A circular clarifier 12 m in diameter and 5.0 m deep treats 3,884 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

   $$A=(\pi /4)(12{)}^2=113.1m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 3884/113.1 = 34.34 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =565.5/3884=3.49h$$

6. Formula

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

7. Substituting — 3884/(π × 12) = 103.0 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

    $$v_s=497.2m/d>SOR=34.34m/d\to particleisremoved$$

Why the other options are there

• SOR = 20.6 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A stack gas stream of 17 m³/s carries 19 g/m³ of particulate. It passes a cyclone at 86% efficiency followed by a fabric filter at 95.0% 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

   $$C_1=19(1-0.86)=2.660g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=2.660(1-0.950)=0.1330g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.140\times 0.0500=0.99300=99.300$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.1330g/m^3\times 17m^3/s\times 3600s/h÷1000=8.140kg/h$$

Why the other options are there

• η = 181.0% (efficiencies added)
• 1,163 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A community of 158,348 people generates 3.1 kg/person/day of MSW and diverts 30% through recycling. Compacted in place at 503 kg/m³ with 10% additional daily cover volume, find the airspace needed for 20 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

   $$158348(3.1)(1-0.30)=343,615kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=343,615/503=683.1m^3/day$$

5. Annual with cover

   $$683.1\times 365\times 1.10=274,277m^3/yr$$

6. Design life

   $$274,277\times 20=5,485,546m^3$$

Why the other options are there

• 7,124,086 m³ (diversion and cover ignored)
• 2,508,390,668 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A water plant treating 36,517 m³/d applies a chlorine dose of 3.5 mg/L against a demand of 3.6 mg/L, with 109 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 4.0-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

   $$3.5-3.6=0.20mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 0.20(109) = 21.8 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$3.5g/m^3\times 36517m^3/d÷1000=127.8kg/day$$

7. Log removal

   $$N=10^6/10^4.0=1.00e+2organisms/100mL$$

Why the other options are there

• CT = 381.5 (dose used instead of residual)
• 127,810 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A city of 27,736 grows at 2.9% per year. Project the population in 15 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 352 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

   $$P=27736(1+0.029{)}^{1}5=42,587people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P=27736(1+0.029\times 15)=39,801people$$

5. Formula

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

6. Substituting

   $$Q_{a}vg=42,587(352)/1000=14,991m^3/d$$

7. Peak day

   $$Q_{p}eak=2.0\times 14,991=29,981m^3/d$$

Why the other options are there

• 12,065 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

Drinking water contains 0.0040 mg/L of a carcinogen. An adult of 63 kg drinks 2.5 L/day. With a cancer slope factor of 1.50 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 44-year half-life remaining after 11 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 = (0.0040 × 2.5)/63 = 1.587e-4 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=1.587e-4(1.50)=2.38e-4$$

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

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}11/44)=84.090$$

Why the other options are there

• Risk = 0.00600 (body weight and intake ignored)
• 87.5% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A stream flowing 2.5 cfs at 5.5 mg/L receives a discharge of 2.5 cfs at 135.0 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

   $$2.5(5.5)=13.8;2.5(135.0)=337.5$$

3. Total flow

   $$2.5+2.5=5.0cfs$$

4. Solve

   $$C=(13.8+337.5)/5.0=70.25mg/L$$

Why the other options are there

• 70.3 mg/L (unweighted average)
• 140.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A wastewater has ultimate BOD L₀ = 280 mg/L and k = 0.17 /day (base e). How much BOD is exerted in 5 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

   $$kt=0.17(5)=0.85$$

3. Exponential

   $$e^{(}-0.85)=0.4274$$

4. Substituting

   $$BO{D}_t=280(1-0.4274)=160.3mg/L$$

5. Remaining

   $$L_0-BO{D}_t=119.7mg/L$$

Why the other options are there

• 119.7 mg/L (remaining reported as exerted)
• 238.0 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A treatment tank holds 32,827 gal and treats 2.0 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

   $$2.0\times 10^6=2,000,000gal/day$$

3. Substituting

   $$\theta =32827/2,000,000=0.0164day$$

4. Convert

   $$\theta =0.0164(24)=0.39hr$$

Why the other options are there

• 0.016 hr (day/hour conversion missed)
• 60.93 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.

A plant treating 4.6 MGD requires a 10.5 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

   $$lb/day=4.6(10.5)(8.34)$$

3. Evaluate — 402.8 lb/day

4. At 65% available strength — 619.7 lb/day of product

Why the other options are there

• 48.3 lb/day (8.34 omitted)
• 5.79 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Insoluble products are shown in italics.