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📐 Class 1 Water Treatment
Formula Sheet — 29 formulas across 8 categories
O. Reg. 170/03 · O. Reg. 169/03 · OWWCO Class 1 Water Treatment Exam
🌊
Water Sources & Quality
4 formulasTurbidity Removal Efficiency
E (%) = [(Turb_in − Turb_out) ÷ Turb_in] × 100(%)
E — Removal efficiency (%)
Turb_in — Influent turbidity (NTU)
Turb_out — Effluent turbidity (NTU)
Worked Example
Influent turbidity = 40 NTU, effluent = 0.1 NTU. What is the removal efficiency?
E = [(40 − 0.1) ÷ 40] × 100 = (39.9 ÷ 40) × 100
✓ 99.75%
💡 Exam tip: Ontario Regulation 170/03 requires treated water turbidity ≤ 1 NTU at all times and ≤ 0.3 NTU in 95% of samples per month.
Flow Rate Conversion
Q (m³/d) = Q (L/s) × 86.4(m³/d)
Q — Flow rate
86.4 — Conversion factor (86,400 s/d ÷ 1,000 L/m³)
Worked Example
A plant treats 25 L/s. What is the daily flow in m³/d?
Q = 25 × 86.4
✓ 2,160 m³/d
💡 Exam tip: Common conversions: 1 L/s = 86.4 m³/d; 1 m³/d = 0.01157 L/s; 1 MGD = 3,785 m³/d.
Hydraulic Detention Time
HDT (h) = Volume (m³) ÷ Flow (m³/h)(hours)
HDT — Hydraulic detention time (hours)
Volume — Tank or basin volume (m³)
Flow — Flow rate through the basin (m³/h)
Worked Example
A sedimentation basin holds 1,200 m³. Flow = 300 m³/h. What is the HDT?
HDT = 1,200 ÷ 300
✓ 4 hours
💡 Exam tip: Typical HDT for sedimentation basins: 2–8 hours. Flocculation basins: 20–60 minutes.
Population Equivalent
PE = BOD Load (kg/d) ÷ 0.054(PE)
PE — Population equivalent
BOD Load — Total BOD load (kg/d)
0.054 — Per capita BOD generation (54 g/person/day)
Worked Example
A plant receives a BOD load of 540 kg/d. What is the PE?
PE = 540 ÷ 0.054
✓ 10,000 PE
💡 Exam tip: Ontario uses 54 g BOD/person/day as the design standard for population equivalent calculations.
⚗️
Coagulation & Flocculation
4 formulasCoagulant Dose (mg/L to kg/d)
Feed Rate (kg/d) = Dose (mg/L) × Flow (m³/d) × 10⁻³(kg/d)
Dose — Target coagulant dose (mg/L)
Flow — Plant flow rate (m³/d)
10⁻³ — Unit conversion factor
Worked Example
Alum dose = 15 mg/L, flow = 4,000 m³/d. What is the alum feed rate?
Feed Rate = 15 × 4,000 × 0.001
✓ 60 kg/d
💡 Exam tip: Typical alum doses for drinking water: 5–50 mg/L depending on raw water turbidity and NOM.
Velocity Gradient (G value)
G (s⁻¹) = √(P ÷ (μ × V))(s⁻¹)
G — Velocity gradient (s⁻¹)
P — Power input to the basin (W)
μ — Dynamic viscosity of water (N·s/m²) — 0.001 at 20°C
V — Basin volume (m³)
Worked Example
Power = 500 W, μ = 0.001 N·s/m², V = 50 m³. What is G?
G = √(500 ÷ (0.001 × 50)) = √(500 ÷ 0.05) = √10,000
✓ 100 s⁻¹
💡 Exam tip: Rapid mix (coagulation): G = 300–1,000 s⁻¹. Flocculation: G = 10–100 s⁻¹. Gt (camp number) for flocculation: 10⁴–10⁵.
Camp Number (Gt)
Gt = G (s⁻¹) × t (s)(dimensionless)
G — Velocity gradient (s⁻¹)
t — Detention time in flocculation basin (s)
Worked Example
G = 40 s⁻¹, flocculation HDT = 30 min. What is Gt?
