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Class 1 Wastewater Formula Sheet

Ontario Operator Exam Reference — 7 Categories · 25+ Formulas

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Wastewater Characteristics

BOD Removal Efficiency

E (%) = [(BOD_in − BOD_out) ÷ BOD_in] × 100

Units: %

ERemoval efficiency (%)

BOD_inInfluent BOD concentration (mg/L)

BOD_outEffluent BOD concentration (mg/L)

WORKED EXAMPLE

Influent BOD = 220 mg/L, effluent BOD = 22 mg/L. What is the removal efficiency?

E = [(220 − 22) ÷ 220] × 100 = (198 ÷ 220) × 100

Answer: 90%

💡Secondary treatment typically achieves 85–95% BOD removal. Primary treatment alone achieves 25–40%.

Suspended Solids Removal

SS Removal (%) = [(SS_in − SS_out) ÷ SS_in] × 100

Units: %

SS_inInfluent suspended solids (mg/L)

SS_outEffluent suspended solids (mg/L)

WORKED EXAMPLE

Influent SS = 240 mg/L, effluent SS = 18 mg/L. What is the SS removal?

SS Removal = [(240 − 18) ÷ 240] × 100 = (222 ÷ 240) × 100

Answer: 92.5%

💡Ontario effluent limits under O. Reg. 347 typically require ≤ 25 mg/L TSS and ≤ 25 mg/L CBOD₅.

Per Capita BOD Loading

BOD Load (kg/d) = Flow (m³/d) × BOD (mg/L) × 10⁻³

Units: kg/d

FlowDaily flow rate (m³/d)

BODBOD concentration (mg/L)

10⁻³Unit conversion factor (mg/L × m³/d → kg/d)

WORKED EXAMPLE

Flow = 5,000 m³/d, BOD = 200 mg/L. What is the BOD load?

BOD Load = 5,000 × 200 × 0.001

Answer: 1,000 kg/d

💡Typical domestic wastewater: 200–300 mg/L BOD. Per capita load ≈ 54 g BOD/person/day.

Population Equivalent (PE)

PE = BOD Load (kg/d) ÷ 0.054

Units: persons

PEPopulation equivalent (persons)

BOD LoadTotal BOD load (kg/d)

0.054Per capita BOD load (54 g/person/day = 0.054 kg/person/day)

WORKED EXAMPLE

A plant receives a BOD load of 540 kg/d. What is the PE?

PE = 540 ÷ 0.054

Answer: 10,000 persons

💡PE is used to size treatment plants and compare industrial to domestic loads.

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Primary Treatment

Hydraulic Detention Time (HDT)

HDT = V ÷ Q

Units: hours or days

HDTHydraulic detention time

VTank volume (m³ or L)

QFlow rate (m³/h or L/h)

WORKED EXAMPLE

Primary clarifier volume = 1,200 m³, flow = 300 m³/h. What is the HDT?

HDT = 1,200 ÷ 300

Answer: 4 hours

💡Typical primary clarifier HDT: 1.5–2.5 hours. Secondary clarifier HDT: 2–4 hours.

Surface Overflow Rate (SOR)

SOR = Q ÷ A

Units: m³/m²·d or m/d

SORSurface overflow rate (m³/m²·d)

QFlow rate (m³/d)

ASurface area of clarifier (m²)

WORKED EXAMPLE

Flow = 4,800 m³/d, clarifier diameter = 20 m. What is the SOR?

A = π × (10)² = 314.2 m²; SOR = 4,800 ÷ 314.2

Answer: 15.3 m³/m²·d

💡Primary clarifier SOR: 24–48 m³/m²·d. Secondary clarifier SOR: 16–32 m³/m²·d.

Weir Overflow Rate (WOR)

WOR = Q ÷ L_weir

Units: m³/m·d

WORWeir overflow rate (m³/m·d)

QFlow rate (m³/d)

L_weirTotal weir length (m)

WORKED EXAMPLE

Flow = 4,800 m³/d, circular clarifier diameter = 20 m. What is the WOR?

Weir length = π × 20 = 62.8 m; WOR = 4,800 ÷ 62.8

Answer: 76.4 m³/m·d

💡Typical WOR limit: < 125 m³/m·d for primary, < 186 m³/m·d for secondary.

