Echelon Institute
Class 4 Wastewater — Formula Sheet
Advanced Treatment · Equipment · Lab · Biosolids · Plant Management
Advanced Treatment Process Monitoring
Biological Nutrient Removal — Total Nitrogen (TN) Removal Efficiency
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%
Internal Recycle Ratio (IRR)
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%)
Denitrification Rate (DNR)
Phosphorus Removal — Luxury Uptake (EBPR)
Membrane Bioreactor — Flux
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
Trans-Membrane Pressure (TMP)
MBBR — Surface Area Loading Rate (SALR)
SRT (Solids Retention Time) for Nitrification
Equipment Operation & Maintenance
Centrifuge — Centrifugal Force (G-Force)
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
Centrifuge — Solids Recovery
UV Disinfection — UV Dose
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
UV Transmittance (UVT)
UV absorbance = 0.20. What is the UVT?
UVT = 10^(−0.20) × 100 = 0.631 × 100 = 63.1%
Answer: 63.1%
Chemical Feed — Polymer Dose
Pump Efficiency
Laboratory Analysis & Interpretation
Biochemical Oxygen Demand (BOD₅) — Seeded
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
Volatile Suspended Solids (VSS)
Specific Resistance to Filtration (SRF)
Respirometry — Specific Oxygen Uptake Rate (SOUR)
Effluent Toxicity — LC₅₀
Phosphorus — Ortho-P to TP Ratio
Nitrogen — TKN
Biosolids Management & Regulations
Biosolids — Volatile Solids Reduction (VSR)
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%
Biosolids — Specific Gravity Correction for Dry Tonnes
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)
Land Application — Agronomic Rate
Pathogen Reduction — Log Reduction
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