Gt = 40 × (30 × 60) = 40 × 1,800
✓ 72,000
💡 Exam tip: Optimal Gt for flocculation: 10,000–100,000. Values below 10,000 indicate under-mixing; above 100,000 indicates floc breakup.
Alum Dose → Alkalinity Consumed
Alk Consumed (mg/L as CaCO₃) = Alum Dose (mg/L) × 0.50(mg/L as CaCO₃)
Alum Dose — Alum (Al₂(SO₄)₃·18H₂O) dose (mg/L)
0.50 — Alkalinity consumption factor (1 mg/L alum consumes ~0.50 mg/L alkalinity as CaCO₃)
Worked Example
Alum dose = 20 mg/L. How much alkalinity is consumed?
Alk Consumed = 20 × 0.50
✓ 10 mg/L as CaCO₃
💡 Exam tip: If raw water alkalinity is low (< 50 mg/L as CaCO₃), add lime or soda ash to maintain pH 6.5–7.5 for effective coagulation.
🪨
Sedimentation
3 formulasSurface Overflow Rate (SOR)
SOR (m³/m²·d) = Flow (m³/d) ÷ Surface Area (m²)(m³/m²·d)
SOR — Surface overflow rate (m³/m²·d)
Flow — Plant flow rate (m³/d)
Surface Area — Clarifier plan area (m²)
Worked Example
Flow = 6,000 m³/d, clarifier area = 120 m². What is the SOR?
SOR = 6,000 ÷ 120
✓ 50 m³/m²·d
💡 Exam tip: Typical SOR for conventional clarifiers: 20–60 m³/m²·d. Lower SOR = better solids removal.
Weir Overflow Rate (WOR)
WOR (m³/m·d) = Flow (m³/d) ÷ Weir Length (m)(m³/m·d)
WOR — Weir overflow rate (m³/m·d)
Flow — Plant flow rate (m³/d)
Weir Length — Total effluent weir length (m)
Worked Example
Flow = 6,000 m³/d, weir length = 40 m. What is the WOR?
WOR = 6,000 ÷ 40
✓ 150 m³/m·d
💡 Exam tip: Typical WOR: 125–250 m³/m·d. Exceeding 250 m³/m·d can cause turbulence and carry-over of settled floc.
Sludge Volume
Sludge Vol (m³/d) = [Flow (m³/d) × (TSS_in − TSS_out) × 10⁻³] ÷ Sludge Conc (kg/m³)(m³/d)
TSS_in — Influent TSS (mg/L)
TSS_out — Effluent TSS (mg/L)
Sludge Conc — Sludge concentration (kg/m³, typically 10–50 kg/m³)
Worked Example
Flow = 5,000 m³/d, TSS_in = 80 mg/L, TSS_out = 5 mg/L, sludge = 20 kg/m³.
Sludge Vol = [5,000 × (80−5) × 0.001] ÷ 20 = [5,000 × 0.075] ÷ 20 = 375 ÷ 20
✓ 18.75 m³/d
💡 Exam tip: Primary sludge is typically 2–6% solids (20–60 kg/m³). Thickened sludge: 4–10% solids.
🔬
Filtration
4 formulasFiltration Rate
Filtration Rate (m³/m²·h) = Flow (m³/h) ÷ Filter Area (m²)(m³/m²·h)
Flow — Flow through the filter (m³/h)
Filter Area — Total active filter surface area (m²)
Worked Example
Flow = 500 m³/h, 4 filters each 25 m². What is the filtration rate?
Filtration Rate = 500 ÷ (4 × 25) = 500 ÷ 100
✓ 5 m³/m²·h
💡 Exam tip: Typical filtration rates: 5–15 m³/m²·h for rapid sand filters. Ontario Reg. 170/03 limits: ≤ 15 m/h for dual-media, ≤ 12 m/h for sand-only.
