A circular clarifier 20 m in diameter and 4.0 m deep treats 18,101 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 90 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

   $$A=(\pi /4)(20{)}^2=314.2m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 18101/314.2 = 57.62 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =1,257/18101=1.67h$$

6. Formula

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

7. Substituting — 18101/(π × 20) = 288.1 m³/m·d

8. Formula

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

9. Substituting

   $$v_s=9.81(1650)(9.0e-5{)}^2/(18\times 0.001)=629.3m/d$$

10. Compare

    $$v_s=629.3m/d>SOR=57.62m/d\to particleisremoved$$

Why the other options are there

• SOR = 72.0 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A stack gas stream of 57 m³/s carries 7 g/m³ of particulate. It passes a cyclone at 87% efficiency followed by a fabric filter at 93.0% 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

   $$C_1=7(1-0.87)=0.910g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=0.910(1-0.930)=0.0637g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.130\times 0.0700=0.99090=99.090$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.0637g/m^3\times 57m^3/s\times 3600s/h÷1000=13.071kg/h$$

Why the other options are there

• η = 180.0% (efficiencies added)
• 1,436 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A community of 51,563 people generates 2.3 kg/person/day of MSW and diverts 45% through recycling. Compacted in place at 775 kg/m³ with 25% additional daily cover volume, find the airspace needed for 19 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

   $$51563(2.3)(1-0.45)=65,227kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=65,227/775=84.2m^3/day$$

5. Annual with cover

   $$84.2\times 365\times 1.25=38,400m^3/yr$$

6. Design life

   $$38,400\times 19=729,598m^3$$

Why the other options are there

• 1,061,233 m³ (diversion and cover ignored)
• 452,350,597 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A water plant treating 17,513 m³/d applies a chlorine dose of 2.0 mg/L against a demand of 1.4 mg/L, with 58 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

   $$2.0-1.4=0.60mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 0.60(58) = 34.8 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$2.0g/m^3\times 17513m^3/d÷1000=35.03kg/day$$

7. Log removal

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

Why the other options are there

• CT = 116.0 (dose used instead of residual)
• 35,026 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A city of 52,249 grows at 0.9% per year. Project the population in 22 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 274 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

   $$P=52249(1+0.009{)}^{2}2=63,633people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P=52249(1+0.009\times 22)=62,594people$$

5. Formula

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

6. Substituting

   $$Q_{a}vg=63,633(274)/1000=17,435m^3/d$$

7. Peak day

   $$Q_{p}eak=2.2\times 17,435=38,358m^3/d$$

Why the other options are there

• 10,345 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

Drinking water contains 0.0160 mg/L of a carcinogen. An adult of 66 kg drinks 1.5 L/day. With a cancer slope factor of 1.53 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 43-year half-life remaining after 100 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 = (0.0160 × 1.5)/66 = 3.636e-4 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=3.636e-4(1.53)=5.56e-4$$

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

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}100/43)=19.949$$

Why the other options are there

• Risk = 0.02448 (body weight and intake ignored)
• -16.3% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A stream flowing 38.0 cfs at 16.0 mg/L receives a discharge of 17.0 cfs at 260.0 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

   $$38.0(16.0)=608.0;17.0(260.0)=4,420$$

3. Total flow

   $$38.0+17.0=55.0cfs$$

4. Solve

   $$C=(608.0+4,420)/55.0=91.42mg/L$$

Why the other options are there

• 138.0 mg/L (unweighted average)
• 276.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A wastewater has ultimate BOD L₀ = 204 mg/L and k = 0.35 /day (base e). How much BOD is exerted in 8 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

   $$kt=0.35(8)=2.80$$

3. Exponential

   $$e^{(}-2.80)=0.0608$$

4. Substituting

   $$BO{D}_t=204(1-0.0608)=191.6mg/L$$

5. Remaining

   $$L_0-BO{D}_t=12.4mg/L$$

Why the other options are there

• 12 mg/L (remaining reported as exerted)
• 571.2 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A treatment tank holds 123,879 gal and treats 4.4 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

   $$4.4\times 10^6=4,400,000gal/day$$

3. Substituting

   $$\theta =123879/4,400,000=0.0282day$$

4. Convert

   $$\theta =0.0282(24)=0.68hr$$

Why the other options are there

• 0.028 hr (day/hour conversion missed)
• 35.52 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities

A plant treating 5.1 MGD requires a 2.0 mg/L dose. What is the daily chemical demand in pounds?

Start with the thinking

• The 8.34 factor converts mg/L·MGD to lb/day.
• Adjust for purity if the product is not 100% active.

Step-by-step solution

1. Feed

   $$lb/day=Q(MGD)\times dose(mg/L)\times 8.34$$

2. Substituting

   $$lb/day=5.1(2.0)(8.34)$$

3. Evaluate — 85.1 lb/day

4. At 65% available strength — 130.9 lb/day of product

Why the other options are there

• 10.2 lb/day (8.34 omitted)
• 1.22 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Horizontal Velocities