tank volume required for a target detention time in a storage tank Given flow rate (Q) = 17,300 m^3/day; hydraulic detention time (theta) = 15.2500 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 17,300 m^3/day, hydraulic detention time (theta) = 15.2500 day.

4. Step 4 — Substitute the given values:

   $$V = 17300\,15.2500$$

5. Step 5 — Evaluate:

   V = 263825\ \text{m^3}

6. Step 6 — Check: returning V = 263,825 m^3 to

   $$V = Q \times \theta$$

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

Why the other options are there

• 527,650 — kept a factor of two that cancels in the correct rearrangement.
• 131,913 — dropped that same factor in the other direction.
• 290,208 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume sizing for an equalization basin Given hydraulic detention time (theta) = 3.8000 day; tank volume (V) = 184,669 m^3, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{V}{\theta}$$

3. Step 3 — List the givens: hydraulic detention time (theta) = 3.8000 day, tank volume (V) = 184,669 m^3.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{184669}{3.8000}$$

5. Step 5 — Evaluate:

   Q = 48597\ \text{m^3/day}

6. Step 6 — Check: returning Q = 48,597 m^3/day to

   $$V = Q \times \theta$$

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

Why the other options are there

• 97,194 — kept a factor of two that cancels in the correct rearrangement.
• 24,299 — dropped that same factor in the other direction.
• 53,457 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

detention tank volume calculated from plant flow rate Given flow rate (Q) = 17,990 m^3/day; tank volume (V) = 245,191 m^3, determine the hydraulic detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except theta is given, so isolate theta 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for theta:

   $$\theta = \dfrac{V}{Q}$$

3. Step 3

   $$List the givens: flow rate (Q) = 17,990 m^3/day, tank volume (V) = 245,191 m^3$$

4. Step 4 — Substitute the given values:

   $$\theta = \dfrac{245191}{17990}$$

5. Step 5 — Evaluate:

   $$\theta = 13.6293\ \text{day}$$

6. Step 6 — Check: returning theta = 13.6293 day to

   $$V = Q \times \theta$$

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

Why the other options are there

• 27.2586 — kept a factor of two that cancels in the correct rearrangement.
• 6.8146 — dropped that same factor in the other direction.
• 14.9922 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume required for a target detention time in a storage tank Given flow rate (Q) = 12,900 m^3/day; hydraulic detention time (theta) = 0.5500 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 12,900 m^3/day, hydraulic detention time (theta) = 0.5500 day.

4. Step 4 — Substitute the given values:

   $$V = 12900\,0.5500$$

5. Step 5 — Evaluate:

   V = 7095\ \text{m^3}

6. Step 6 — Check: returning V = 7,095 m^3 to

   $$V = Q \times \theta$$

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

Why the other options are there

• 14,190 — kept a factor of two that cancels in the correct rearrangement.
• 3,548 — dropped that same factor in the other direction.
• 7,805 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume sizing for an equalization basin Given hydraulic detention time (theta) = 3.2500 day; tank volume (V) = 40,693 m^3, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{V}{\theta}$$

3. Step 3 — List the givens: hydraulic detention time (theta) = 3.2500 day, tank volume (V) = 40,693 m^3.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{40693}{3.2500}$$

5. Step 5 — Evaluate:

   Q = 12521\ \text{m^3/day}

6. Step 6 — Check: returning Q = 12,521 m^3/day to

   $$V = Q \times \theta$$

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

Why the other options are there

• 25,042 — kept a factor of two that cancels in the correct rearrangement.
• 6,260 — dropped that same factor in the other direction.
• 13,773 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

detention tank volume calculated from plant flow rate Given flow rate (Q) = 14,580 m^3/day; tank volume (V) = 192,478 m^3, determine the hydraulic detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except theta is given, so isolate theta 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for theta:

   $$\theta = \dfrac{V}{Q}$$

3. Step 3

   $$List the givens: flow rate (Q) = 14,580 m^3/day, tank volume (V) = 192,478 m^3$$

