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Canadian Water & Wastewater

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WPI Class IV Wastewater Formula Sheet

Advanced process control · BNR · Biogas · Biosolids · Asset management · Emergency response

⚙️

Advanced Process Control

Sludge Volume Index (SVI)

SVI = (Settled sludge volume mL/L × 1000) / MLSS mg/L

• Settled sludge volume: mL/L after 30-min settling

• MLSS: Mixed Liquor Suspended Solids (mg/L)

Example: Settled volume = 250 mL/L, MLSS = 2,500 mg/L

→ SVI = (250 × 1000) / 2500 = 100 mL/g

💡 Good settling: SVI 50–150 mL/g. Bulking: SVI > 150 mL/g.

Food-to-Microorganism Ratio (F:M)

F:M = BOD_applied (kg/d) / MLVSS_in_aeration (kg)

• BOD_applied: kg/d = Flow (m³/d) × BOD (mg/L) / 1000

• MLVSS: Mixed Liquor Volatile Suspended Solids (kg)

Example: BOD = 300 mg/L, Flow = 10,000 m³/d, MLVSS = 2,000 mg/L, Volume = 3,000 m³

→ F:M = (10,000 × 300/1000) / (2,000 × 3,000/1000) = 3,000 / 6,000 = 0.5 kg BOD/kg MLVSS·d

💡 Conventional: 0.2–0.4. Extended aeration: 0.05–0.15.

Solids Retention Time (SRT / MCRT)

SRT = MLSS_in_system (kg) / (WAS_rate + Effluent_SS) (kg/d)

• MLSS_in_system: Total solids in aeration basin (kg)

• WAS_rate: Waste activated sludge rate (kg/d)

• Effluent_SS: Solids leaving in effluent (kg/d)

Example: MLSS = 5,000 kg, WAS = 400 kg/d, Effluent SS = 50 kg/d

→ SRT = 5,000 / (400 + 50) = 11.1 days

💡 Nitrification requires SRT > 10 days at 15°C. Denitrification: 15–30 days.

Hydraulic Retention Time (HRT)

HRT (h) = Volume (m³) / Flow (m³/h)

• Volume: Aeration basin volume (m³)

• Flow: Influent flow rate (m³/h)

Example: Volume = 3,000 m³, Flow = 500 m³/h

→ HRT = 3,000 / 500 = 6 hours

💡 Typical activated sludge HRT: 4–8 hours. Extended aeration: 18–36 hours.

Return Activated Sludge (RAS) Rate

RAS ratio = Q_RAS / Q_influent

• Q_RAS: Return sludge flow (m³/d)

• Q_influent: Influent flow (m³/d)

Example: Q_RAS = 4,000 m³/d, Q_influent = 10,000 m³/d

→ RAS ratio = 4,000 / 10,000 = 0.4 (40%)

💡 Typical RAS ratio: 25–75% of influent flow.

♻️

Biological Nutrient Removal (BNR)

Nitrogen Mass Balance

TN_removed = TN_in − TN_effluent

• TN_in: Total nitrogen in influent (mg/L)

• TN_effluent: Total nitrogen in effluent (mg/L)

Example: TN_in = 40 mg/L, TN_effluent = 8 mg/L

→ TN_removed = 40 − 8 = 32 mg/L (80% removal)

💡 Typical effluent TN target: ≤ 10 mg/L for advanced treatment.

Nitrification Rate

Nitrification = NH₄⁺_in − NH₄⁺_effluent (mg/L)

• NH₄⁺_in: Influent ammonia (mg/L as N)

• NH₄⁺_effluent: Effluent ammonia (mg/L as N)

Example: NH₄⁺_in = 30 mg/L, NH₄⁺_effluent = 1 mg/L

→ Nitrification = 30 − 1 = 29 mg/L NH₄⁺ oxidized

💡 Oxygen demand: 4.57 g O₂ per g NH₄⁺-N nitrified.

Denitrification Rate

NO₃⁻_removed = NO₃⁻_in − NO₃⁻_effluent (mg/L)

• NO₃⁻_in: Nitrate entering anoxic zone (mg/L)

• NO₃⁻_effluent: Effluent nitrate (mg/L)

Example: NO₃⁻_in = 25 mg/L, NO₃⁻_effluent = 5 mg/L

→ NO₃⁻_removed = 25 − 5 = 20 mg/L

💡 Carbon source required: ~4 g BOD per g NO₃⁻-N removed.

Phosphorus Removal (Bio-P)

TP_removed = TP_in − TP_effluent (mg/L)

• TP_in: Total phosphorus influent (mg/L)

• TP_effluent: Total phosphorus effluent (mg/L)

Example: TP_in = 8 mg/L, TP_effluent = 0.5 mg/L

→ TP_removed = 8 − 0.5 = 7.5 mg/L (94% removal)

💡 Enhanced Bio-P (EBPR) can achieve TP < 1 mg/L without chemicals.

