An activated sludge plant treats 6,042 m³/d with an influent BOD of 373 mg/L, effluent BOD 20 mg/L, MLSS 3,557 mg/L, and an aeration basin volume of 5,772 m³. With Y = 0.55 mg VSS/mg BOD, k_d = 0.08 d⁻¹ and an SRT of 5 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

   $$F/M=6042(373)/(5772\times 3557)=0.110d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =5772/6042=0.9553d=22.93h$$

5. Formula

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

6. Substituting

   $$P_x=0.55(6042)(373-20)/(1+0.08\times 5)mg/d$$

7. Evaluate

   $$P_x=837.9kgVSS/day$$

Why the other options are there

• F/M = 633.6 (basin volume omitted)
• P_x = 1,173 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A circular clarifier 16 m in diameter and 4.5 m deep treats 15,987 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 80 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

   $$A=(\pi /4)(16{)}^2=201.1m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 15987/201.1 = 79.51 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =904.8/15987=1.36h$$

6. Formula

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

7. Substituting — 15987/(π × 16) = 318.1 m³/m·d

8. Formula

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

9. Substituting

   $$v_s=9.81(1650)(8.0e-5{)}^2/(18\times 0.001)=497.2m/d$$

10. Compare

    $$v_s=497.2m/d>SOR=79.51m/d\to particleisremoved$$

Why the other options are there

• SOR = 70.7 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A stack gas stream of 29 m³/s carries 8 g/m³ of particulate. It passes a cyclone at 81% efficiency followed by a fabric filter at 96.0% 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

   $$C_1=8(1-0.81)=1.520g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=1.520(1-0.960)=0.0608g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.190\times 0.0400=0.99240=99.240$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.0608g/m^3\times 29m^3/s\times 3600s/h÷1000=6.348kg/h$$

Why the other options are there

• η = 177.0% (efficiencies added)
• 835.2 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A community of 249,377 people generates 2.1 kg/person/day of MSW and diverts 35% through recycling. Compacted in place at 892 kg/m³ with 20% additional daily cover volume, find the airspace needed for 11 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

   $$249377(2.1)(1-0.35)=340,400kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=340,400/892=381.6m^3/day$$

5. Annual with cover

   $$381.6\times 365\times 1.20=167,147m^3/yr$$

6. Design life

   $$167,147\times 11=1,838,616m^3$$

Why the other options are there

• 2,357,200 m³ (diversion and cover ignored)
• 1,366,704,414 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A water plant treating 4,351 m³/d applies a chlorine dose of 3.0 mg/L against a demand of 3.0 mg/L, with 54 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 2.5-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

   $$3.0-3.0=0.20mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 0.20(54) = 10.8 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$3.0g/m^3\times 4351m^3/d÷1000=13.05kg/day$$

7. Log removal

   $$N=10^6/10^2.5=3.16e+3organisms/100mL$$

Why the other options are there

• CT = 162.0 (dose used instead of residual)
• 13,053 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A city of 129,062 grows at 2.4% per year. Project the population in 29 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 253 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

   $$P=129062(1+0.024{)}^{2}9=256,742people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P=129062(1+0.024\times 29)=218,889people$$

5. Formula

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

6. Substituting

   $$Q_{a}vg=256,742(253)/1000=64,956m^3/d$$

7. Peak day

   $$Q_{p}eak=3.0\times 64,956=194,867m^3/d$$

Why the other options are there

• 89,827 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

Drinking water contains 0.0240 mg/L of a carcinogen. An adult of 66 kg drinks 2.3 L/day. With a cancer slope factor of 0.07 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 50-year half-life remaining after 98 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 = (0.0240 × 2.3)/66 = 8.364e-4 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=8.364e-4(0.07)=5.85e-5$$

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

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}98/50)=25.703$$

Why the other options are there

• Risk = 0.00168 (body weight and intake ignored)
• 2.0% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A stream flowing 20.5 cfs at 10.0 mg/L receives a discharge of 14.0 cfs at 225.0 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

   $$20.5(10.0)=205.0;14.0(225.0)=3,150$$

3. Total flow

   $$20.5+14.0=34.5cfs$$

4. Solve

   $$C=(205.0+3,150)/34.5=97.25mg/L$$

Why the other options are there

• 117.5 mg/L (unweighted average)
• 235.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A wastewater has ultimate BOD L₀ = 271 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

   $$BO{D}_t=271(1-0.5827)=113.1mg/L$$

5. Remaining

   $$L_0-BO{D}_t=157.9mg/L$$

Why the other options are there

• 157.9 mg/L (remaining reported as exerted)
• 146.3 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature

A treatment tank holds 97,794 gal and treats 0.9 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

   $$0.9\times 10^6=900,000gal/day$$

3. Substituting

   $$\theta =97794/900,000=0.1087day$$

4. Convert

   $$\theta =0.1087(24)=2.61hr$$

Why the other options are there

• 0.109 hr (day/hour conversion missed)
• 9.20 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Nomenclature