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

A rectangular clarifier is 25 m long and 8 m wide, treating 6,000 m³/day. Find the surface overflow rate and, for a 3.0 m depth, the detention time.

Start with the thinking

• Overflow rate uses plan area only — depth does not appear.
• Detention time uses the full volume.

Step-by-step solution

1. Plan area

   $$A = 25(8) = 200 m^{2}$$

2. Overflow rate

   $$v_o = Q/A = 6,000/200 = 30.0 m/day$$

3. Volume

   $$V = 200(3.0) = 600 m^{3}$$

4. Detention time

   $$t = V/Q = 600/6,000 = 0.100 day$$

5. Result — v_o = 30 m³/(m²·day), t = 2.4 hours

Why the other options are there

• 10 m/day (depth divided in)
• t = 24 h (day-to-hour conversion missed)

Reference: FE Reference Handbook — Environmental — Sedimentation

water treatment plant sedimentation basin surface loading rate design Given plan area (A) = 2,398 m^2; surface loading (overflow) rate (v) = 46.0000 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) = 2,398 m^2, surface loading (overflow) rate (v) = 46.0000 m/day.

4. Step 4 — Substitute the given values:

   $$Q = 2398\,46.0000$$

5. Step 5 — Evaluate:

   Q = 110308\ \text{m^3/day}

6. Step 6 — Check: returning Q = 110,308 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

• 220,616 — kept a factor of two that cancels in the correct rearrangement.
• 55,154 — dropped that same factor in the other direction.
• 121,339 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

water treatment settling basin sizing from flow and overflow rate Given surface loading (overflow) rate (v) = 56.0000 m/day; flow rate (Q) = 97,060 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) = 56.0000 m/day, flow rate (Q) = 97,060 m^3/day.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{97060}{56.0000}$$

5. Step 5 — Evaluate:

   A = 1733\ \text{m^2}

6. Step 6 — Check: returning A = 1,733 m^2 to

   $$Q = A \times v$$

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

Why the other options are there

• 3,466 — kept a factor of two that cancels in the correct rearrangement.
• 866.6 — dropped that same factor in the other direction.
• 1,907 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

water treatment clarifier surface loading rate calculation Given plan area (A) = 683.0 m^2; flow rate (Q) = 291,220 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

   $$List the givens: plan area (A) = 683.0 m^2, flow rate (Q) = 291,220 m^3/day$$

4. Step 4 — Substitute the given values:

   $$v = \dfrac{291220}{683.0}$$

5. Step 5 — Evaluate:

   $$v = 426.4\ \text{m/day}$$

6. Step 6 — Check: returning v = 426.4 m/day to

   $$Q = A \times v$$

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

Why the other options are there

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

Reference: FE Handbook — Water Treatment (surface loading)

water treatment plant sedimentation basin surface loading rate design Given plan area (A) = 1,834 m^2; surface loading (overflow) rate (v) = 12.5000 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) = 1,834 m^2, surface loading (overflow) rate (v) = 12.5000 m/day.

4. Step 4 — Substitute the given values:

   $$Q = 1834\,12.5000$$

5. Step 5 — Evaluate:

   Q = 22925\ \text{m^3/day}

6. Step 6 — Check: returning Q = 22,925 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

• 45,850 — kept a factor of two that cancels in the correct rearrangement.
• 11,463 — dropped that same factor in the other direction.
• 25,218 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

water treatment settling basin sizing from flow and overflow rate Given surface loading (overflow) rate (v) = 39.5000 m/day; flow rate (Q) = 108,220 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) = 39.5000 m/day, flow rate (Q) = 108,220 m^3/day.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{108220}{39.5000}$$

5. Step 5 — Evaluate:

   A = 2740\ \text{m^2}

6. Step 6 — Check: returning A = 2,740 m^2 to

   $$Q = A \times v$$

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

Why the other options are there

• 5,479 — kept a factor of two that cancels in the correct rearrangement.
• 1,370 — dropped that same factor in the other direction.
• 3,014 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

water treatment clarifier surface loading rate calculation Given plan area (A) = 3,580 m^2; flow rate (Q) = 141,320 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

   $$List the givens: plan area (A) = 3,580 m^2, flow rate (Q) = 141,320 m^3/day$$

4. Step 4 — Substitute the given values:

   $$v = \dfrac{141320}{3580}$$

5. Step 5 — Evaluate:

   $$v = 39.4749\ \text{m/day}$$

6. Step 6 — Check: returning v = 39.4749 m/day to

   $$Q = A \times v$$

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

Why the other options are there

• 78.9497 — kept a factor of two that cancels in the correct rearrangement.
• 19.7374 — dropped that same factor in the other direction.
• 43.4223 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

water treatment plant sedimentation basin surface loading rate design Given plan area (A) = 540.0 m^2; surface loading (overflow) rate (v) = 20.5000 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) = 540.0 m^2, surface loading (overflow) rate (v) = 20.5000 m/day.

4. Step 4 — Substitute the given values:

   $$Q = 540.0\,20.5000$$

5. Step 5 — Evaluate:

   Q = 11070\ \text{m^3/day}

6. Step 6 — Check: returning Q = 11,070 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

• 22,140 — kept a factor of two that cancels in the correct rearrangement.
• 5,535 — dropped that same factor in the other direction.
• 12,177 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)

water treatment settling basin sizing from flow and overflow rate Given surface loading (overflow) rate (v) = 49.5000 m/day; flow rate (Q) = 53,290 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) = 49.5000 m/day, flow rate (Q) = 53,290 m^3/day.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{53290}{49.5000}$$

5. Step 5 — Evaluate:

   A = 1077\ \text{m^2}

6. Step 6 — Check: returning A = 1,077 m^2 to

   $$Q = A \times v$$

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

Why the other options are there

• 2,153 — kept a factor of two that cancels in the correct rearrangement.
• 538.3 — dropped that same factor in the other direction.
• 1,184 — rounded an intermediate value before the final step.

Reference: FE Handbook — Water Treatment (surface loading)