Volumetric Organic Loading Rate
Worked example: 864 kg BOD/d into 1200 m3 → 0.72 kg/(m3.d) — press Try an example to run it live, then adjust anything.
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Volumetric Organic Loading Rate explained
Volumetric loading answers the question a designer actually asks: how big does the tank have to be? Take the pounds — or kilograms — of BOD arriving each day and divide by the reactor volume. A flow of 4320 m³/d at 200 mg/L delivers 864 kg BOD/d, and in a 1200 m³ aeration basin that is 0.72 kg BOD/(m³·d). Conventional activated sludge sits at 0.3–1.0, extended aeration well below 0.3, high-rate systems above 1.5, a rock trickling filter around 0.2–0.5, and a mesophilic anaerobic digester on volatile solids at 1.6–3.2.
The value of the number is that it collapses two variables — flow and strength — into one, so it exposes what a hydraulic detention time hides. Two plants with identical two-hour detention times are not comparable if one treats 150 mg/L domestic sewage and the other 900 mg/L cannery waste. Push the loading past design and the symptoms are textbook: dissolved oxygen collapses, filamentous organisms take over because they out-compete floc-formers at low DO, sludge stops settling, and solids leave over the clarifier weir. The classic mistake is loading a digester by volume of sludge pumped rather than by pounds of volatile solids — a thin, watery feed can double the pumping and still starve the bugs, while a good thickener can overload the same digester at half the flow.
Volumetric Organic Loading Rate formula
- = Volumetric loading (kg BOD/m³·d) (kg BOD/(m³·d))
- = Flow rate (L/min)
- = Influent BOD concentration (%)
- = Reactor volume (L)
Missing one of these? Work it out first, then come back
- Flow rate — Hydraulic Detention Time, Surface Overflow Rate
- Influent BOD concentration — Food-to-Microorganism (F/M) Ratio, BOD Removal Efficiency
- Reactor volume — Space Time and Space Velocity, CSTR Design Equation (First Order)