Filter Run Volume
Run Volume (m³) = Filtration Rate (m³/m²·h) × Filter Area (m²) × Run Time (h)(m³)
Filtration Rate — Hydraulic loading rate (m³/m²·h)
Filter Area — Filter surface area (m²)
Run Time — Time between backwashes (h)
Worked Example
Rate = 8 m³/m²·h, area = 30 m², run time = 24 h. What is the run volume?
Run Volume = 8 × 30 × 24
✓ 5,760 m³
💡 Exam tip: Typical filter run times: 24–72 hours. Shorter runs indicate high turbidity or biological growth.
Backwash Rate
Backwash Rate (m³/m²·h) = Backwash Flow (m³/h) ÷ Filter Area (m²)(m³/m²·h)
Backwash Flow — Backwash pump flow rate (m³/h)
Filter Area — Filter surface area (m²)
Worked Example
Backwash flow = 600 m³/h, filter area = 30 m². What is the backwash rate?
Backwash Rate = 600 ÷ 30
✓ 20 m³/m²·h
💡 Exam tip: Typical backwash rates: 36–60 m³/m²·h (10–17 m/h). Backwash should expand the bed 20–30% to release trapped solids.
Filter Bed Expansion
Expansion (%) = [(Expanded Depth − Original Depth) ÷ Original Depth] × 100(%)
Expanded Depth — Filter media depth during backwash (m)
Original Depth — Filter media depth at rest (m)
Worked Example
Original depth = 0.75 m, expanded depth = 0.97 m. What is the expansion?
Expansion = [(0.97 − 0.75) ÷ 0.75] × 100 = (0.22 ÷ 0.75) × 100
✓ 29.3%
💡 Exam tip: Target bed expansion: 20–30%. Less than 20% may not adequately clean the media; more than 40% may wash media out of the filter.
🧪
Disinfection & CT Values
4 formulasCT Value
CT (mg·min/L) = C (mg/L) × T (min)(mg·min/L)
C — Residual disinfectant concentration (mg/L)
T — Contact time (min) — T₁₀ for design (time for 10% of water to pass through)
Worked Example
Chlorine residual = 0.8 mg/L, T₁₀ = 25 min. What is the CT?
CT = 0.8 × 25
✓ 20 mg·min/L
💡 Exam tip: Ontario Reg. 170/03 requires CT ≥ 6 mg·min/L for Giardia inactivation (2-log) at 15°C, pH 7. Required CT increases at lower temperatures.
Chlorine Dose
Dose (mg/L) = Demand (mg/L) + Residual (mg/L)(mg/L)
Dose — Total chlorine applied (mg/L)
Demand — Chlorine consumed by organic matter, ammonia, etc. (mg/L)
Residual — Free chlorine remaining after contact time (mg/L)
Worked Example
Chlorine demand = 1.5 mg/L, target residual = 0.5 mg/L. What dose is needed?
Dose = 1.5 + 0.5
✓ 2.0 mg/L
💡 Exam tip: Ontario requires a minimum free chlorine residual of 0.2 mg/L throughout the distribution system at all times.
Chlorine Feed Rate
Feed Rate (kg/d) = Dose (mg/L) × Flow (m³/d) × 10⁻³(kg/d)
Dose — Chlorine dose (mg/L)
Flow — Plant flow rate (m³/d)
Worked Example
Dose = 2.0 mg/L, flow = 3,500 m³/d. What is the chlorine feed rate?
Feed Rate = 2.0 × 3,500 × 0.001
✓ 7.0 kg/d
💡 Exam tip: For gas chlorine: 1 kg/d = 1,000 g/d. For sodium hypochlorite (12.5% solution): divide kg/d by 0.125 to get L/d.
Log Inactivation
Log Inactivation = log₁₀(N₀ ÷ N)(log units)
N₀ — Initial pathogen concentration
N — Final pathogen concentration after treatment
Worked Example
Initial Giardia cysts = 1,000/L, after treatment = 1/L. What is the log inactivation?
Log Inactivation = log₁₀(1,000 ÷ 1) = log₁₀(1,000)
✓ 3-log (99.9% removal)
💡 Exam tip: Ontario requires 3-log (99.9%) removal/inactivation of Giardia and 4-log (99.99%) for viruses through combined filtration and disinfection.
