A stack gas contains 10.0 ppm of a compound with molecular weight 28 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Start with the thinking

• ppm by volume needs the molar volume to become a mass concentration.
• 24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

1. Conversion

   $$mg/m^3=ppm\times MW/24.45$$

2. Substituting

   $$mg/m^3=10.0(28)/24.45$$

3. Evaluate — 11.45 mg/m³

4. Reverse check

   $$ppm=11.45(24.45)/28=10.0✓$$

Why the other options are there

• 8.7 mg/m³ (ratio inverted)
• 12.5 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A reactor of volume 4,964 m³ treats 1.65 m³/s carrying 361 mg/L of a contaminant that decays first-order with k = 0.60 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Start with the thinking

• The mass balance for steady state is: in − out − reaction = 0.
• For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

1. Formula

   $$\tau =V/Q$$

2. Substituting

   $$\tau =4964/1.65=3,008s=0.836h$$

3. Formula (CSTR)

   $$C=C_0/(1+k\tau )$$

4. Substituting

   $$C=361/(1+0.60\times 0.836)=240.4mg/L$$

5. Formula (PFR)

   $$C=C_0{e}^{(}-k\tau )$$

6. Substituting

   $$C=361e^{(}-0.60\times 0.836)=218.6mg/L$$

7. Comparison — plug flow removes 39.4% versus 33.4% for the CSTR

Why the other options are there

• 180.0 mg/L (linear decay assumed)
• 240.4 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

An activated sludge plant treats 14,625 m³/d with an influent BOD of 230 mg/L, effluent BOD 5 mg/L, MLSS 3,867 mg/L, and an aeration basin volume of 2,698 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.07 d⁻¹ and an SRT of 20 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

   $$F/M=14625(230)/(2698\times 3867)=0.322d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =2698/14625=0.1845d=4.43h$$

5. Formula

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

6. Substituting

   $$P_x=0.65(14625)(230-5)/(1+0.07\times 20)mg/d$$

7. Evaluate

   $$P_x=891.2kgVSS/day$$

Why the other options are there

• F/M = 869.9 (basin volume omitted)
• P_x = 2,139 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A circular clarifier 23 m in diameter and 3.5 m deep treats 28,904 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 70 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

   $$A=(\pi /4)(23{)}^2=415.5m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 28904/415.5 = 69.57 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =1,454/28904=1.21h$$

6. Formula

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

7. Substituting — 28904/(π × 23) = 400.0 m³/m·d

8. Formula

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

9. Substituting

   $$v_s=9.81(1650)(7.0e-5{)}^2/(18\times 0.001)=380.7m/d$$

10. Compare

    $$v_s=380.7m/d>SOR=69.57m/d\to particleisremoved$$

Why the other options are there

• SOR = 114.3 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A stack gas stream of 59 m³/s carries 28 g/m³ of particulate. It passes a cyclone at 76% efficiency followed by a fabric filter at 92.0% 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

   $$C_1=28(1-0.76)=6.720g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=6.720(1-0.920)=0.5376g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.240\times 0.0800=0.98080=98.080$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.5376g/m^3\times 59m^3/s\times 3600s/h÷1000=114.2kg/h$$

Why the other options are there

• η = 168.0% (efficiencies added)
• 5,947 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A community of 31,723 people generates 1.9 kg/person/day of MSW and diverts 20% through recycling. Compacted in place at 514 kg/m³ with 20% additional daily cover volume, find the airspace needed for 25 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

   $$31723(1.9)(1-0.20)=48,219kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=48,219/514=93.8m^3/day$$

5. Annual with cover

   $$93.8\times 365\times 1.20=41,089m^3/yr$$

6. Design life

   $$41,089\times 25=1,027,233m^3$$

Why the other options are there

• 1,070,034 m³ (diversion and cover ignored)
• 439,998,010 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A water plant treating 12,817 m³/d applies a chlorine dose of 5.0 mg/L against a demand of 2.0 mg/L, with 78 minutes of contact time. Compute the free residual, the CT value, the daily chlorine mass required, and the concentration remaining after a 3.5-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

   $$5.0-2.0=3.00mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 3.00(78) = 234.0 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$5.0g/m^3\times 12817m^3/d÷1000=64.09kg/day$$

7. Log removal

   $$N=10^6/10^3.5=3.16e+2organisms/100mL$$

Why the other options are there

• CT = 390.0 (dose used instead of residual)
• 64,085 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A city of 64,437 grows at 3.4% per year. Project the population in 30 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 495 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

   $$P=64437(1+0.034{)}^{3}0=175,692people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P=64437(1+0.034\times 30)=130,163people$$

5. Formula

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

6. Substituting

   $$Q_{a}vg=175,692(495)/1000=86,967m^3/d$$

7. Peak day

   $$Q_{p}eak=2.6\times 86,967=226,115m^3/d$$

Why the other options are there

• 65,726 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

Drinking water contains 0.0100 mg/L of a carcinogen. An adult of 75 kg drinks 1.7 L/day. With a cancer slope factor of 0.59 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 25-year half-life remaining after 34 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 = (0.0100 × 1.7)/75 = 2.267e-4 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=2.267e-4(0.59)=1.34e-4$$

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

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}34/25)=38.958$$

Why the other options are there

• Risk = 0.00590 (body weight and intake ignored)
• 32.0% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion

A stream flowing 32.5 cfs at 15.5 mg/L receives a discharge of 18.0 cfs at 160.0 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

   $$32.5(15.5)=503.8;18.0(160.0)=2,880$$

3. Total flow

   $$32.5+18.0=50.5cfs$$

4. Solve

   $$C=(503.8+2,880)/50.5=67.00mg/L$$

Why the other options are there

• 87.8 mg/L (unweighted average)
• 175.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Anaerobic Digestion