62,776 m^3/day to

   $$Q = A \times v$$

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

Why the other options are there

• 125,552 — kept a factor of two that cancels in the correct rearrangement.
• 31,388 — dropped that same factor in the other direction.
• 69,054 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

wastewater treatment secondary process efficiency calculation Given influent BOD (S_0) = 305.0 mg/L; effluent BOD (S) = 79.0000 mg/L, determine the removal efficiency (E) in %.

Start with the thinking

• The governing relation printed in this handbook section is Wastewater Treatment.
• Everything except E is given, so isolate E 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.
• Wastewater treatment plant performance is quantified by the BOD removal efficiency between influent and effluent streams.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{S_0-S}{S_0}\times 100$$

3. Step 3

   $$List the givens: influent BOD (S_0) = 305.0 mg/L, effluent BOD (S) = 79.0000 mg/L$$

4. Step 4 — Substitute the given values:

   $$E = \dfrac{S_0-79.0000}{S_0}\times 100$$

5. Step 5 — Evaluate:

   E = 74.0984\ \text{%}

6. Step 6 — Check: returning E = 74.0984 % to

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

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

Why the other options are there

• 148.2 — kept a factor of two that cancels in the correct rearrangement.
• 37.0492 — dropped that same factor in the other direction.
• 81.5082 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastewater Treatment Efficiency

water treatment settling basin sizing from flow and overflow rate Given surface loading (overflow) rate (v) = 25.0000 m/day; flow rate (Q) = 190,250 m^3/day, determine the plan area (A) in m^2.

Start with the thinking

• The governing relation printed in this handbook section is Water Treatment.
• Everything except A is given, so isolate A 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.
• Water treatment plant sedimentation basins are sized using surface loading rate relating flow, area, and overflow velocity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = A \times v$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{Q}{v}$$

3. Step 3 — List the givens: surface loading (overflow) rate (v) = 25.0000 m/day, flow rate (Q) = 190,250 m^3/day.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{190250}{25.0000}$$

5. Step 5 — Evaluate:

   A = 7610\ \text{m^2}

6. Step 6 — Check: returning A = 7,610 m^2 to

   $$Q = A \times v$$

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

Why the other options are there

• 15,220 — kept a factor of two that cancels in the correct rearrangement.
• 3,805 — dropped that same factor in the other direction.
• 8,371 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

wastewater treatment plant performance efficiency between influent and effluent Given influent BOD (S_0) = 309.0 mg/L; removal efficiency (E) = 90.0000 %, determine the effluent BOD (S) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Wastewater Treatment.
• Everything except S is given, so isolate S 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.
• Wastewater treatment plant performance is quantified by the BOD removal efficiency between influent and effluent streams.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

2. Step 2 — Rearrange symbolically for S:

   $$S = S_0\left(1-\dfrac{E}{100}\right)$$

3. Step 3

   $$List the givens: influent BOD (S_0) = 309.0 mg/L, removal efficiency (E) = 90.0000$$

4. Step 4 — Substitute the given values:

   $$S = S_0\left(1-\dfrac{90.0000}{100}\right)$$

5. Step 5 — Evaluate:

   $$S = 30.9000\ \text{mg/L}$$

6. Step 6 — Check: returning S = 30.9000 mg/L to

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

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

Why the other options are there

• 61.8000 — kept a factor of two that cancels in the correct rearrangement.
• 15.4500 — dropped that same factor in the other direction.
• 33.9900 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastewater Treatment Efficiency

water treatment clarifier surface loading rate calculation Given plan area (A) = 4,137 m^2; flow rate (Q) = 221,710 m^3/day, determine the surface loading (overflow) rate (v) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Water Treatment.
• Everything except v is given, so isolate v 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.
• Water treatment plant sedimentation basins are sized using surface loading rate relating flow, area, and overflow velocity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = A \times v$$

2. Step 2 — Rearrange symbolically for v:

   $$v = \dfrac{Q}{A}$$

3. Step 3

   $$List the givens: plan area (A) = 4,137 m^2, flow rate (Q) = 221,710 m^3/day$$

4. Step 4 — Substitute the given values:

   $$v = \dfrac{221710}{4137}$$

5. Step 5 — Evaluate:

   $$v = 53.5920\ \text{m/day}$$

6. Step 6 — Check: returning v = 53.5920 m/day to

   $$Q = A \times v$$

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

Why the other options are there

• 107.2 — kept a factor of two that cancels in the correct rearrangement.
• 26.7960 — dropped that same factor in the other direction.
• 58.9512 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

wastewater treatment plant BOD removal efficiency evaluation Given effluent BOD (S) = 15.0000 mg/L; removal efficiency (E) = 65.2000 %, determine the influent BOD (S_0) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Wastewater Treatment.
• Everything except S_0 is given, so isolate S_0 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.
• Wastewater treatment plant performance is quantified by the BOD removal efficiency between influent and effluent streams.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

