A reactor of volume 718 m³ treats 1.20 m³/s carrying 355 mg/L of a contaminant that decays first-order with k = 0.10 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

   $$\tau =718/1.20=598.3s=0.166h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=355/(1+0.10\times 0.166)=349.2mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=355e^{(}-0.10\times 0.166)=349.1mg/L$$

7. Comparison — plug flow removes 1.6% versus 1.6% for the CSTR

Why the other options are there

• 349.1 mg/L (linear decay assumed)
• 349.2 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Wind

A reactor of volume 3,702 m³ treats 1.45 m³/s carrying 162 mg/L of a contaminant that decays first-order with k = 0.20 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

   $$\tau =3702/1.45=2,553s=0.709h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=162/(1+0.20\times 0.709)=141.9mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=162e^{(}-0.20\times 0.709)=140.6mg/L$$

7. Comparison — plug flow removes 13.2% versus 12.4% for the CSTR

Why the other options are there

• 139.0 mg/L (linear decay assumed)
• 141.9 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Wind

A reactor of volume 4,543 m³ treats 1.00 m³/s carrying 196 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

   $$\tau =4543/1.00=4,543s=1.262h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=196/(1+0.25\times 1.262)=149.0mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=196e^{(}-0.25\times 1.262)=143.0mg/L$$

7. Comparison — plug flow removes 27.1% versus 24.0% for the CSTR

Why the other options are there

• 134.2 mg/L (linear decay assumed)
• 149.0 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Wind

A reactor of volume 482 m³ treats 1.35 m³/s carrying 122 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

   $$\tau =482/1.35=357.0s=0.099h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=122/(1+0.25\times 0.099)=119.0mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=122e^{(}-0.25\times 0.099)=119.0mg/L$$

7. Comparison — plug flow removes 2.4% versus 2.4% for the CSTR

Why the other options are there

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

Reference: FE Reference Handbook — Environmental Engineering → Wind

A treatment tank holds 94,793 gal and treats 4.8 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

   $$4.8\times 10^6=4,800,000gal/day$$

3. Substituting

   $$\theta =94793/4,800,000=0.0197day$$

4. Convert

   $$\theta =0.0197(24)=0.47hr$$

Why the other options are there

• 0.020 hr (day/hour conversion missed)
• 50.64 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Wind

A plant treating 7.3 MGD requires a 7.5 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

   $$lb/day=7.3(7.5)(8.34)$$

3. Evaluate — 456.6 lb/day

4. At 65% available strength — 702.5 lb/day of product

Why the other options are there

• 54.8 lb/day (8.34 omitted)
• 6.56 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Wind

A stack gas contains 51.0 ppm of a compound with molecular weight 64 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

   $$mg/m^3=51.0(64)/24.45$$

3. Evaluate — 133.5 mg/m³

4. Reverse check

   $$ppm=133.5(24.45)/64=51.0✓$$

Why the other options are there

• 19.5 mg/m³ (ratio inverted)
• 145.7 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Wind

An activated sludge plant treats 11,901 m³/d with an influent BOD of 356 mg/L, effluent BOD 15 mg/L, MLSS 1,967 mg/L, and an aeration basin volume of 3,703 m³. With Y = 0.50 mg VSS/mg BOD, k_d = 0.04 d⁻¹ and an SRT of 11 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

   $$F/M=11901(356)/(3703\times 1967)=0.582d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =3703/11901=0.3112d=7.47h$$

5. Formula

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

6. Substituting

   $$P_x=0.50(11901)(356-15)/(1+0.04\times 11)mg/d$$

7. Evaluate

   $$P_x=1,409kgVSS/day$$

Why the other options are there

• F/M = 2,154 (basin volume omitted)
• P_x = 2,029 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Wind

A circular clarifier 22 m in diameter and 5.0 m deep treats 5,880 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 30 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

   $$A=(\pi /4)(22{)}^2=380.1m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 5880/380.1 = 15.47 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =1,901/5880=7.76h$$

6. Formula

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

7. Substituting — 5880/(π × 22) = 85.1 m³/m·d

8. Formula

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

9. Substituting

   $$v_s=9.81(1650)(3.0e-5{)}^2/(18\times 0.001)=69.93m/d$$

10. Compare

    $$v_s=69.93m/d>SOR=15.47m/d\to particleisremoved$$

Why the other options are there

• SOR = 17.0 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Wind

A stack gas stream of 5 m³/s carries 14 g/m³ of particulate. It passes a cyclone at 90% efficiency followed by a fabric filter at 93.5% 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

   $$C_1=14(1-0.90)=1.400g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=1.400(1-0.935)=0.0910g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.100\times 0.0650=0.99350=99.350$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.0910g/m^3\times 5m^3/s\times 3600s/h÷1000=1.638kg/h$$

Why the other options are there

• η = 183.5% (efficiencies added)
• 252.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Wind