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Standard rate trickling filter, 3 MLD, 300 mg/l

Short answer

Volume = 3000 m³, depth = 3.00 m, area = 1000 m²; NRC efficiency = 80.47%.

The effluent from PST is applied to a standard rate trickling filter at the rate of 3 million liters/day with BOD5 300 mg/l. Calculate the depth and volume of filter considering the surface loading of 3000 liters/m²day and organic loading of 300 gm/m³day.

Data: Q=3×106Q=3\times10^6 l/day =3000=3000 m³/day, settled-sewage BOD5_5 =300=300 mg/l, surface loading =3000=3000 l/m²/day, organic loading =300=300 g/m³/day. (The NRC efficiency is also given below for completeness.)

Step 1: BOD load

W=Q×BOD=3000×300×10−3=900.0 kg/dayW=Q\times BOD=3000\times 300\times10^{-3}=900.0\ \text{kg/day}

Step 2: Volume (organic loading 300 g/m³/day)

V=Worganic loading=900.00.300=3000.0 m3V=\frac{W}{\text{organic loading}}=\frac{900.0}{0.300}=3000.0\ \text{m}^3

Step 3: Area (hydraulic loading 3000 l/m²/day) and depth

A=Qsurface loading=3000×10003000=1000.0 m2,Depth=VA=3000.01000.0=3.00 mA=\frac{Q}{\text{surface loading}}=\frac{3000\times1000}{3000}=1000.0\ \text{m}^2,\qquad \text{Depth}=\frac{V}{A}=\frac{3000.0}{1000.0}=3.00\ \text{m}

Diameter of the circular filter: D=4A/π=35.7D=\sqrt{4A/\pi}=35.7 m.

Step 4: Efficiency by the NRC equation (no recirculation, F=1F=1)

E=1001+0.4432W/V=1001+0.4432900.0/3000.0=80.47%E=\frac{100}{1+0.4432\sqrt{W/V}}=\frac{100}{1+0.4432\sqrt{900.0/3000.0}}=80.47\%

Effluent BOD =300 (1−80.47/100)=58.6=300\,(1-80.47/100)=58.6 mg/l.

Answer: volume =3000.0=3000.0 m³; depth =3.00=3.00 m; area =1000.0=1000.0 m²; NRC efficiency =80.47%=80.47\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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