A environmental engineering problem uses Hydraulic detention time. Given tank volume (V) = 26,000 gal; flow rate (Q) = 565,000 gal/day, determine the detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic detention time.
• 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.
• Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta = V / Q$$

2. Step 2 — Rearrange the relation so that theta stands alone on the left-hand side.

3. Step 3

   $$List the givens: tank volume (V) = 26,000 gal, flow rate (Q) = 565,000 gal/day$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\theta = 0.0460\ \text{day}$$

6. Step 6 — Check: returning theta = 0.0460 day to

   $$\theta = V / Q$$

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

Why the other options are there

• 0.0920 — kept a factor of two that cancels in the correct rearrangement.
• 0.0230 — dropped that same factor in the other direction.
• 0.0506 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

density of a soil or waste material sample Given mass (m) = 320.0 kg; volume (V) = 0.7200 m^3, determine the density (rho) in kg/m^3.

Start with the thinking

• The governing relation printed in this handbook section is Density.
• Everything except rho is given, so isolate rho 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.
• Density of a material sample is mass per unit volume, a fundamental property used throughout environmental engineering calculations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\rho = \dfrac{m}{V}$$

2. Step 2 — Rearrange symbolically for rho:

   $$\rho = \dfrac{m}{V}$$

3. Step 3

   $$List the givens: mass (m) = 320.0 kg, volume (V) = 0.7200 m^3$$

4. Step 4 — Substitute the given values:

   $$\rho = \dfrac{320.0}{0.7200}$$

5. Step 5 — Evaluate:

   \rho = 444.4\ \text{kg/m^3}

6. Step 6 — Check: returning rho = 444.4 kg/m^3 to

   $$\rho = \dfrac{m}{V}$$

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

Why the other options are there

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

Reference: FE Handbook — Density

A environmental engineering problem uses Hydraulic detention time. Given flow rate (Q) = 4,899,000 gal/day; detention time (theta) = 1.2200 day, determine the tank volume (V) in gal.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic detention time.
• 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.
• Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta = V / Q$$

2. Step 2 — Rearrange the relation so that V stands alone on the left-hand side.

3. Step 3

   $$List the givens: flow rate (Q) = 4,899,000 gal/day, detention time (theta) = 1.2200 day$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$V = 5976780\ \text{gal}$$

6. Step 6 — Check: returning V = 5,976,780 gal to

   $$\theta = V / Q$$

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

Why the other options are there

• 11,953,560 — kept a factor of two that cancels in the correct rearrangement.
• 2,988,390 — dropped that same factor in the other direction.
• 6,574,458 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

density calculation from measured mass and volume of a liquid sample Given volume (V) = 0.0210 m^3; density (rho) = 1,313 kg/m^3, determine the mass (m) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Density.
• Everything except m is given, so isolate m 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.
• Density of a material sample is mass per unit volume, a fundamental property used throughout environmental engineering calculations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\rho = \dfrac{m}{V}$$

2. Step 2 — Rearrange symbolically for m:

   $$m = \rho V$$

3. Step 3

   $$List the givens: volume (V) = 0.0210 m^3, density (rho) = 1,313 kg/m^3$$

4. Step 4 — Substitute the given values:

   $$m = 1313 0.0210$$

5. Step 5 — Evaluate:

   $$m = 27.5730\ \text{kg}$$

6. Step 6 — Check: returning m = 27.5730 kg to

   $$\rho = \dfrac{m}{V}$$

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

Why the other options are there

• 55.1460 — kept a factor of two that cancels in the correct rearrangement.
• 13.7865 — dropped that same factor in the other direction.
• 30.3303 — rounded an intermediate value before the final step.

Reference: FE Handbook — Density

A environmental engineering problem uses Hydraulic detention time. Given tank volume (V) = 455,000 gal; detention time (theta) = 1.2300 day, determine the flow rate (Q) in gal/day.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic detention time.
• 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.
• Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta = V / Q$$

2. Step 2 — Rearrange the relation so that Q stands alone on the left-hand side.

3. Step 3

   $$List the givens: tank volume (V) = 455,000 gal, detention time (theta) = 1.2300 day$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$Q = 369919\ \text{gal/day}$$

6. Step 6 — Check: returning Q = 369,919 gal/day to

   $$\theta = V / Q$$

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

Why the other options are there

• 739,837 — kept a factor of two that cancels in the correct rearrangement.
• 184,959 — dropped that same factor in the other direction.
• 406,911 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

density of a sludge sample measured in the laboratory Given mass (m) = 361.0 kg; density (rho) = 2,789 kg/m^3, determine the volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Density.
• 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.
• Density of a material sample is mass per unit volume, a fundamental property used throughout environmental engineering calculations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\rho = \dfrac{m}{V}$$

2. Step 2 — Rearrange symbolically for V:

   $$V = \dfrac{m}{\rho}$$

3. Step 3

   $$List the givens: mass (m) = 361.0 kg, density (rho) = 2,789 kg/m^3$$

4. Step 4 — Substitute the given values:

   $$V = \dfrac{361.0}{2789}$$

5. Step 5 — Evaluate:

   V = 0.1294\ \text{m^3}

6. Step 6 — Check: returning V = 0.1294 m^3 to

   $$\rho = \dfrac{m}{V}$$

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

Why the other options are there

• 0.2589 — kept a factor of two that cancels in the correct rearrangement.
• 0.0647 — dropped that same factor in the other direction.
• 0.1424 — rounded an intermediate value before the final step.