💊
Chemical Feed & Dosing
3 formulasChemical Feed Rate (Solution)
Feed Rate (L/d) = [Dose (mg/L) × Flow (m³/d)] ÷ [Conc (%) × 10,000](L/d)
Dose — Target chemical dose (mg/L)
Flow — Plant flow rate (m³/d)
Conc (%) — Chemical solution concentration (%, e.g. 12.5 for 12.5%)
10,000 — Conversion factor (% × 10,000 = mg/L)
Worked Example
Dose = 3 mg/L fluoride, flow = 2,000 m³/d, solution = 25% (250,000 mg/L). Feed rate?
Feed Rate = (3 × 2,000) ÷ (25 × 10,000) = 6,000 ÷ 250,000
✓ 0.024 m³/d = 24 L/d
💡 Exam tip: Always verify solution density when converting between volume and mass for concentrated chemicals like sodium hypochlorite (SG ≈ 1.17).
Fluoride Dose Calculation
Volume (L) = [Target F (mg/L) × Vol Treated (L)] ÷ Stock Conc (mg/L)(L)
Target F — Target fluoride concentration (mg/L) — Ontario: 0.7 mg/L
Vol Treated — Volume of water to be treated (L)
Stock Conc — Fluoride stock solution concentration (mg/L)
Worked Example
Target = 0.7 mg/L F, treat 1,000,000 L/d, stock = 18,000 mg/L. Volume needed?
Volume = (0.7 × 1,000,000) ÷ 18,000 = 700,000 ÷ 18,000
✓ 38.9 L/d
💡 Exam tip: Ontario's optimal fluoride level is 0.7 mg/L (Health Canada, 2010). Maximum acceptable concentration (MAC) is 1.5 mg/L.
Lime Dose for pH Adjustment
Lime Dose (mg/L) = [Target Alk − Current Alk] × 0.74(mg/L as Ca(OH)₂)
Target Alk — Target alkalinity (mg/L as CaCO₃)
Current Alk — Current alkalinity (mg/L as CaCO₃)
0.74 — Ratio of Ca(OH)₂ to CaCO₃ molecular weights (74/100)
Worked Example
Current alkalinity = 30 mg/L, target = 80 mg/L. What lime dose is needed?
Lime Dose = (80 − 30) × 0.74 = 50 × 0.74
✓ 37 mg/L Ca(OH)₂
💡 Exam tip: Lime (Ca(OH)₂) raises both pH and alkalinity. Soda ash (Na₂CO₃) raises alkalinity with less pH increase. Use soda ash when pH is already adequate.
🔩
Iron & Manganese Removal
3 formulasOxidant Demand for Iron
Cl₂ Required (mg/L) = Fe²⁺ (mg/L) × 0.64(mg/L Cl₂)
Fe²⁺ — Dissolved iron concentration (mg/L)
0.64 — Stoichiometric ratio (0.64 mg Cl₂ per mg Fe²⁺)
Worked Example
Raw water iron = 3.0 mg/L. How much chlorine is needed to oxidize it?
Cl₂ Required = 3.0 × 0.64
✓ 1.92 mg/L Cl₂
💡 Exam tip: Ontario MAC for iron: 0.3 mg/L (aesthetic). Oxidized iron precipitates as Fe(OH)₃ and is removed by sedimentation and filtration.
Oxidant Demand for Manganese
Cl₂ Required (mg/L) = Mn²⁺ (mg/L) × 1.29(mg/L Cl₂)
Mn²⁺ — Dissolved manganese concentration (mg/L)
1.29 — Stoichiometric ratio (1.29 mg Cl₂ per mg Mn²⁺)
Worked Example
Raw water manganese = 0.5 mg/L. How much chlorine is needed?
Cl₂ Required = 0.5 × 1.29
✓ 0.645 mg/L Cl₂
💡 Exam tip: Ontario MAC for manganese: 0.05 mg/L (health-based, 2019). Mn oxidation requires pH > 8.0 with chlorine; KMnO₄ is more effective at lower pH.
