A environmental engineering problem uses First-order BOD decay. Given ultimate BOD (L0) = 315.0 mg/L; rate constant (k) = 0.0700 1/day; time (t) = 4.5000 day, determine the 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:

   $$L_t = L_0 e^{-k t}$$

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

3. Step 3

   $$List the givens: ultimate BOD (L_{0}) = 315.0 mg/L, rate constant (k) = 0.0700 1/day, time (t) = 4.5000 day$$

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

5. Step 5 — Evaluate:

   $$Lt = 229.9\ \text{mg/L}$$

6. Step 6 — Check: returning Lt = 229.9 mg/L to

   $$L_t = L_0 e^{-k t}$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Environmental Engineering → Activated Sludge

A environmental engineering problem uses Steady-state mass balance. Given flow 1 (Q1) = 2.0000 MGD; concentration 1 (C1) = 8.0000 mg/L; flow 2 (Q2) = 10.0000 MGD; concentration 2 (C2) = 29.0000 mg/L, determine the blended concentration (C) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Steady-state mass balance.
• Everything except C is given, so isolate C 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:

   $$C = (Q_1 C_1 + Q_2 C_2) / (Q_1 + Q_2)$$

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

3. Step 3 — List the givens: flow 1 (Q1) = 2.0000 MGD, concentration 1 (C1) = 8.0000 mg/L, flow 2 (Q2) = 10.0000 MGD, concentration 2 (C2) = 29.0000 mg/L.

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

5. Step 5 — Evaluate:

   $$C = 25.5000\ \text{mg/L}$$

6. Step 6 — Check: returning C = 25.5000 mg/L to

   $$C = (Q_1 C_1 + Q_2 C_2) / (Q_1 + Q_2)$$

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

Why the other options are there

• 51.0000 — kept a factor of two that cancels in the correct rearrangement.
• 12.7500 — dropped that same factor in the other direction.
• 28.0500 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Activated Sludge

A environmental engineering problem uses Hydraulic detention time. Given tank volume (V) = 1,206,000 gal; flow rate (Q) = 3,865,000 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

   $$List the givens: tank volume (V) = 1,206,000 gal, flow rate (Q) = 3,865,000 gal/day$$

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

5. Step 5 — Evaluate:

   $$\theta = 0.3120\ \text{day}$$

6. Step 6 — Check: returning theta = 0.3120 day to

   $$\theta = V / Q$$

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

Why the other options are there

• 0.6241 — kept a factor of two that cancels in the correct rearrangement.
• 0.1560 — dropped that same factor in the other direction.
• 0.3432 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Activated Sludge

density of a sludge sample measured in the laboratory Given mass (m) = 879.0 kg; volume (V) = 0.9360 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

   $$List the givens: mass (m) = 879.0 kg, volume (V) = 0.9360 m^3$$

4. Step 4 — Substitute the given values:

   $$\rho = \dfrac{879.0}{0.9360}$$

5. Step 5 — Evaluate:

   \rho = 939.1\ \text{kg/m^3}

6. Step 6 — Check: returning rho = 939.1 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

• 1,878 — kept a factor of two that cancels in the correct rearrangement.
• 469.6 — dropped that same factor in the other direction.
• 1,033 — rounded an intermediate value before the final step.

Reference: FE Handbook — Density

activated sludge process mean cell residence time and F/M ratio Given aeration tank volume (V) = 17,220 m^3; MLSS concentration (X) = 3,760 mg/L; waste sludge flow (Q_w) = 47.0000 m^3/day; waste sludge concentration (X_w) = 7,100 mg/L; effluent flow (Q_e) = 5,970 m^3/day; effluent suspended solids (X_e) = 10.5000 mg/L, determine the mean cell residence time (MCRT) (theta_c) in day.

Start with the thinking

• The governing relation printed in this handbook section is Activated Sludge (MCRT, F/M, SVI).
• Everything except theta_c is given, so isolate theta_c 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.
• Activated sludge process control uses mean cell residence time (MCRT), the F/M ratio, and SVI to monitor the aeration basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta_c = \dfrac{V X}{Q_w X_w + Q_e X_e}$$

2. Step 2 — Rearrange symbolically for theta_c:

   $$\theta_{c} = \dfrac{VX}{Q_w X_w+Q_e X_e}$$

3. Step 3 — List the givens: aeration tank volume (V) = 17,220 m^3, MLSS concentration (X) = 3,760 mg/L, waste sludge flow (Q_w) = 47.0000 m^3/day, waste sludge concentration (X_w) = 7,100 mg/L, effluent flow (Q_e) = 5,970 m^3/day, effluent suspended solids (X_e) = 10.5000 mg/L.

