Wastewater Calculator

Model the activated sludge process for municipal and industrial wastewater treatment. Calculate key parameters including F/M ratio, SRT, reactor volume, oxygen requirements, and sludge production to optimize your treatment plant design and operation.

m³/day

Daily average flow rate (m³/day)

mg/L

BOD concentration in influent (mg/L)

mg/L

Target BOD in effluent (mg/L)

mg/L

Typical: 2000-4000 mg/L

m³

Aeration tank volume (m³)

m³/day

Sludge wastage (m³/day)

m³/day

Return Activated Sludge flow (m³/day)

mg/L

RAS suspended solids (mg/L)

F/M = (Flow × BOD) ÷ (MLSS × Volume) SRT = (MLSS × V) ÷ [RAS Conc × (Wastage + RAS Flow)] O₂ Required = 0.9 × Flow × BODremoved ÷ 1000 + 4.0 × Flow ÷ 1000 Sludge = 0.4 × Flow × BODremoved ÷ 1000 HRT = (Volume ÷ Flow) × 24 hours
F/M: Flow=1000 m³/day, BOD=200 mg/L, MLSS=3000 mg/L, V=500 m³ Food Load = 1000×200/1000 = 200 kg/day, Biomass = 3×500 = 1500 kg F/M = 200/1500 = 0.13 (Low, good nitrification) SRT: MLSS=3000, V=500, Wastage=50, RAS=500, RAS Conc=8000 SRT = (3000×500/1000) ÷ (8×550) = 0.34 days (Very short - adjust!) O₂: Flow=1000, BODin=200, BODout=10 O₂ = 0.9×1000×190/1000 + 4×1000/1000 = 175 kg O₂/day

What is the activated sludge process?

The activated sludge process is a biological wastewater treatment method where air is pumped into a tank containing wastewater and microorganisms (bacteria). The microorganisms consume organic matter (BOD/COD), converting it into carbon dioxide, water, and new cell biomass. The process effectively removes organic pollutants and nutrients from municipal and industrial wastewater.

What is a good F/M ratio for activated sludge?

F/M (Food to Microorganism) ratio measures the balance between organic load and biomass. Optimal ranges: 0.2-0.5 kg BOD/kg MLSS·day for conventional activated sludge, 0.05-0.15 for extended aeration (better nitrification), and 0.5-1.0 for high-rate systems. Too high causes poor treatment; too low wastes tank capacity.

What is SRT and why is it important?

Sludge Retention Time (SRT) is the average time biomass stays in the system. Longer SRT (10-20 days) promotes nitrification and reduces sludge production but increases tank size. Shorter SRT (3-8 days) produces more sludge but requires smaller tanks. SRT directly affects treatment efficiency, sludge characteristics, and nutrient removal capability.

How much oxygen does an activated sludge plant need?

Oxygen requirements are typically 0.8-1.2 kg O₂ per kg BOD removed, plus additional oxygen for nitrification (about 4.0 kg O₂ per kg NH₃-N oxidized). A typical municipal plant needs 1.0-1.5 kg O₂ per kg BOD. Proper aeration is critical—insufficient oxygen causes process failure, while excess wastes energy.

What is MLSS and how is it controlled?

Mixed Liquor Suspended Solids (MLSS) measures the concentration of microorganisms in the aeration tank, typically 2000-4000 mg/L. MLSS is controlled by adjusting wastage rate and Return Activated Sludge (RAS) flow. Too high MLSS causes poor settling; too low reduces treatment capacity. Regular measurement and control are essential for stable operation.

How do I calculate reactor volume for a new plant?

Reactor volume depends on flow rate, BOD load, target F/M ratio, and desired MLSS. Use formula: Volume = (Flow × BOD × SRT) / (MLSS × (1 + kd×SRT)). For quick estimation: Volume ≈ Flow × HRT / 24, where HRT is typically 4-12 hours depending on treatment goals. Our calculator handles these calculations automatically.

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