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Echelon Institute

Class 4 Wastewater — Formula Sheet

Advanced Treatment · Equipment · Lab · Biosolids · Plant Management

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Advanced Treatment Process Monitoring

Biological Nutrient Removal — Total Nitrogen (TN) Removal Efficiency

TN Removal (%) = (TN_in − TN_eff) ÷ TN_in × 100
Units: %
Variables
TN_inInfluent total nitrogen (mg/L)
TN_effEffluent total nitrogen (mg/L)
Worked Example

Influent TN = 42 mg/L, effluent TN = 6 mg/L. What is the TN removal efficiency?

TN Removal = (42 − 6) ÷ 42 × 100 = 36 ÷ 42 × 100 = 85.7%

Answer: 85.7%

💡 Exam tip: Class 4 plants typically target TN < 10 mg/L. Effluent TN rising with stable MLSS often indicates anoxic zone DO intrusion or insufficient carbon for denitrification.

Internal Recycle Ratio (IRR)

IRR = Q_recycle ÷ Q_influent
Units: dimensionless
Variables
Q_recycleInternal mixed liquor recycle flow rate (m³/d)
Q_influentInfluent flow rate (m³/d)
Worked Example

Influent flow = 10,000 m³/d, internal recycle = 30,000 m³/d. What is the IRR?

IRR = 30,000 ÷ 10,000 = 3.0

Answer: IRR = 3.0 (300%)

💡 Exam tip: IRR of 2–4 is typical for 4-stage Bardenpho and MLE processes. Higher IRR improves TN removal but increases pumping energy. IRR > 5 rarely improves performance.

Denitrification Rate (DNR)

DNR = (NO₃_in − NO₃_out) × Q ÷ (V_anoxic × MLVSS)
Units: g NO₃-N/g VSS·d
Variables
NO₃_inNitrate entering anoxic zone (mg/L)
NO₃_outNitrate leaving anoxic zone (mg/L)
QFlow through anoxic zone (m³/d)
V_anoxicAnoxic zone volume (m³)
MLVSSMixed liquor volatile suspended solids (g/m³)
💡 Exam tip: Typical DNR = 0.03–0.11 g NO₃-N/g VSS·d at 20°C. Low DNR indicates insufficient carbon (BOD:TKN < 5:1) or temperature inhibition. Add methanol or acetate as supplemental carbon.

Phosphorus Removal — Luxury Uptake (EBPR)

P_release = MLVSS × f_PAO × q_P
Units: mg P/L
Variables
MLVSSMixed liquor volatile suspended solids (mg/L)
f_PAOFraction of PAOs in biomass (typically 0.15–0.35)
q_PSpecific P release rate (mg P/g VSS·h)
💡 Exam tip: EBPR requires a true anaerobic zone (DO = 0, NO₃ = 0). Nitrate intrusion into the anaerobic zone is the #1 cause of EBPR failure. Monitor VFA:TP ratio; target > 4 mg VFA/mg TP.

Membrane Bioreactor — Flux

J = Q_permeate ÷ A_membrane
Units: L/m²·h (LMH)
Variables
Q_permeatePermeate flow rate (L/h)
A_membraneTotal membrane area (m²)
Worked Example

Permeate flow = 500,000 L/h, total membrane area = 20,000 m². What is the flux?

J = 500,000 ÷ 20,000 = 25 LMH

Answer: 25 LMH

💡 Exam tip: Typical MBR flux = 15–30 LMH at design conditions. Flux > 30 LMH increases fouling rate. TMP rising at constant flux = membrane fouling. Reduce flux and increase backwash frequency.

Trans-Membrane Pressure (TMP)

TMP = P_feed − P_permeate
Units: kPa
Variables
P_feedFeed side pressure (kPa)
P_permeatePermeate side pressure (kPa)
💡 Exam tip: TMP rising > 5 kPa/day at constant flux indicates fouling. Initiate maintenance clean (citric acid or sodium hypochlorite). Recovery clean required when TMP > 50 kPa.

MBBR — Surface Area Loading Rate (SALR)

SALR = BOD_load ÷ A_media
Units: g BOD/m²·d
Variables
BOD_loadApplied BOD load (g/d)
A_mediaTotal effective media surface area (m²)
💡 Exam tip: Typical MBBR SALR for carbonaceous removal = 5–15 g BOD/m²·d. For nitrification = 0.5–2.0 g NH₄-N/m²·d. Media fill fraction typically 40–70% of reactor volume.