Chemical Phosphorus Removal — Alum Dose

Al:P molar ratio = Al dose (mg/L as Al) / P to remove (mg/L as P) × (27/31)

• Al dose: Aluminum sulfate dose (mg/L as Al)

• P: Phosphorus to remove (mg/L as P)

• Molar masses: Al = 27, P = 31

Example: Remove 5 mg/L P, Al:P ratio = 1.5

→ Al dose = 1.5 × 5 × (27/31) = 6.5 mg/L as Al → Alum dose = 6.5 × (342/54) = 41 mg/L

💡 Al:P molar ratio of 1.5–2.0 typical for effluent TP < 1 mg/L.

⚡

Biogas & Energy Recovery

Biogas Production from Digestion

Biogas (m³/d) = VS_destroyed (kg/d) × Specific gas yield (m³/kg VS)

• VS_destroyed: Volatile solids destroyed in digester (kg/d)

• Specific gas yield: typically 0.75–1.12 m³/kg VS

Example: VS_destroyed = 1,000 kg/d, yield = 0.85 m³/kg VS

→ Biogas = 1,000 × 0.85 = 850 m³/d

💡 Biogas is typically 60–70% methane (CH₄). Energy content: ~22 MJ/m³ biogas.

Methane Energy Content

Energy (kWh/d) = CH₄ volume (m³/d) × 9.97 kWh/m³

• CH₄ volume: Methane produced per day (m³/d)

• 9.97 kWh/m³: Energy content of methane at STP

Example: CH₄ = 500 m³/d

→ Energy = 500 × 9.97 = 4,985 kWh/d

💡 CHP (combined heat and power) efficiency: ~35% electrical, ~45% thermal.

Volatile Solids Reduction

VS_reduction (%) = (VS_in − VS_out) / VS_in × 100

• VS_in: Volatile solids entering digester (kg/d)

• VS_out: Volatile solids leaving digester (kg/d)

Example: VS_in = 2,000 kg/d, VS_out = 900 kg/d

→ VS_reduction = (2,000 − 900) / 2,000 × 100 = 55%

💡 Mesophilic digestion target: ≥ 38% VS reduction. Thermophilic: ≥ 40%.

Digester Loading Rate

OLR (kg VS/m³·d) = VS_fed (kg/d) / Digester volume (m³)

• VS_fed: Volatile solids fed to digester daily (kg/d)

• Digester volume: Active digester volume (m³)

Example: VS_fed = 2,000 kg/d, Volume = 2,500 m³

→ OLR = 2,000 / 2,500 = 0.8 kg VS/m³·d

💡 Mesophilic OLR: 1.6–4.8 kg VS/m³·d. Overloading causes VFA accumulation.

📋

Effluent Quality & Regulatory

BOD Removal Efficiency

BOD removal (%) = (BOD_in − BOD_out) / BOD_in × 100

• BOD_in: Influent BOD (mg/L)

• BOD_out: Effluent BOD (mg/L)

Example: BOD_in = 250 mg/L, BOD_out = 10 mg/L

→ BOD removal = (250 − 10) / 250 × 100 = 96%

💡 Secondary treatment target: ≥ 85% BOD removal or ≤ 25 mg/L effluent BOD.

TSS Removal Efficiency

TSS removal (%) = (TSS_in − TSS_out) / TSS_in × 100

• TSS_in: Influent total suspended solids (mg/L)

• TSS_out: Effluent TSS (mg/L)

Example: TSS_in = 220 mg/L, TSS_out = 12 mg/L

→ TSS removal = (220 − 12) / 220 × 100 = 94.5%

💡 Secondary treatment target: ≥ 85% TSS removal or ≤ 25 mg/L effluent TSS.

Effluent Dilution Factor

DF = (Q_river + Q_effluent) / Q_effluent

• Q_river: Receiving water flow (m³/s or L/s)

• Q_effluent: Effluent discharge flow (m³/s or L/s)

Example: Q_river = 10 m³/s, Q_effluent = 0.5 m³/s

→ DF = (10 + 0.5) / 0.5 = 21

💡 Higher dilution factor = less stringent effluent limits required.

Chlorine Residual (CT Value)

CT = Concentration (mg/L) × Contact time (min)

• Concentration: Free chlorine residual (mg/L)

• Contact time: T₁₀ (min) — time for 10% of water to pass

Example: Cl₂ = 1.5 mg/L, T₁₀ = 30 min

→ CT = 1.5 × 30 = 45 mg·min/L

💡 CT for 4-log Giardia inactivation at 15°C, pH 7: ~65 mg·min/L.

🌱

Biosolids & Solids Management

Biosolids Production Rate

Biosolids (kg/d) = TSS_removed (kg/d) + Biomass_growth (kg/d)

• TSS_removed: Solids removed in primary + secondary (kg/d)

• Biomass_growth: Net biological growth (kg/d)

Example: TSS_removed = 800 kg/d, Biomass = 400 kg/d

→ Biosolids = 800 + 400 = 1,200 kg/d dry solids

💡 Typical: 0.8–1.2 kg dry solids per kg BOD removed.

