OpenSpecEngineering

Works out the loading figures for the aeration basin of an activated sludge plant: the BOD it receives each day, the load per cubic meter, the hydraulic retention time and the food-to-microorganism (F/M) ratio. It also gives the BOD removal. Plant operators and process engineers use these to check a design or a running plant against typical ranges.

Equations

L = Q·S₀/1000     L_v = L/V

HRT = 24·V/Q     F/M = Q·S₀/(V·X)

E = 100·(1 − S/S₀)

Q is the flow, in m³/d. S₀ and S are the influent and effluent BOD, in mg/L (the same as g/m³). V is the aeration basin volume, in m³, and X the mixed liquor volatile suspended solids (MLVSS), in mg/L. L is the BOD load, in kg/d, and L_v the volumetric loading. HRT is in hours, F/M in d⁻¹, and E is the removal in percent [1], [2].

Inputs

• The effluent BOD can come from the US secondary treatment limits: 30 mg/L as a 30-day average or 45 mg/L as a 7-day average [3]. The same rule also needs 85 % removal, so the page tells you when a weak influent needs a lower effluent BOD.
• Enter 0 for the effluent BOD to skip the removal, and 0 for MLVSS to skip F/M.

For conventional activated sludge, typical values are an HRT of about 4 to 8 h, an F/M of 0.2 to 0.4 d⁻¹ and a volumetric loading of 0.3 to 0.6 kg BOD/m³·d [1], [2].

Limits and common mistakes

• Don’t enter MLSS for MLVSS. The volatile part is often only 70 to 80 % of the total, so F/M comes out too low.
• Use the aeration basin volume alone, without the clarifiers.
• HRT is based on the influent flow. It ignores the return sludge, so the actual time in the basin is shorter.
• Some texts work F/M on the BOD removed rather than the BOD applied. This page uses the BOD applied.
• Use the flow and BOD for the case you are checking: annual average, maximum month or peak day.

Worked example

A town’s activated sludge plant treats 6,000 m³/d with an influent BOD of 200 mg/L. Its aeration basin holds 2,000 m³ and runs at 2,000 mg/L MLVSS. Is the basin loaded in the normal range, and can it meet the 30-day limit?

1. Enter Flow Rate 6,000 m³/d and Influent BOD 200 mg/L.
2. Choose Effluent BOD Basis US secondary treatment, 30-day average. The effluent BOD fills in as 30 mg/L.
3. Enter Aeration Basin Volume 2,000 m³ and Mixed Liquor VSS 2,000 mg/L, and press Solve. The Solution shows: BOD mass load: 1,200 kg/d
   Volumetric (organic) loading: 0.6 kg BOD/m³·d
   Hydraulic retention time (HRT): 8 h
   Food-to-microorganism ratio F/M: 0.3 d⁻¹
   BOD removal efficiency: 85 % with a note that the 30 mg/L limit also gives the 85 % removal required.

Check by hand: 6,000 × 200/1,000 = 1,200 kg/d, and 1,200/2,000 = 0.6 kg/m³·d. The basin holds 2,000/6,000 of a day, which is 8 h. F/M = 1,200/(2,000 × 2.0) = 0.30 d⁻¹. All three sit within the typical ranges, the volumetric loading at its top end.