Sludge Volume (Primary)

Sludge Volume (m³/d) = [Q × SS_removed (mg/L)] ÷ [ρ_sludge × %solids × 10⁶]

Units: m³/d

QFlow rate (m³/d)

SS_removedSuspended solids removed (mg/L)

ρ_sludgeSludge density (≈ 1,000 kg/m³ for dilute sludge)

%solidsSludge solids content (decimal, e.g. 0.04 for 4%)

WORKED EXAMPLE

Q = 5,000 m³/d, SS removed = 150 mg/L, sludge is 4% solids. What volume of sludge is produced?

Sludge = (5,000 × 150) ÷ (1,000 × 0.04 × 10⁶) = 750,000 ÷ 40,000,000

Answer: 18.75 m³/d

💡Primary sludge is typically 3–8% solids. Raw primary sludge has high putrescibility — pump frequently.

🦠

Secondary Treatment

Food-to-Microorganism Ratio (F/M)

F/M = (Q × BOD_in) ÷ (V × MLVSS)

Units: kg BOD/kg MLVSS·d

QFlow rate (m³/d)

BOD_inInfluent BOD (mg/L)

VAeration basin volume (m³)

MLVSSMixed Liquor Volatile Suspended Solids (mg/L)

WORKED EXAMPLE

Q = 3,000 m³/d, BOD_in = 200 mg/L, V = 1,500 m³, MLVSS = 2,000 mg/L. What is F/M?

F/M = (3,000 × 200) ÷ (1,500 × 2,000) = 600,000 ÷ 3,000,000

Answer: 0.2 kg BOD/kg MLVSS·d

💡Conventional activated sludge F/M: 0.2–0.4. Extended aeration: 0.05–0.15. High F/M → filamentous bulking risk.

Sludge Volume Index (SVI)

SVI = (SV₃₀ × 1,000) ÷ MLSS

Units: mL/g

SV₃₀Settled sludge volume after 30 min (mL/L)

MLSSMixed Liquor Suspended Solids (mg/L)

WORKED EXAMPLE

SV₃₀ = 320 mL/L, MLSS = 2,500 mg/L. What is the SVI?

SVI = (320 × 1,000) ÷ 2,500 = 320,000 ÷ 2,500

Answer: 128 mL/g

💡Good settling: SVI 80–150 mL/g. Bulking: SVI > 200 mL/g. Pin floc: SVI < 50 mL/g.

Hydraulic Retention Time (HRT)

HRT = V ÷ Q

Units: hours

VAeration basin volume (m³)

QInfluent flow rate (m³/h)

WORKED EXAMPLE

Aeration basin = 2,400 m³, flow = 200 m³/h. What is the HRT?

HRT = 2,400 ÷ 200

Answer: 12 hours

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

Return Activated Sludge (RAS) Rate

RAS (%) = [SV₃₀ ÷ (1,000 − SV₃₀)] × 100

Units: %

SV₃₀Settled sludge volume after 30 min (mL/L)

WORKED EXAMPLE

SV₃₀ = 250 mL/L. What RAS rate is required?

RAS = [250 ÷ (1,000 − 250)] × 100 = (250 ÷ 750) × 100

Answer: 33%

💡Typical RAS rate: 25–75% of influent flow. Higher SVI requires higher RAS to maintain MLSS.

Oxygen Demand (Theoretical)

O₂ Required (kg/d) = BOD Load removed (kg/d) × 1.5

Units: kg O₂/d

BOD Load removedBOD removed per day (kg/d)

1.5Approximate O₂:BOD ratio for carbonaceous oxidation

WORKED EXAMPLE

BOD removed = 800 kg/d. Estimate oxygen demand.

O₂ = 800 × 1.5

Answer: 1,200 kg O₂/d

💡Actual O₂ demand depends on MLSS, temperature, and nitrification. Nitrification adds ~4.6 kg O₂/kg NH₃-N.