4. Step 4 — Substitute the given values:

   $$\theta = \dfrac{192478}{14580}$$

5. Step 5 — Evaluate:

   $$\theta = 13.2015\ \text{day}$$

6. Step 6 — Check: returning theta = 13.2015 day to

   $$V = Q \times \theta$$

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

Why the other options are there

• 26.4030 — kept a factor of two that cancels in the correct rearrangement.
• 6.6008 — dropped that same factor in the other direction.
• 14.5217 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume required for a target detention time in a storage tank Given flow rate (Q) = 13,270 m^3/day; hydraulic detention time (theta) = 3.4000 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 13,270 m^3/day, hydraulic detention time (theta) = 3.4000 day.

4. Step 4 — Substitute the given values:

   $$V = 13270\,3.4000$$

5. Step 5 — Evaluate:

   V = 45118\ \text{m^3}

6. Step 6 — Check: returning V = 45,118 m^3 to

   $$V = Q \times \theta$$

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

Why the other options are there

• 90,236 — kept a factor of two that cancels in the correct rearrangement.
• 22,559 — dropped that same factor in the other direction.
• 49,630 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume sizing for an equalization basin Given hydraulic detention time (theta) = 0.7000 day; tank volume (V) = 185,687 m^3, determine the flow rate (Q) in m^3/day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for Q:

   $$Q = \dfrac{V}{\theta}$$

3. Step 3 — List the givens: hydraulic detention time (theta) = 0.7000 day, tank volume (V) = 185,687 m^3.

4. Step 4 — Substitute the given values:

   $$Q = \dfrac{185687}{0.7000}$$

5. Step 5 — Evaluate:

   Q = 265267\ \text{m^3/day}

6. Step 6 — Check: returning Q = 265,267 m^3/day to

   $$V = Q \times \theta$$

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

Why the other options are there

• 530,534 — kept a factor of two that cancels in the correct rearrangement.
• 132,634 — dropped that same factor in the other direction.
• 291,794 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

detention tank volume calculated from plant flow rate Given flow rate (Q) = 9,670 m^3/day; tank volume (V) = 354,469 m^3, determine the hydraulic detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• Everything except theta is given, so isolate theta 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for theta:

   $$\theta = \dfrac{V}{Q}$$

3. Step 3

   $$List the givens: flow rate (Q) = 9,670 m^3/day, tank volume (V) = 354,469 m^3$$

4. Step 4 — Substitute the given values:

   $$\theta = \dfrac{354469}{9670}$$

5. Step 5 — Evaluate:

   $$\theta = 36.6566\ \text{day}$$

6. Step 6 — Check: returning theta = 36.6566 day to

   $$V = Q \times \theta$$

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

Why the other options are there

• 73.3131 — kept a factor of two that cancels in the correct rearrangement.
• 18.3283 — dropped that same factor in the other direction.
• 40.3222 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume

tank volume required for a target detention time in a storage tank Given flow rate (Q) = 19,130 m^3/day; hydraulic detention time (theta) = 14.2500 day, determine the tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Tank Volume.
• 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.
• Tank volume for a treatment basin is the product of flow rate and required hydraulic detention time.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$V = Q \times \theta$$

2. Step 2 — Rearrange symbolically for V:

   $$V = Q\,\theta$$

3. Step 3 — List the givens: flow rate (Q) = 19,130 m^3/day, hydraulic detention time (theta) = 14.2500 day.

4. Step 4 — Substitute the given values:

   $$V = 19130\,14.2500$$

5. Step 5 — Evaluate:

   V = 272603\ \text{m^3}

6. Step 6 — Check: returning V = 272,603 m^3 to

   $$V = Q \times \theta$$

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

Why the other options are there

• 545,205 — kept a factor of two that cancels in the correct rearrangement.
• 136,301 — dropped that same factor in the other direction.
• 299,863 — rounded an intermediate value before the final step.

Reference: FE Handbook — Tank Volume