SRT (Solids Retention Time) for Nitrification

SRT_min = 1 ÷ (μ_max_AOB × (NH₄ ÷ (K_s + NH₄)) − b_AOB)
Units: days
Variables
μ_max_AOBMaximum growth rate of AOB at 20°C (≈ 0.9 d⁻¹)
K_sHalf-saturation constant for NH₄ (≈ 0.5–1.0 mg/L)
b_AOBDecay rate of AOB (≈ 0.05 d⁻¹)
NH₄Effluent ammonia concentration (mg/L)
💡 Exam tip: At 20°C, minimum SRT for nitrification ≈ 4–6 days. Apply safety factor of 1.5–2.0. At 10°C, minimum SRT doubles. Nitrification failure at SRT < safety factor × SRT_min.
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Equipment Operation & Maintenance

Centrifuge — Centrifugal Force (G-Force)

G = 1.118 × 10⁻⁵ × r × N²
Units: × g (relative centrifugal force)
Variables
rRadius from centre to liquid surface (cm)
NRotational speed (RPM)
Worked Example

Centrifuge bowl radius = 25 cm, speed = 2,000 RPM. What is the G-force?

G = 1.118 × 10⁻⁵ × 25 × 2,000² = 1.118 × 10⁻⁵ × 25 × 4,000,000 = 1,118 × g

Answer: 1,118 × g

💡 Exam tip: Higher G-force improves solids capture but increases wear and energy. Typical decanter centrifuge G-force = 1,500–3,000 × g for biosolids dewatering.

Centrifuge — Solids Recovery

Solids Recovery (%) = (TS_cake × Q_cake) ÷ (TS_feed × Q_feed) × 100
Units: %
Variables
TS_cakeTotal solids in dewatered cake (%)
Q_cakeCake production rate (kg/h)
TS_feedTotal solids in feed sludge (%)
Q_feedFeed sludge flow rate (kg/h)
💡 Exam tip: Target solids recovery > 90% for Class A biosolids. Polymer dose optimization: increase dose if recovery < 90%, reduce if centrate TSS is very low (over-dosing wastes polymer).

UV Disinfection — UV Dose

UV Dose = I × t
Units: mJ/cm²
Variables
IUV intensity (mW/cm²)
tExposure time (seconds)
Worked Example

UV intensity = 30 mW/cm², exposure time = 10 seconds. What is the UV dose?

UV Dose = 30 × 10 = 300 mJ/cm²

Answer: 300 mJ/cm²

💡 Exam tip: Ontario Reg. 170/03 requires UV dose ≥ 40 mJ/cm² for drinking water. For wastewater reuse, target ≥ 100 mJ/cm². UV transmittance (UVT) < 55% requires dose correction. Clean lamps regularly.

UV Transmittance (UVT)

UVT (%) = 10^(−A) × 100
Units: %
Variables
AUV absorbance at 254 nm (dimensionless)
Worked Example

UV absorbance = 0.20. What is the UVT?

UVT = 10^(−0.20) × 100 = 0.631 × 100 = 63.1%

Answer: 63.1%

💡 Exam tip: UVT < 55% severely reduces UV effectiveness. High TSS, iron, humic acids, and nitrite reduce UVT. Pre-treatment (filtration, coagulation) improves UVT before UV disinfection.

Chemical Feed — Polymer Dose

Polymer Dose (kg/d) = Q_sludge × TS_feed × Dose_specific
Units: kg/d
Variables
Q_sludgeSludge feed flow rate (m³/d)
TS_feedFeed sludge total solids concentration (kg/m³)
Dose_specificSpecific polymer dose (kg polymer/tonne dry solids)
💡 Exam tip: Typical polymer dose for biosolids dewatering = 5–15 kg/tonne DS. Jar testing determines optimal dose. Over-dosing is expensive and can reduce cake solids. Under-dosing reduces capture efficiency.

Pump Efficiency

η_pump = (Q × H × ρ × g) ÷ P_shaft × 100
Units: %
Variables
QFlow rate (m³/s)
HTotal dynamic head (m)
ρFluid density (kg/m³, water ≈ 1,000)
gGravitational acceleration (9.81 m/s²)
P_shaftShaft power input (W)
💡 Exam tip: Pump efficiency typically 65–85%. Efficiency drops with wear (impeller erosion), off-design operation, or cavitation. Monitor pump curves and compare to design point quarterly.
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Laboratory Analysis & Interpretation

Biochemical Oxygen Demand (BOD₅) — Seeded

BOD₅ = [(DO_i − DO_f) − (B_i − B_f) × P] ÷ f
Units: mg/L
Variables
DO_iInitial DO of diluted sample (mg/L)
DO_fFinal DO of diluted sample after 5 days (mg/L)
B_iInitial DO of seed control (mg/L)
B_fFinal DO of seed control (mg/L)
PFraction of seed in sample bottle
fDecimal fraction of sample used (dilution factor)
Worked Example

DO_i = 8.2, DO_f = 2.1, B_i = 8.0, B_f = 6.5, P = 0.10, f = 0.02. What is BOD₅?