🧪

Disinfection

Chlorine Dose

Dose (mg/L) = Demand (mg/L) + Residual (mg/L)

Units: mg/L

DoseTotal chlorine applied (mg/L)

DemandChlorine consumed by organics/ammonia (mg/L)

ResidualRemaining chlorine after contact time (mg/L)

WORKED EXAMPLE

Chlorine demand = 3.5 mg/L, required residual = 0.5 mg/L. What dose is needed?

Dose = 3.5 + 0.5

Answer: 4.0 mg/L

💡Ontario O. Reg. 170/03 requires minimum 0.2 mg/L free chlorine residual in distribution. Wastewater effluent residual requirements vary by MOECP permit.

CT Value

CT = C × t

Units: mg·min/L

CDisinfectant residual concentration (mg/L)

tContact time (minutes)

WORKED EXAMPLE

Chlorine residual = 1.0 mg/L, contact time = 30 minutes. What is the CT?

CT = 1.0 × 30

Answer: 30 mg·min/L

💡CT values are used to verify inactivation of Giardia and Cryptosporidium. Higher CT = more effective disinfection.

Chlorine Feed Rate

Feed Rate (kg/d) = Dose (mg/L) × Flow (m³/d) × 10⁻³

Units: kg/d

DoseChlorine dose (mg/L)

FlowFlow rate (m³/d)

WORKED EXAMPLE

Dose = 5 mg/L, flow = 6,000 m³/d. What is the daily chlorine feed rate?

Feed Rate = 5 × 6,000 × 0.001

Answer: 30 kg/d

💡Convert to cylinder usage: 1 kg Cl₂ = 1 kg from 100% gas cylinders. For 12.5% sodium hypochlorite: 1 kg Cl₂ ≈ 8 L solution.

♻️

Solids Handling & Biosolids

Sludge Mass Balance

Sludge Produced (kg/d) = Q × (SS_in − SS_out) × 10⁻³

Units: kg/d

QFlow rate (m³/d)

SS_inInfluent suspended solids (mg/L)

SS_outEffluent suspended solids (mg/L)

WORKED EXAMPLE

Q = 4,000 m³/d, SS_in = 220 mg/L, SS_out = 20 mg/L. How much sludge is produced?

Sludge = 4,000 × (220 − 20) × 0.001 = 4,000 × 200 × 0.001

Answer: 800 kg/d

💡Net sludge production in activated sludge: typically 0.3–0.8 kg VSS/kg BOD removed, depending on SRT.

Volatile Solids Reduction (Digestion)

VS Reduction (%) = [(VS_in − VS_out) ÷ VS_in] × 100

Units: %

VS_inVolatile solids fed to digester (kg/d)

VS_outVolatile solids leaving digester (kg/d)

WORKED EXAMPLE

VS fed = 500 kg/d, VS out = 200 kg/d. What is the VS reduction?

VS Reduction = [(500 − 200) ÷ 500] × 100 = (300 ÷ 500) × 100

Answer: 60%

💡Ontario Reg. 267/03 (Biosolids) requires ≥ 38% VS reduction for Class B biosolids, ≥ 38% for Class A (with pathogen reduction).

Sludge Thickening — Solids Loading Rate

Solids Loading (kg/m²·d) = Sludge Mass (kg/d) ÷ Thickener Area (m²)

Units: kg/m²·d

Sludge MassTotal solids applied (kg/d)

Thickener AreaSurface area of gravity thickener (m²)

WORKED EXAMPLE

Sludge = 1,200 kg/d, thickener diameter = 10 m. What is the solids loading rate?

A = π × 5² = 78.5 m²; Loading = 1,200 ÷ 78.5

Answer: 15.3 kg/m²·d

💡Gravity thickener design loading: 25–80 kg/m²·d for primary sludge; 20–40 kg/m²·d for WAS.

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Collection Systems & Hydraulics

Manning's Equation (Pipe Flow)

Q = (1/n) × A × R^(2/3) × S^(1/2)

Units: m³/s

QFlow rate (m³/s)

nManning's roughness coefficient (0.013 for concrete)

ACross-sectional area of flow (m²)

RHydraulic radius = A ÷ wetted perimeter (m)

SSlope of pipe (m/m)

WORKED EXAMPLE

A 600 mm concrete pipe (n = 0.013) flows full at S = 0.002. What is the flow?

