A stack gas contains 29.5 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

   $$mg/m^3=29.5(44)/24.45$$

3. Evaluate — 53.09 mg/m³

4. Reverse check

   $$ppm=53.09(24.45)/44=29.5✓$$

Why the other options are there

• 16.4 mg/m³ (ratio inverted)
• 57.9 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A reactor of volume 1,018 m³ treats 1.65 m³/s carrying 215 mg/L of a contaminant that decays first-order with k = 0.45 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

   $$\tau =1018/1.65=617.0s=0.171h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=215/(1+0.45\times 0.171)=199.6mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=215e^{(}-0.45\times 0.171)=199.0mg/L$$

7. Comparison — plug flow removes 7.4% versus 7.2% for the CSTR

Why the other options are there

• 198.4 mg/L (linear decay assumed)
• 199.6 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

An activated sludge plant treats 22,351 m³/d with an influent BOD of 283 mg/L, effluent BOD 16 mg/L, MLSS 2,599 mg/L, and an aeration basin volume of 2,158 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 17 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

   $$F/M=22351(283)/(2158\times 2599)=1.128d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =2158/22351=0.0966d=2.32h$$

5. Formula

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

6. Substituting

   $$P_x=0.45(22351)(283-16)/(1+0.05\times 17)mg/d$$

7. Evaluate

   $$P_x=1,452kgVSS/day$$

Why the other options are there

• F/M = 2,434 (basin volume omitted)
• P_x = 2,685 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A circular clarifier 29 m in diameter and 4.0 m deep treats 16,735 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 40 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

   $$A=(\pi /4)(29{)}^2=660.5m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 16735/660.5 = 25.34 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =2,642/16735=3.79h$$

6. Formula

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

7. Substituting — 16735/(π × 29) = 183.7 m³/m·d

8. Formula

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

9. Substituting

   $$v_s=9.81(1650)(4.0e-5{)}^2/(18\times 0.001)=124.3m/d$$

10. Compare

    $$v_s=124.3m/d>SOR=25.34m/d\to particleisremoved$$

Why the other options are there

• SOR = 45.9 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A stack gas stream of 26 m³/s carries 10 g/m³ of particulate. It passes a cyclone at 88% efficiency followed by a fabric filter at 94.0% 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

   $$C_1=10(1-0.88)=1.200g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=1.200(1-0.940)=0.0720g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.120\times 0.0600=0.99280=99.280$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.0720g/m^3\times 26m^3/s\times 3600s/h÷1000=6.739kg/h$$

Why the other options are there

• η = 182.0% (efficiencies added)
• 936.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A community of 210,133 people generates 1.9 kg/person/day of MSW and diverts 30% through recycling. Compacted in place at 565 kg/m³ with 15% additional daily cover volume, find the airspace needed for 29 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

   $$210133(1.9)(1-0.30)=279,477kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=279,477/565=494.6m^3/day$$

5. Annual with cover

   $$494.6\times 365\times 1.15=207,629m^3/yr$$

6. Design life

   $$207,629\times 29=6,021,243m^3$$

Why the other options are there

• 7,479,805 m³ (diversion and cover ignored)
• 2,958,262,881 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A water plant treating 29,822 m³/d applies a chlorine dose of 5.0 mg/L against a demand of 3.0 mg/L, with 87 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

   $$5.0-3.0=2.00mg/L$$

3. Formula

   $$CT=C\times t$$

4. Substituting — CT = 2.00(87) = 174.0 mg·min/L

5. Formula

   $$mass=dose\times Q$$

6. Substituting

   $$5.0g/m^3\times 29822m^3/d÷1000=149.1kg/day$$

7. Log removal

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

Why the other options are there

• CT = 435.0 (dose used instead of residual)
• 149,110 kg/day (unit conversion missed)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A city of 76,575 grows at 2.3% per year. Project the population in 28 years by both geometric and arithmetic growth, then compute the average and peak day water demand at 302 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

   $$P=76575(1+0.023{)}^{2}8=144,745people$$

3. Formula (arithmetic)

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

4. Substituting

   $$P=76575(1+0.023\times 28)=125,889people$$

5. Formula

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

6. Substituting

   $$Q_{a}vg=144,745(302)/1000=43,713m^3/d$$

7. Peak day

   $$Q_{p}eak=2.7\times 43,713=118,025m^3/d$$

Why the other options are there

• 49,314 (growth increment reported as population)
• Q sized on the average day only

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

Drinking water contains 0.0360 mg/L of a carcinogen. An adult of 73 kg drinks 2.5 L/day. With a cancer slope factor of 1.41 (mg/kg·d)⁻¹, compute the chronic daily intake and the incremental lifetime cancer risk. Also determine the fraction of a radionuclide with a 36-year half-life remaining after 62 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 = (0.0360 × 2.5)/73 = 1.233e-3 mg/kg·d

3. Formula

   $$Risk=CDI\times CSF$$

4. Substituting

   $$Risk=1.233e-3(1.41)=1.74e-3$$

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

6. Formula

   $$N/N_0=(½{)}^{(}t/t½)$$

7. Substituting

   $$(½{)}^{(}62/36)=30.308$$

Why the other options are there

• Risk = 0.05076 (body weight and intake ignored)
• 13.9% remaining (linear decay)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion

A stream flowing 23.0 cfs at 1.5 mg/L receives a discharge of 18.5 cfs at 220.0 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

   $$23.0(1.5)=34.5;18.5(220.0)=4,070$$

3. Total flow

   $$23.0+18.5=41.5cfs$$

4. Solve

   $$C=(34.5+4,070)/41.5=98.90mg/L$$

Why the other options are there

• 110.8 mg/L (unweighted average)
• 221.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Bed Expansion