4. Step 4 — Substitute the given values:

   $$\theta_{c} = \dfrac{V3760}{Q_w X_w+Q_e X_e}$$

5. Step 5 — Evaluate:

   $$\theta_{c} = 163.3\ \text{day}$$

6. Step 6 — Check: returning theta_c = 163.3 day to

   $$\theta_c = \dfrac{V X}{Q_w X_w + Q_e X_e}$$

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

Why the other options are there

• 326.7 — kept a factor of two that cancels in the correct rearrangement.
• 81.6721 — dropped that same factor in the other direction.
• 179.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Activated Sludge MCRT

A environmental engineering problem uses First-order BOD decay. Given rate constant (k) = 0.2600 1/day; time (t) = 4.0000 day; remaining BOD (Lt) = 94.7000 mg/L, determine the 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:

   $$L_t = L_0 e^{-k t}$$

2. Step 2 — Rearrange the relation so that L0 stands alone on the left-hand side.

3. Step 3

   $$List the givens: rate constant (k) = 0.2600 1/day, time (t) = 4.0000 day, remaining BOD (Lt) = 94.7000 mg/L$$

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

5. Step 5 — Evaluate:

   $$L_{0} = 267.9\ \text{mg/L}$$

6. Step 6 — Check: returning L0 = 267.9 mg/L to

   $$L_t = L_0 e^{-k t}$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Environmental Engineering → Activated Sludge

A environmental engineering problem uses Steady-state mass balance. Given flow 1 (Q1) = 6.0000 MGD; flow 2 (Q2) = 13.5000 MGD; concentration 2 (C2) = 16.0000 mg/L; blended concentration (C) = 23.7600 mg/L, determine the concentration 1 (C1) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Steady-state mass balance.
• Everything except C1 is given, so isolate C1 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:

   $$C = (Q_1 C_1 + Q_2 C_2) / (Q_1 + Q_2)$$

2. Step 2 — Rearrange the relation so that C1 stands alone on the left-hand side.

3. Step 3 — List the givens: flow 1 (Q1) = 6.0000 MGD, flow 2 (Q2) = 13.5000 MGD, concentration 2 (C2) = 16.0000 mg/L, blended concentration (C) = 23.7600 mg/L.

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

5. Step 5 — Evaluate:

   $$C_{1} = 41.2200\ \text{mg/L}$$

6. Step 6 — Check: returning C1 = 41.2200 mg/L to

   $$C = (Q_1 C_1 + Q_2 C_2) / (Q_1 + Q_2)$$

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

Why the other options are there

• 82.4400 — kept a factor of two that cancels in the correct rearrangement.
• 20.6100 — dropped that same factor in the other direction.
• 45.3420 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Activated Sludge

A environmental engineering problem uses Hydraulic detention time. Given flow rate (Q) = 4,456,000 gal/day; detention time (theta) = 2.6400 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

   $$List the givens: flow rate (Q) = 4,456,000 gal/day, detention time (theta) = 2.6400 day$$

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

5. Step 5 — Evaluate:

   $$V = 11763840\ \text{gal}$$

6. Step 6 — Check: returning V = 11,763,840 gal to

   $$\theta = V / Q$$

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

Why the other options are there

• 23,527,680 — kept a factor of two that cancels in the correct rearrangement.
• 5,881,920 — dropped that same factor in the other direction.
• 12,940,224 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Environmental Engineering → Activated Sludge

density of a soil or waste material sample Given volume (V) = 0.3610 m^3; density (rho) = 828.0 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

   $$List the givens: volume (V) = 0.3610 m^3, density (rho) = 828.0 kg/m^3$$

4. Step 4 — Substitute the given values:

   $$m = 828.0 0.3610$$

5. Step 5 — Evaluate:

   $$m = 298.9\ \text{kg}$$

6. Step 6 — Check: returning m = 298.9 kg to

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

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

Why the other options are there

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

Reference: FE Handbook — Density

activated sludge SVI and MCRT monitoring for sludge settleability Given MLSS concentration (X) = 3,270 mg/L; waste sludge flow (Q_w) = 89.0000 m^3/day; waste sludge concentration (X_w) = 7,520 mg/L; effluent flow (Q_e) = 22,820 m^3/day; effluent suspended solids (X_e) = 29.0000 mg/L; mean cell residence time (MCRT) (theta_c) = 17.3000 day, determine the aeration tank volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Activated Sludge (MCRT, F/M, SVI).
• 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.
• Activated sludge process control uses mean cell residence time (MCRT), the F/M ratio, and SVI to monitor the aeration basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\theta_c = \dfrac{V X}{Q_w X_w + Q_e X_e}$$

2. Step 2 — Rearrange symbolically for V:

   $$V = \dfrac{\theta_c\left(Q_w X_w+Q_e X_e\right)}{X}$$

3. Step 3 — List the givens: MLSS concentration (X) = 3,270 mg/L, waste sludge flow (Q_w) = 89.0000 m^3/day, waste sludge concentration (X_w) = 7,520 mg/L, effluent flow (Q_e) = 22,820 m^3/day, effluent suspended solids (X_e) = 29.0000 mg/L, mean cell residence time (MCRT) (theta_c) = 17.3000 day.

4. Step 4 — Substitute the given values:

   $$V = \dfrac{\theta_c\left(Q_w X_w+Q_e X_e\right)}{3270}$$

5. Step 5 — Evaluate:

   V = 7042\ \text{m^3}

6. Step 6 — Check: returning V = 7,042 m^3 to

   $$\theta_c = \dfrac{V X}{Q_w X_w + Q_e X_e}$$

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

Why the other options are there

• 14,084 — kept a factor of two that cancels in the correct rearrangement.
• 3,521 — dropped that same factor in the other direction.
• 7,746 — rounded an intermediate value before the final step.

Reference: FE Handbook — Activated Sludge MCRT