An activated sludge plant treats 8,706 m³/d with an influent BOD of 379 mg/L, effluent BOD 20 mg/L, MLSS 2,754 mg/L, and an aeration basin volume of 2,579 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 15 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Start with the thinking

• F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
• Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

1. Formula

   $$F/M=Q{S}_0/(VX)$$

2. Substituting

   $$F/M=8706(379)/(2579\times 2754)=0.465d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =2579/8706=0.2962d=7.11h$$

5. Formula

   $$P_x=YQ(S_0-S)/(1+k_d\cdot SRT)$$

6. Substituting

   $$P_x=0.65(8706)(379-20)/(1+0.06\times 15)mg/d$$

7. Evaluate

   $$P_x=1,069kgVSS/day$$

Why the other options are there

• F/M = 1,198 (basin volume omitted)
• P_x = 2,032 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A circular clarifier 28 m in diameter and 3.5 m deep treats 13,887 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 40 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A=(\pi /4)(28{)}^2=615.8m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 13887/615.8 = 22.55 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =2,155/13887=3.72h$$

6. Formula

   $$weirloading=Q/(\pi D)$$

7. Substituting — 13887/(π × 28) = 157.9 m³/m·d

8. Formula

   $$v_s=g({\rho }_s-\rho )d^2/(18\mu )$$

9. Substituting

   $$v_s=9.81(1650)(4.0e-5{)}^2/(18\times 0.001)=124.3m/d$$

10. Compare

    $$v_s=124.3m/d>SOR=22.55m/d\to particleisremoved$$

Why the other options are there

• SOR = 45.1 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A stack gas stream of 5 m³/s carries 23 g/m³ of particulate. It passes a cyclone at 90% efficiency followed by a fabric filter at 98.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Start with the thinking

• Efficiencies in series multiply as penetrations (1 − η), they never simply add.
• The overall penetration is the product of the individual penetrations.

Step-by-step solution

1. Formula

   $$C_1=C_{i}n(1-{\eta }_1)$$

2. Substituting

   $$C_1=23(1-0.90)=2.300g/m^3$$

3. Formula

   $$C_2=C_1(1-{\eta }_2)$$

4. Substituting

   $$C_2=2.300(1-0.985)=0.0345g/m^3$$

5. Formula

   $${\eta }_{o}verall=1-(1-{\eta }_1)(1-{\eta }_2)$$

6. Substituting

   $$\eta =1-0.100\times 0.0150=0.99850=99.850$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.0345g/m^3\times 5m^3/s\times 3600s/h÷1000=0.621kg/h$$

Why the other options are there

• η = 188.5% (efficiencies added)
• 414.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A community of 161,353 people generates 3.0 kg/person/day of MSW and diverts 20% through recycling. Compacted in place at 529 kg/m³ with 20% additional daily cover volume, find the airspace needed for 29 years.

Start with the thinking

• Only the landfilled fraction consumes airspace — diverted material is subtracted first.
• Daily cover soil is real volume and must be added to the waste volume.

Step-by-step solution

1. Formula

   $$landfilledmass=population\times rate\times (1-diversion)$$

2. Substituting

   $$161353(3.0)(1-0.20)=387,247kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=387,247/529=732.0m^3/day$$

5. Annual with cover

   $$732.0\times 365\times 1.20=320,632m^3/yr$$

6. Design life

   $$320,632\times 29=9,298,325m^3$$

Why the other options are there

• 9,685,755 m³ (diversion and cover ignored)
• 4,099,011,612 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A water plant treating 31,023 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 3.2 mg/L, with 45 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.0-log pathogen reduction of an initial 10⁶ organisms/100 mL.

Start with the thinking

• Residual = dose − demand; the residual, not the dose, drives disinfection credit.
• Each log of removal divides the surviving organism count by ten.

Step-by-step solution

1. Formula

   $$residual=dose-demand$$

2. Substituting

   $$2.0-3.2=0.20mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 0.20(45) = 9.0 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$2.0g/m^3\times 31023m^3/d÷1000=62.05kg/day$$

7. Log removal

   $$N=10^6/10^3.0=1.00e+3organisms/100mL$$

Why the other options are there

• CT = 90 (dose used instead of residual)
• 62,046 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A city of 177,983 grows at 3.1% per year. Project the population in 19 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 333 L/capita/day.