BOD₅ = [(8.2 − 2.1) − (8.0 − 6.5) × 0.10] ÷ 0.02 = [6.1 − 0.15] ÷ 0.02 = 5.95 ÷ 0.02 = 297.5 mg/L

Answer: 297.5 mg/L

💡 Exam tip: Valid BOD test: DO depletion 2–7 mg/L, DO_f > 1 mg/L. Seed correction critical for industrial wastewater. Nitrification inhibitor (TCMP) added to measure only carbonaceous BOD.

Volatile Suspended Solids (VSS)

VSS = (W_dried − W_ignited) ÷ V_sample × 1000
Units: mg/L
Variables
W_driedWeight of dried residue at 105°C (g)
W_ignitedWeight after ignition at 550°C (g)
V_sampleSample volume (mL)
💡 Exam tip: VSS/TSS ratio (volatile fraction) should be 0.75–0.85 for healthy activated sludge. Low VSS/TSS = inert inorganic accumulation (grit, precipitates). High VSS/TSS = young sludge or industrial input.

Specific Resistance to Filtration (SRF)

SRF = 2 × b × A² × ΔP ÷ (μ × C)
Units: m/kg
Variables
bSlope of t/V vs V plot (s/m⁶)
AFilter area (m²)
ΔPApplied pressure (Pa)
μFiltrate viscosity (Pa·s)
CMass of dry solids per unit volume of filtrate (kg/m³)
💡 Exam tip: SRF < 10¹² m/kg = good dewaterability. SRF > 10¹³ m/kg = poor dewaterability, requires polymer conditioning. SRF test predicts centrifuge and belt press performance.

Respirometry — Specific Oxygen Uptake Rate (SOUR)

SOUR = OUR × 1000 ÷ MLVSS
Units: mg O₂/g VSS·h
Variables
OUROxygen uptake rate (mg/L·min)
MLVSSMixed liquor volatile suspended solids (mg/L)
💡 Exam tip: SOUR 8–20 = active aerobic conditions. SOUR < 5 = low activity (old sludge, toxicity, low temperature). SOUR > 25 = overloaded. Used to diagnose bulking, toxicity, and nitrification.

Effluent Toxicity — LC₅₀

LC₅₀ = Concentration causing 50% mortality in test organisms
Units: % effluent (v/v)
💡 Exam tip: Ontario ECA may require whole effluent toxicity (WET) testing using rainbow trout or Daphnia magna. LC₅₀ > 100% means effluent is not acutely toxic at 100% concentration. Chronic toxicity (IC₂₅) is more sensitive.

Phosphorus — Ortho-P to TP Ratio

Ortho-P:TP = Ortho-P ÷ TP × 100
Units: %
Variables
Ortho-PSoluble reactive phosphorus (mg/L)
TPTotal phosphorus (mg/L)
💡 Exam tip: Ortho-P:TP > 80% in effluent indicates most P is in dissolved form — chemical precipitation (alum, ferric) or EBPR needed. Ortho-P:TP < 30% indicates P is mainly particulate — improve solids removal.

Nitrogen — TKN

TKN = Organic N + NH₄⁺-N
Units: mg/L
💡 Exam tip: TKN measures organic nitrogen + ammonia. Does NOT include nitrate or nitrite. TKN ÷ BOD ratio > 0.1 indicates nitrogen-limited conditions. Typical municipal wastewater TKN = 35–60 mg/L.
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Biosolids Management & Regulations

Biosolids — Volatile Solids Reduction (VSR)

VSR (%) = (VS_in − VS_out) ÷ VS_in × 100
Units: %
Variables
VS_inVolatile solids entering digester (kg/d)
VS_outVolatile solids leaving digester (kg/d)
Worked Example

VS_in = 5,000 kg/d, VS_out = 2,750 kg/d. What is the VSR?

VSR = (5,000 − 2,750) ÷ 5,000 × 100 = 2,250 ÷ 5,000 × 100 = 45%

Answer: 45%

💡 Exam tip: Ontario Reg. 267/03 requires VSR ≥ 38% for Class B biosolids via anaerobic digestion. Class A (EQ) requires additional pathogen reduction (e.g., thermophilic digestion, lime stabilization).

Biosolids — Specific Gravity Correction for Dry Tonnes

Dry Mass (kg/d) = Q_sludge (m³/d) × TS (%) × SG × 10
Units: kg/d
Variables
Q_sludgeSludge volume (m³/d)
TSTotal solids content (%)
SGSpecific gravity of sludge (typically 1.02–1.06)
Worked Example

Sludge flow = 200 m³/d, TS = 4%, SG = 1.03. What is the dry mass?