KMnO₄ Dose for Manganese
KMnO₄ (mg/L) = Mn²⁺ (mg/L) × 1.92(mg/L KMnO₄)
Mn²⁺ — Dissolved manganese (mg/L)
1.92 — Stoichiometric ratio (1.92 mg KMnO₄ per mg Mn²⁺)
Worked Example
Manganese = 0.4 mg/L. What KMnO₄ dose is needed?
KMnO₄ = 0.4 × 1.92
✓ 0.768 mg/L
💡 Exam tip: KMnO₄ is a strong oxidant — overdosing causes pink water (MAC = 0.05 mg/L KMnO₄ in finished water). Always jar test before full-scale application.
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Water Quality & Regulations
4 formulasLangelier Saturation Index (LSI)
LSI = pH − pHs(dimensionless)
pH — Actual measured pH of the water
pHs — pH at which water is saturated with CaCO₃ (calculated from temp, TDS, Ca²⁺, alkalinity)
Worked Example
Measured pH = 7.8, pHs = 7.5. What is the LSI?
LSI = 7.8 − 7.5
✓ LSI = +0.3 (slightly scale-forming)
💡 Exam tip: LSI > 0: scale-forming (protective CaCO₃ coating). LSI < 0: corrosive. LSI = 0: balanced. Target LSI: 0 to +0.5 for distribution system corrosion control.
Hardness (as CaCO₃)
Hardness (mg/L as CaCO₃) = [Ca²⁺ (mg/L) ÷ 40.1] × 100 + [Mg²⁺ (mg/L) ÷ 24.3] × 100(mg/L as CaCO₃)
Ca²⁺ — Calcium concentration (mg/L)
Mg²⁺ — Magnesium concentration (mg/L)
40.1 / 24.3 — Atomic masses of Ca and Mg (g/mol)
Worked Example
Ca²⁺ = 80 mg/L, Mg²⁺ = 12 mg/L. What is the total hardness?
Hardness = (80/40.1 × 100) + (12/24.3 × 100) = 199.5 + 49.4
✓ 248.9 mg/L as CaCO₃
💡 Exam tip: Hardness classification: 0–60 soft, 61–120 moderately hard, 121–180 hard, >180 very hard (mg/L as CaCO₃). Ontario has no MAC for hardness.
Alkalinity (as CaCO₃)
Alkalinity (mg/L as CaCO₃) = Titrant Vol (mL) × N × 50,000 ÷ Sample Vol (mL)(mg/L as CaCO₃)
Titrant Vol — Volume of H₂SO₄ used in titration (mL)
N — Normality of H₂SO₄ titrant (e.g. 0.02 N)
50,000 — Equivalent weight of CaCO₃ × 1,000
Sample Vol — Volume of water sample (mL)
Worked Example
Titrant = 8.5 mL of 0.02 N H₂SO₄, sample = 100 mL. What is the alkalinity?
Alkalinity = 8.5 × 0.02 × 50,000 ÷ 100 = 8,500 ÷ 100
✓ 85 mg/L as CaCO₃
💡 Exam tip: Minimum alkalinity for corrosion control: 30–50 mg/L as CaCO₃. Alkalinity also buffers pH during chlorination and coagulation.
SUVA (Specific UV Absorbance)
SUVA (L/mg·m) = UV₂₅₄ (cm⁻¹) × 100 ÷ DOC (mg/L)(L/mg·m)
UV₂₅₄ — UV absorbance at 254 nm (cm⁻¹)
DOC — Dissolved organic carbon (mg/L)
100 — Converts cm⁻¹ to m⁻¹
Worked Example
UV₂₅₄ = 0.15 cm⁻¹, DOC = 5.0 mg/L. What is the SUVA?
SUVA = 0.15 × 100 ÷ 5.0 = 15 ÷ 5.0
✓ 3.0 L/mg·m
💡 Exam tip: SUVA > 4: high NOM, hydrophobic — responds well to coagulation. SUVA < 2: low NOM, hydrophilic — coagulation less effective. SUVA used to predict DBP formation potential.