Oxygen Requirement (Aeration)
Calculate the oxygen requirement of an aerobic treatment system, O₂ = Q × ΔS ÷ 1000 × f, multiplying the flow (m³/day) by the BOD removed (mg/L) and an oxygen-demand factor (typically 1.0–1.5 kg O₂/kg BOD). The result, in kg O₂/day, sizes blowers and aerators in activated sludge and aerated lagoons, ensuring enough oxygen for the biological oxidation of organic matter. Enter the flow, the BOD removed and the oxygenation factor.
Result
—
Oxygen requirement (aeration)
In aerobic treatment, bacteria oxidize organic matter using oxygen — and supplying that oxygen (through blowers and diffusers or through mechanical aerators) is usually the single largest energy consumer in a wastewater treatment plant. Estimating how much O₂ per day the system needs is therefore essential. The simplified form is O₂ = Q × ΔS ÷ 1000 × f: the flow Q (m³/day) is multiplied by the BOD removed ΔS (mg/L, divided by 1000 to become kg/m³) to obtain the daily load of organic matter oxidized, and a factor f (kg O₂ per kg BOD, typically 1.0–1.5) is applied to cover both carbon oxidation and the endogenous respiration of the biomass. The result, in kg O₂/day, is the starting point for sizing blower capacity, later corrected by transfer factors (α, β) and by altitude and temperature. Undersizing the aeration wrecks treatment efficiency and generates odors; oversizing it wastes energy every hour of every day. Enter the flow, the BOD removed and the oxygenation factor.
Related Tools
Sludge Production
Calculate the biological sludge production of a plant, P_x = Y × ΔS ÷ 1000 × Q, multiplying the cell yield coefficient (Y, kg VSS/kg BOD), the BOD removed (mg/L) and the flow (m³/day). The result, in kg/day, estimates the excess sludge mass generated by biomass growth, key to sizing wasting, thickening, dewatering and final disposal — a step that often drives much of a treatment plant's operating cost. Enter the yield Y, the BOD removed and the flow.
Solids Loading Rate (Clarifier)
Calculate the solids loading rate (SLR) of a secondary clarifier, SLR = Q × X ÷ A, multiplying the flow (m³/day) by the mixed-liquor solids concentration (mg/L, converted to kg/m³) and dividing by the surface area (m²). The result, in kg/(m²·day), is a design criterion independent of the surface overflow (hydraulic) rate: an activated-sludge secondary clarifier must satisfy both the hydraulic limit and the solids loading limit, since it receives a concentrated mixed liquor that must thicken at the bottom. Excessive solids loading causes sludge to wash out with the effluent. Enter the flow, the solids concentration and the clarifier area.
Sludge Recycle Ratio
Calculate the sludge recycle ratio (R) of an activated-sludge system by mass balance, R = X ÷ (X_r − X), from the mixed-liquor suspended solids (MLSS) and the return sludge concentration. The result (dimensionless, or ×100%) gives the fraction of influent flow that must be recycled from the secondary clarifier to keep the desired biomass in the reactor. Typical ratios range from 0.25 to 1.0. Enter the reactor MLSS and the return sludge concentration.
Volumetric Organic Loading Rate
Calculate the volumetric organic loading rate (OLR) of a biological reactor, OLR = BOD load ÷ volume, dividing the influent organic load (kg BOD/day) by the reactor's working volume (m³). The result, in kg BOD/(m³·day), shows how much organic matter is applied per unit volume and is central to sizing lagoons, trickling filters, UASB and activated sludge: high loads demand more biomass and oxygen, while low loads indicate an oversized reactor. Enter the daily BOD load and the reactor volume.
On-Demand Breastfeeding Guide
Understand on-demand breastfeeding: estimate the hours per day and how long to keep nursing based on your baby age. A guide for first-time mothers.
Leaching Requirement (Irrigation)
Computes the leaching requirement of an irrigated field — the fraction of the applied depth that must pass through the root zone and drain away to flush out the salts the irrigation water leaves behind: LR = water EC ÷ (5 × tolerable saturation extract EC − water EC), with both electrical conductivities in decisiemens per metre. The tolerable EC comes from the crop salt tolerance table — beans sit near 1 dS/m, maize near 1.7 and barley above 8. The result, as a percentage, feeds the gross depth calculation, which is the net depth divided by (1 − LR): a requirement of 13.6%, for instance, forces you to apply about 16% more water than the crop consumes. The saltier the water relative to what the crop tolerates, the larger the fraction; and when the water EC approaches five times the tolerable EC the value blows up, a sign that this water is unusable for that crop without artificial drainage or a change of species. Enter the electrical conductivity of the irrigation water and the tolerable electrical conductivity of the soil saturation extract.
The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.