Start with the thinking

• Geometric growth compounds; arithmetic growth adds a fixed increment and always gives a smaller value over long horizons.
• Water systems are sized on peak day (and peak hour) flow, never on the average.

Step-by-step solution

1. Formula (geometric)

   $$P=P_0(1+i{)}^n$$

2. Substituting

   $$P=177983(1+0.031{)}^{1}9=317,902people$$

3. Formula (arithmetic)

   $$P=P_0(1+in)$$

4. Substituting

   $$P=177983(1+0.031\times 19)=282,815people$$

5. Formula

   $$Q_{a}vg=P\times percapitause$$

6. Substituting

   $$Q_{a}vg=317,902(333)/1000=105,861m^3/d$$

7. Peak day

   $$Q_{p}eak=2.9\times 105,861=306,998m^3/d$$

Why the other options are there

• 104,832 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

Drinking water contains 0.0460 mg/L of a carcinogen. An adult of 80 kg drinks 2.0 L/day. With a cancer slope factor of 0.98 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 8-year half-life remaining after 24 years.

Start with the thinking

• CDI normalises exposure to body weight so a dose-response slope can be applied.
• A risk above 10⁻⁶ to 10⁻⁴ is typically the regulatory action range.

Step-by-step solution

1. Formula

   $$CDI=(C\times IR)/BW$$

2. Substituting — CDI = (0.0460 × 2.0)/80 = 1.150e-3 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=1.150e-3(0.98)=1.13e-3$$

5. Interpretation — above the 10⁻⁶–10⁻⁴ risk range

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}24/8)=12.500$$

Why the other options are there

• Risk = 0.04508 (body weight and intake ignored)
• -50.0% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A stream flowing 8.0 cfs at 17.0 mg/L receives a discharge of 6.5 cfs at 120.0 mg/L. Find the fully mixed concentration.

Start with the thinking

• Mass in equals mass out at steady state.
• Weight by flow, never a simple average.

Step-by-step solution

1. Mass balance

   $$Q_1{C}_1+Q_2{C}_2=(Q_1+Q_2)C$$

2. Loads

   $$8.0(17.0)=136.0;6.5(120.0)=780.0$$

3. Total flow

   $$8.0+6.5=14.5cfs$$

4. Solve

   $$C=(136.0+780.0)/14.5=63.17mg/L$$

Why the other options are there

• 68.5 mg/L (unweighted average)
• 137.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A wastewater has ultimate BOD L₀ = 274 mg/L and k = 0.18 /day (base e). How much BOD is exerted in 3 days?

Start with the thinking

• BOD exertion is first-order and approaches L₀ asymptotically.
• Check whether k is base e or base 10.

Step-by-step solution

1. First-order

   $$BO{D}_t=L_0(1-e^{(}-kt))$$

2. Exponent

   $$kt=0.18(3)=0.54$$

3. Exponential

   $$e^{(}-0.54)=0.5827$$

4. Substituting

   $$BO{D}_t=274(1-0.5827)=114.3mg/L$$

5. Remaining

   $$L_0-BO{D}_t=159.7mg/L$$

Why the other options are there

• 159.7 mg/L (remaining reported as exerted)
• 148.0 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements

A treatment tank holds 197,035 gal and treats 1.9 MGD. Find the hydraulic detention time.

Start with the thinking

• θ = V/Q with consistent volume units.
• Express the answer in hours for design comparison.

Step-by-step solution

1. Detention

   $$\theta =V/Q$$

2. Flow in gal/day

   $$1.9\times 10^6=1,900,000gal/day$$

3. Substituting

   $$\theta =197035/1,900,000=0.1037day$$

4. Convert

   $$\theta =0.1037(24)=2.49hr$$

Why the other options are there

• 0.104 hr (day/hour conversion missed)
• 9.64 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Removal and Inactivation Requirements