Reference: FE Handbook — Density

A environmental engineering problem uses Hydraulic detention time. Given tank volume (V) = 935,000 gal; flow rate (Q) = 1,952,000 gal/day, determine the detention time (theta) in day.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic detention time.
• 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.
• Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta = V / Q$$

2. Step 2 — Rearrange the relation so that theta stands alone on the left-hand side.

3. Step 3

   $$List the givens: tank volume (V) = 935,000 gal, flow rate (Q) = 1,952,000 gal/day$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\theta = 0.4790\ \text{day}$$

6. Step 6 — Check: returning theta = 0.4790 day to

   $$\theta = V / Q$$

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

Why the other options are there

• 0.9580 — kept a factor of two that cancels in the correct rearrangement.
• 0.2395 — dropped that same factor in the other direction.
• 0.5269 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

density of a soil or waste material sample Given mass (m) = 52.0000 kg; volume (V) = 0.6380 m^3, determine the density (rho) in kg/m^3.

Start with the thinking

• The governing relation printed in this handbook section is Density.
• Everything except rho is given, so isolate rho 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.
• Density of a material sample is mass per unit volume, a fundamental property used throughout environmental engineering calculations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\rho = \dfrac{m}{V}$$

2. Step 2 — Rearrange symbolically for rho:

   $$\rho = \dfrac{m}{V}$$

3. Step 3

   $$List the givens: mass (m) = 52.0000 kg, volume (V) = 0.6380 m^3$$

4. Step 4 — Substitute the given values:

   $$\rho = \dfrac{52.0000}{0.6380}$$

5. Step 5 — Evaluate:

   \rho = 81.5047\ \text{kg/m^3}

6. Step 6 — Check: returning rho = 81.5047 kg/m^3 to

   $$\rho = \dfrac{m}{V}$$

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

Why the other options are there

• 163.0 — kept a factor of two that cancels in the correct rearrangement.
• 40.7524 — dropped that same factor in the other direction.
• 89.6552 — rounded an intermediate value before the final step.

Reference: FE Handbook — Density

A environmental engineering problem uses Hydraulic detention time. Given flow rate (Q) = 2,033,000 gal/day; detention time (theta) = 8.9400 day, determine the tank volume (V) in gal.

Start with the thinking

• The governing relation printed in this handbook section is Hydraulic detention time.
• 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.
• Environmental Engineering items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta = V / Q$$

2. Step 2 — Rearrange the relation so that V stands alone on the left-hand side.

3. Step 3

   $$List the givens: flow rate (Q) = 2,033,000 gal/day, detention time (theta) = 8.9400 day$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$V = 18175020\ \text{gal}$$

6. Step 6 — Check: returning V = 18,175,020 gal to

   $$\theta = V / Q$$

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

Why the other options are there

• 36,350,040 — kept a factor of two that cancels in the correct rearrangement.
• 9,087,510 — dropped that same factor in the other direction.
• 19,992,522 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Steady-State Reactor Parameters (Constant Density Systems)

density calculation from measured mass and volume of a liquid sample Given volume (V) = 0.4750 m^3; density (rho) = 2,096 kg/m^3, determine the mass (m) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Density.
• Everything except m is given, so isolate m 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.
• Density of a material sample is mass per unit volume, a fundamental property used throughout environmental engineering calculations.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\rho = \dfrac{m}{V}$$

2. Step 2 — Rearrange symbolically for m:

   $$m = \rho V$$

3. Step 3

   $$List the givens: volume (V) = 0.4750 m^3, density (rho) = 2,096 kg/m^3$$

4. Step 4 — Substitute the given values:

   $$m = 2096 0.4750$$

5. Step 5 — Evaluate:

   $$m = 995.6\ \text{kg}$$

6. Step 6 — Check: returning m = 995.6 kg to

   $$\rho = \dfrac{m}{V}$$

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

Why the other options are there

• 1,991 — kept a factor of two that cancels in the correct rearrangement.
• 497.8 — dropped that same factor in the other direction.
• 1,095 — rounded an intermediate value before the final step.

Reference: FE Handbook — Density