Sludge Thickening — Gravity

Solids loading (kg/m²·d) = Solids fed (kg/d) / Thickener area (m²)

• Solids fed: Dry solids to thickener (kg/d)

• Thickener area: Surface area of gravity thickener (m²)

Example: Solids = 1,200 kg/d, Area = 60 m²

→ Solids loading = 1,200 / 60 = 20 kg/m²·d

💡 Gravity thickener design loading: 25–80 kg/m²·d for primary sludge.

Centrifuge Capture Efficiency

Capture (%) = (Solids_in − Solids_centrate) / Solids_in × 100

• Solids_in: Solids fed to centrifuge (kg/d)

• Solids_centrate: Solids in centrate (kg/d)

Example: Solids_in = 1,000 kg/d, Solids_centrate = 30 kg/d

→ Capture = (1,000 − 30) / 1,000 × 100 = 97%

💡 Target capture efficiency: ≥ 95% for biosolids dewatering.

Biosolids Land Application — Loading Rate

N loading (kg/ha) = Biosolids applied (t/ha) × TKN (kg/t) × Availability factor

• Biosolids applied: Wet tonnes per hectare

• TKN: Total Kjeldahl Nitrogen in biosolids (kg/t)

• Availability factor: 0.5 for Class B, 0.7 for Class A

Example: Applied = 10 t/ha, TKN = 50 kg/t, Class B (factor = 0.5)

→ N loading = 10 × 50 × 0.5 = 250 kg N/ha

💡 Typical agronomic N rate: 150–300 kg N/ha/yr depending on crop.

🚨

Emergency Response & Risk

Overflow Volume Calculation

Overflow volume (m³) = (Inflow − Capacity) × Duration (h) × 3,600 / 1,000

• Inflow: Peak wet weather flow (L/s)

• Capacity: Treatment capacity (L/s)

• Duration: Duration of overflow event (h)

Example: Inflow = 500 L/s, Capacity = 350 L/s, Duration = 2 h

→ Overflow = (500 − 350) × 2 × 3,600 / 1,000 = 1,080 m³

💡 Report all SSOs (Sanitary Sewer Overflows) per regulatory requirements.

Chlorine Demand for Emergency Disinfection

Cl₂ demand (mg/L) = Cl₂ applied − Cl₂ residual

• Cl₂ applied: Chlorine dose added (mg/L)

• Cl₂ residual: Remaining free chlorine after contact (mg/L)

Example: Cl₂ applied = 8 mg/L, Cl₂ residual = 0.5 mg/L

→ Cl₂ demand = 8 − 0.5 = 7.5 mg/L

💡 Emergency disinfection target: ≥ 0.5 mg/L free chlorine residual after 30 min.

Pump Station Wet Well Volume

Wet well volume (m³) = Peak flow (m³/min) × Cycle time (min) / 4

• Peak flow: Maximum inflow to wet well (m³/min)

• Cycle time: Desired pump on/off cycle time (min)

Example: Peak flow = 2 m³/min, Cycle time = 10 min

→ Wet well volume = 2 × 10 / 4 = 5 m³

💡 Minimum wet well volume prevents excessive pump cycling (< 6 starts/hr).

🏛️

Asset Management & Economics

Net Present Value (NPV)

NPV = Σ [CF_t / (1 + r)^t] − Initial Cost

• CF_t: Cash flow in year t ($)

• r: Discount rate (decimal)

• t: Year number

• Initial Cost: Capital investment ($)

Example: Annual savings = $50,000, r = 5%, 20 years, Initial cost = $500,000

→ NPV = $50,000 × [(1−(1.05)⁻²⁰)/0.05] − $500,000 = $623,111 − $500,000 = +$123,111

💡 Positive NPV = project is economically viable.

Life Cycle Cost

LCC = Capital cost + PV(O&M costs) + PV(Replacement costs) − PV(Salvage value)

• Capital cost: Initial investment ($)

• PV: Present value of future costs ($)

• O&M: Operations and maintenance costs ($/yr)

Example: Capital = $1M, PV(O&M) = $800K, PV(Replacement) = $200K, PV(Salvage) = $50K

→ LCC = $1,000,000 + $800,000 + $200,000 − $50,000 = $1,950,000

💡 Use LCC to compare alternatives over their full service life.

Energy Intensity

Energy intensity (kWh/m³) = Total energy use (kWh/d) / Flow treated (m³/d)

• Total energy use: Facility-wide energy consumption (kWh/d)

• Flow treated: Average daily flow (m³/d)

Example: Energy = 5,000 kWh/d, Flow = 10,000 m³/d

→ Energy intensity = 5,000 / 10,000 = 0.5 kWh/m³

💡 Benchmark: 0.3–0.6 kWh/m³ for secondary treatment. Advanced: 0.6–1.0 kWh/m³.

Unit Cost of Treatment

Unit cost ($/m³) = Total annual cost ($) / Annual volume treated (m³)

• Total annual cost: All operating + capital costs ($)

• Annual volume: Total flow treated per year (m³)

Example: Annual cost = $2,000,000, Annual volume = 3,650,000 m³

→ Unit cost = $2,000,000 / 3,650,000 = $0.548/m³

💡 Typical range: $0.20–$1.50/m³ depending on treatment level and plant size.

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