A = π(0.3)² = 0.283 m²; R = 0.3/2 = 0.15 m; Q = (1/0.013) × 0.283 × 0.15^(2/3) × 0.002^(1/2) ≈ 0.283 m³/s

Answer: ≈ 0.28 m³/s (280 L/s)

💡For sewer design, minimum velocity = 0.6 m/s (self-cleaning). Maximum velocity = 3 m/s (erosion limit).

Pump Head (Total Dynamic Head)

TDH = Static Head + Friction Head + Minor Losses

Units: m

Static HeadElevation difference between suction and discharge (m)

Friction HeadHead loss due to pipe friction (m)

Minor LossesHead loss from fittings, valves, bends (m)

WORKED EXAMPLE

Static head = 12 m, friction losses = 3.5 m, minor losses = 0.5 m. What is TDH?

TDH = 12 + 3.5 + 0.5

Answer: 16 m

💡Pump must be selected to deliver required flow at the calculated TDH. Check pump curve for operating point.

Pump Power

Power (kW) = (Q × ρ × g × TDH) ÷ (η × 1000)

Units: kW

QFlow rate (m³/s)

ρFluid density (1,000 kg/m³ for water)

gGravitational acceleration (9.81 m/s²)

TDHTotal dynamic head (m)

ηPump efficiency (decimal, e.g. 0.75 for 75%)

WORKED EXAMPLE

Q = 0.05 m³/s, TDH = 20 m, efficiency = 75%. What is the pump power?

Power = (0.05 × 1,000 × 9.81 × 20) ÷ (0.75 × 1,000) = 9,810 ÷ 750

Answer: 13.1 kW

💡Motor nameplate power should exceed calculated power by 10–20% safety factor.

Wet Well Volume (Pump Cycling)

V = (Q_in × T_on) or V ≥ (Q_pump − Q_in) × T_off

Units: m³

Q_inInflow rate (m³/min)

Q_pumpPump capacity (m³/min)

T_onPump run time (min)

T_offPump off time (min)

WORKED EXAMPLE

Inflow = 2 m³/min, pump capacity = 6 m³/min. What minimum wet well volume allows 5-min off time?

V = (6 − 2) × 5 = 4 × 5

Answer: 20 m³

💡Minimum cycle time protects pump motors. Typical minimum off time: 5–10 minutes for submersible pumps.

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Regulations & Safety

Confined Space Atmospheric Testing

Safe Entry: O₂ = 19.5–23%, LEL < 10%, H₂S < 10 ppm, CO < 25 ppm

O₂Oxygen content (19.5–23% is safe range)

LELLower Explosive Limit (< 10% of LEL required)

H₂SHydrogen sulphide (< 10 ppm for entry)

COCarbon monoxide (< 25 ppm TWA)

💡Ontario O. Reg. 632/05 (Confined Spaces) requires atmospheric testing before and during entry. Continuous monitoring required when hazards exist.

Effluent Limits (O. Reg. 347 — Municipal)

CBOD₅ ≤ 25 mg/L | TSS ≤ 25 mg/L | pH 6.0–9.5

💡These are minimum secondary treatment standards. Individual Environmental Compliance Approvals (ECAs) may be more stringent. Phosphorus limits (e.g. 1.0 mg/L TP) apply to many Ontario plants.

Biosolids Classification (O. Reg. 267/03)

Class B: ≥ 38% VS reduction OR fecal coliform < 2×10⁶ MPN/g TS
Class A: ≥ 38% VS reduction AND fecal coliform < 1,000 MPN/g TS

💡Class A biosolids can be applied to agricultural land without site restrictions. Class B requires 30-day buffer from application to harvest for food crops.

Chlorination Safety — TLV/TWA

Cl₂: TLV-TWA = 0.5 ppm | IDLH = 10 ppm
H₂S: TLV-TWA = 1 ppm | IDLH = 50 ppm

TLV-TWAThreshold Limit Value — Time Weighted Average (8-hr day)

IDLHImmediately Dangerous to Life or Health

💡WHMIS 2015 requires SDS for all hazardous chemicals. Chlorine gas detectors must be installed in chlorination rooms. Self-contained breathing apparatus (SCBA) required for Cl₂ IDLH response.

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