A stream flowing 29.0 cfs at 9.5 mg/L receives a discharge of 18.5 cfs at 195.0 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

   $$29.0(9.5)=275.5;18.5(195.0)=3,608$$

3. Total flow

   $$29.0+18.5=47.5cfs$$

4. Solve

   $$C=(275.5+3,608)/47.5=81.75mg/L$$

Why the other options are there

• 102.3 mg/L (unweighted average)
• 204.5 mg/L (concentrations added)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A wastewater has ultimate BOD L₀ = 174 mg/L and k = 0.18 /day (base e). How much BOD is exerted in 3 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

   $$kt=0.18(3)=0.54$$

3. Exponential

   $$e^{(}-0.54)=0.5827$$

4. Substituting

   $$BO{D}_t=174(1-0.5827)=72.6mg/L$$

5. Remaining

   $$L_0-BO{D}_t=101.4mg/L$$

Why the other options are there

• 101.4 mg/L (remaining reported as exerted)
• 94 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A treatment tank holds 63,811 gal and treats 5.1 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

   $$5.1\times 10^6=5,100,000gal/day$$

3. Substituting

   $$\theta =63811/5,100,000=0.0125day$$

4. Convert

   $$\theta =0.0125(24)=0.30hr$$

Why the other options are there

• 0.013 hr (day/hour conversion missed)
• 79.92 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A plant treating 2.2 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

   $$lb/day=2.2(7.5)(8.34)$$

3. Evaluate — 137.6 lb/day

4. At 65% available strength — 211.7 lb/day of product

Why the other options are there

• 16.5 lb/day (8.34 omitted)
• 1.98 lb/day (factor divided)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A stack gas contains 16.0 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

   $$mg/m^3=16.0(28)/24.45$$

3. Evaluate — 18.32 mg/m³

4. Reverse check

   $$ppm=18.32(24.45)/28=16.0✓$$

Why the other options are there

• 14.0 mg/m³ (ratio inverted)
• 20.0 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A reactor of volume 3,364 m³ treats 1.65 m³/s carrying 242 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

   $$\tau =3364/1.65=2,039s=0.566h$$

3. Formula (CSTR)

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

4. Substituting

   $$C=242/(1+0.45\times 0.566)=192.9mg/L$$

5. Formula (PFR)

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

6. Substituting

   $$C=242e^{(}-0.45\times 0.566)=187.6mg/L$$

7. Comparison — plug flow removes 22.5% versus 20.3% for the CSTR

Why the other options are there

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

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

An activated sludge plant treats 19,653 m³/d with an influent BOD of 173 mg/L, effluent BOD 24 mg/L, MLSS 2,057 mg/L, and an aeration basin volume of 4,644 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 19 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

   $$F/M=19653(173)/(4644\times 2057)=0.356d^{-1}$$

3. Formula

   $$\tau =V/Q$$

4. Substituting

   $$\tau =4644/19653=0.2363d=5.67h$$

5. Formula

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

6. Substituting

   $$P_x=0.45(19653)(173-24)/(1+0.06\times 19)mg/d$$

7. Evaluate

   $$P_x=615.8kgVSS/day$$

Why the other options are there

• F/M = 1,653 (basin volume omitted)
• P_x = 1,318 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A circular clarifier 14 m in diameter and 5.0 m deep treats 4,540 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

   $$A=(\pi /4)(14{)}^2=153.9m^2$$

2. Formula

   $$SOR=Q/A$$

3. Substituting — SOR = 4540/153.9 = 29.49 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau =769.7/4540=4.07h$$

6. Formula

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

7. Substituting — 4540/(π × 14) = 103.2 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

    $$v_s=629.3m/d>SOR=29.49m/d\to particleisremoved$$

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 → Greenhouse Gases: Global Warming Potential

A stack gas stream of 18 m³/s carries 27 g/m³ of particulate. It passes a cyclone at 85% efficiency followed by a fabric filter at 97.0% 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

   $$C_1=27(1-0.85)=4.050g/m^3$$

3. Formula

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

4. Substituting

   $$C_2=4.050(1-0.970)=0.1215g/m^3$$

5. Formula

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

6. Substituting

   $$\eta =1-0.150\times 0.0300=0.99550=99.550$$

7. Formula

   $$emission=C_{2}Q$$

8. Substituting

   $$0.1215g/m^3\times 18m^3/s\times 3600s/h÷1000=7.873kg/h$$

Why the other options are there

• η = 182.0% (efficiencies added)
• 1,750 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential

A community of 103,287 people generates 2.6 kg/person/day of MSW and diverts 20% through recycling. Compacted in place at 853 kg/m³ with 20% 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

   $$103287(2.6)(1-0.20)=214,837kg/day$$

3. Formula

   $$V=mass/density$$

4. Substituting

   $$V=214,837/853=251.9m^3/day$$

5. Annual with cover

   $$251.9\times 365\times 1.20=110,315m^3/yr$$

6. Design life

   $$110,315\times 26=2,868,187m^3$$

Why the other options are there

• 2,987,695 m³ (diversion and cover ignored)
• 2,038,802,750 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Greenhouse Gases: Global Warming Potential