Dry Mass = 200 × 4 × 1.03 × 10 = 8,240 kg/d

Answer: 8,240 kg/d (8.24 tonnes/d)

💡 Exam tip: Accurate dry mass calculation is essential for Reg. 267/03 nutrient management planning. Land application rates are based on agronomic nitrogen loading, not volume.

Land Application — Agronomic Rate

Rate (kg/ha) = Crop N Requirement ÷ (TKN × Availability Factor)
Units: kg dry solids/ha
Variables
Crop N RequirementNitrogen uptake by crop (kg N/ha)
TKNTotal Kjeldahl nitrogen in biosolids (kg N/tonne DS)
Availability FactorFraction of TKN available to crop (0.3–0.8 depending on form)
💡 Exam tip: Ontario Reg. 267/03 limits land application based on agronomic nitrogen need. Phosphorus accumulation in soil may be the limiting factor on some fields. Maintain 100 m setback from water bodies.

Pathogen Reduction — Log Reduction

Log Reduction = log₁₀(N_in ÷ N_out)
Units: log₁₀ units
Variables
N_inPathogen concentration entering process (CFU/g or MPN/g)
N_outPathogen concentration leaving process (CFU/g or MPN/g)
Worked Example

Fecal coliforms: N_in = 10⁷ MPN/g, N_out = 10³ MPN/g. What is the log reduction?

Log Reduction = log₁₀(10⁷ ÷ 10³) = log₁₀(10⁴) = 4 log

Answer: 4 log reduction

💡 Exam tip: Class A EQ biosolids: fecal coliforms < 1,000 MPN/g or Salmonella < 3 MPN/4g. Class B: fecal coliforms < 2 × 10⁶ MPN/g. Class A requires additional vector attraction reduction (VAR).

Biogas Production Rate

V_biogas = VS_destroyed × Y_biogas
Units: m³/d
Variables
VS_destroyedVolatile solids destroyed in digester (kg/d)
Y_biogasBiogas yield (typically 0.75–1.12 m³ biogas/kg VS destroyed)
💡 Exam tip: Biogas is typically 60–70% methane, 30–40% CO₂. Energy content ≈ 22 MJ/m³ biogas. Biogas production dropping at constant loading = digester upset (pH, temperature, toxic inhibition).
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Plant Management, Safety & Administration

Energy Intensity

Energy Intensity = Total Energy (kWh/d) ÷ Flow Treated (m³/d)
Units: kWh/m³
💡 Exam tip: Typical municipal WWTP energy intensity = 0.3–0.6 kWh/m³. Aeration accounts for 50–60% of energy. MBR plants use 0.8–1.5 kWh/m³. Benchmark against similar plants and target 10–20% reduction.

Cost Per Tonne of Biosolids Processed

Unit Cost = Total Annual Biosolids Cost ($) ÷ Dry Tonnes Produced (t/yr)
Units: $/tonne DS
💡 Exam tip: Typical biosolids management cost = $100–$400/tonne DS depending on treatment level and disposal method. Land application is usually lowest cost; incineration highest. Class A biosolids may have market value.

Permit Compliance Rate

Compliance Rate (%) = (Compliant Samples ÷ Total Samples) × 100
Units: %
💡 Exam tip: Ontario ECA requires 100% compliance with effluent limits. Exceedances must be reported to MECP within 2 hours (acute) or 2 days (chronic). Root cause analysis and corrective action required for all exceedances.

Confined Space — Atmospheric Testing Order

Test Order: O₂ → Flammable Gas → Toxic Gas (H₂S, CO)
💡 Exam tip: O₂ must be tested first: 19.5–23.0% is safe range. < 19.5% = oxygen deficient (IDLH). > 23.0% = oxygen enriched (fire hazard). H₂S IDLH = 50 ppm. LEL for methane = 5%. Never enter without continuous monitoring.

SCADA — Alarm Priority Classification

Priority 1 (Critical): Immediate response required Priority 2 (High): Response within 15 min Priority 3 (Medium): Response within 1 hour Priority 4 (Low): Response within shift
💡 Exam tip: Alarm rationalization reduces nuisance alarms. Target < 1 alarm per 10 minutes during steady-state operation. Flooding alarms, effluent limit exceedance, and equipment failure are always Priority 1.

Staffing Level — Ontario Reg. 128/04

Class 4 WW plant requires: Operator-in-Charge (OIC) holding Class 4 WW licence + sufficient licensed operators for continuous coverage
💡 Exam tip: OIC is responsible for overall plant operation and compliance. Operator of Record (OOR) is responsible for day-to-day operations. Class 4 plants must have a Class 4 OIC on call 24/7. Document all OIC decisions.