A stack gas contains 28.5 ppm of a compound with molecular weight 44 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

   $$mg/m^3=28.5(44)/24.45$$

3. Evaluate — 51.29 mg/m³

4. Reverse check

   $$ppm=51.29(24.45)/44=28.5✓$$

Why the other options are there

• 15.8 mg/m³ (ratio inverted)
• 56.0 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A reactor of volume 903 m³ treats 1.25 m³/s carrying 163 mg/L of a contaminant that decays first-order with k = 0.25 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

   $$\tau =903/1.25=722.4s=0.201h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=163/(1+0.25\times 0.201)=155.2mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=163e^{(}-0.25\times 0.201)=155.0mg/L$$

7. Comparison — plug flow removes 4.9% versus 4.8% for the CSTR

Why the other options are there

• 154.8 mg/L (linear decay assumed)
• 155.2 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

An activated sludge plant treats 8,353 m³/d with an influent BOD of 176 mg/L, effluent BOD 12 mg/L, MLSS 2,755 mg/L, and an aeration basin volume of 1,836 m³. With Y = 0.40 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 12 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

   $$F/M=8353(176)/(1836\times 2755)=0.291d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =1836/8353=0.2198d=5.28h$$

5. Formula

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

6. Substituting

   $$P_x=0.40(8353)(176-12)/(1+0.06\times 12)mg/d$$

7. Evaluate

   $$P_x=318.6kgVSS/day$$

Why the other options are there

• F/M = 533.6 (basin volume omitted)
• P_x = 548.0 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A circular clarifier 33 m in diameter and 3.5 m deep treats 10,537 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

   $$A=(\pi /4)(33{)}^2=855.3m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 10537/855.3 = 12.32 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =2,994/10537=6.82h$$

6. Formula

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

7. Substituting — 10537/(π × 33) = 101.6 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

    $$v_s=629.3m/d>SOR=12.32m/d\to particleisremoved$$

Why the other options are there

• SOR = 29.0 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A stack gas stream of 54 m³/s carries 23 g/m³ of particulate. It passes a cyclone at 77% efficiency followed by a fabric filter at 92.5% 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

   $$C_1=23(1-0.77)=5.290g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=5.290(1-0.925)=0.3967g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.230\times 0.0750=0.98275=98.275$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.3967g/m^3\times 54m^3/s\times 3600s/h÷1000=77.128kg/h$$

Why the other options are there

• η = 169.5% (efficiencies added)
• 4,471 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A community of 154,210 people generates 2.2 kg/person/day of MSW and diverts 35% through recycling. Compacted in place at 660 kg/m³ with 25% additional daily cover volume, find the airspace needed for 26 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

   $$154210(2.2)(1-0.35)=220,520kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=220,520/660=334.1m^3/day$$

5. Annual with cover

   $$334.1\times 365\times 1.25=152,443m^3/yr$$

6. Design life

   $$152,443\times 26=3,963,518m^3$$

Why the other options are there

• 4,878,176 m³ (diversion and cover ignored)
• 2,092,737,647 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A water plant treating 11,846 m³/d applies a chlorine dose of 5.5 mg/L against a demand of 3.6 mg/L, with 73 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

   $$5.5-3.6=1.90mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 1.90(73) = 138.7 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$5.5g/m^3\times 11846m^3/d÷1000=65.15kg/day$$

7. Log removal

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

Why the other options are there

• CT = 401.5 (dose used instead of residual)
• 65,153 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A city of 25,182 grows at 3.0% per year. Project the population in 23 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 278 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

   $$P=25182(1+0.030{)}^{2}3=49,699people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P=25182(1+0.030\times 23)=42,558people$$

5. Formula

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

6. Substituting

   $$Q_{a}vg=49,699(278)/1000=13,816m^3/d$$

7. Peak day

   $$Q_{p}eak=1.9\times 13,816=26,251m^3/d$$

Why the other options are there

• 17,376 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

Drinking water contains 0.0460 mg/L of a carcinogen. An adult of 76 kg drinks 1.9 L/day. With a cancer slope factor of 1.86 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 18-year half-life remaining after 101 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 × 1.9)/76 = 1.150e-3 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=1.150e-3(1.86)=2.14e-3$$

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

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}101/18)=2.046$$

Why the other options are there

• Risk = 0.08556 (body weight and intake ignored)
• -180.6% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer

A stream flowing 29.5 cfs at 2.0 mg/L receives a discharge of 15.5 cfs at 295.0 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

   $$29.5(2.0)=59.0;15.5(295.0)=4,573$$

3. Total flow

   $$29.5+15.5=45.0cfs$$

4. Solve

   $$C=(59.0+4,573)/45.0=102.9mg/L$$

Why the other options are there

• 148.5 mg/L (unweighted average)
• 297.0 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Turbulent Flow Impeller Mixer