2. Step 2 — Rearrange symbolically for S_0:

   $$S_{0} = \dfrac{S}{1-\dfrac{E}{100}}$$

3. Step 3

   $$List the givens: effluent BOD (S) = 15.0000 mg/L, removal efficiency (E) = 65.2000$$

4. Step 4 — Substitute the given values:

   $$S_{0} = \dfrac{15.0000}{1-\dfrac{65.2000}{100}}$$

5. Step 5 — Evaluate:

   $$S_{0} = 43.1034\ \text{mg/L}$$

6. Step 6 — Check: returning S_0 = 43.1034 mg/L to

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

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

Why the other options are there

• 86.2069 — kept a factor of two that cancels in the correct rearrangement.
• 21.5517 — dropped that same factor in the other direction.
• 47.4138 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastewater Treatment Efficiency

water treatment plant sedimentation basin surface loading rate design Given plan area (A) = 4,633 m^2; surface loading (overflow) rate (v) = 12.0000 m/day, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Water Treatment.
• Everything except Q is given, so isolate Q 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.
• Water treatment plant sedimentation basins are sized using surface loading rate relating flow, area, and overflow velocity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = A \times v$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = A\,v$$

3. Step 3 — List the givens: plan area (A) = 4,633 m^2, surface loading (overflow) rate (v) = 12.0000 m/day.

4. Step 4 — Substitute the given values:

   $$Q = 4633\,12.0000$$

5. Step 5 — Evaluate:

   Q = 55596\ \text{m^3/day}

6. Step 6 — Check: returning Q = 55,596 m^3/day to

   $$Q = A \times v$$

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

Why the other options are there

• 111,192 — kept a factor of two that cancels in the correct rearrangement.
• 27,798 — dropped that same factor in the other direction.
• 61,156 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

wastewater treatment secondary process efficiency calculation Given influent BOD (S_0) = 292.0 mg/L; effluent BOD (S) = 75.0000 mg/L, determine the removal efficiency (E) in %.

Start with the thinking

• The governing relation printed in this handbook section is Wastewater Treatment.
• Everything except E is given, so isolate E 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.
• Wastewater treatment plant performance is quantified by the BOD removal efficiency between influent and effluent streams.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{S_0-S}{S_0}\times 100$$

3. Step 3

   $$List the givens: influent BOD (S_0) = 292.0 mg/L, effluent BOD (S) = 75.0000 mg/L$$

4. Step 4 — Substitute the given values:

   $$E = \dfrac{S_0-75.0000}{S_0}\times 100$$

5. Step 5 — Evaluate:

   E = 74.3151\ \text{%}

6. Step 6 — Check: returning E = 74.3151 % to

   $$E = \dfrac{S_0 - S}{S_0} \times 100$$

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

Why the other options are there

• 148.6 — kept a factor of two that cancels in the correct rearrangement.
• 37.1575 — dropped that same factor in the other direction.
• 81.7466 — rounded an intermediate value before the final step.

Reference: FE Handbook — Wastewater Treatment Efficiency

water treatment settling basin sizing from flow and overflow rate Given surface loading (overflow) rate (v) = 16.5000 m/day; flow rate (Q) = 221,780 m^3/day, determine the plan area (A) in m^2.

Start with the thinking

• The governing relation printed in this handbook section is Water Treatment.
• Everything except A is given, so isolate A 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.
• Water treatment plant sedimentation basins are sized using surface loading rate relating flow, area, and overflow velocity.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$Q = A \times v$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{Q}{v}$$

3. Step 3 — List the givens: surface loading (overflow) rate (v) = 16.5000 m/day, flow rate (Q) = 221,780 m^3/day.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{221780}{16.5000}$$

5. Step 5 — Evaluate:

   A = 13441\ \text{m^2}

6. Step 6 — Check: returning A = 13,441 m^2 to

   $$Q = A \times v$$

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

Why the other options are there

• 26,882 — kept a factor of two that cancels in the correct rearrangement.
• 6,721 — dropped that same factor in the other direction.
• 14,785 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)