Standard rate trickling filter, 4 MLD, NRC

Data: $Q=4$ MLD $=4000$ m³/day, settled sewage BOD $=180$ mg/l, hydraulic (surface) loading $=2000$ l/m²/day, organic loading $=150$ g/m³/day.

Step 1: BOD load

$$W=Q\times BOD=4000\times 180\times10^{-3}=720.0\ \text{kg/day}$$

Step 2: Volume (organic loading 150 g/m³/day)

$$V=\frac{W}{\text{organic loading}}=\frac{720.0}{0.150}=4800.0\ \text{m}^3$$

Step 3: Area (hydraulic loading 2000 l/m²/day) and depth

$$A=\frac{Q}{\text{surface loading}}=\frac{4000\times1000}{2000}=2000.0\ \text{m}^2,\qquad \text{Depth}=\frac{V}{A}=\frac{4800.0}{2000.0}=2.40\ \text{m}$$

Diameter of the circular filter: $D=\sqrt{4A/\pi}=50.5$ m.

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

$$E=\frac{100}{1+0.4432\sqrt{W/V}}=\frac{100}{1+0.4432\sqrt{720.0/4800.0}}=85.35\%$$

Effluent BOD $=180\,(1-85.35/100)=26.4$ mg/l.

Answer: volume $=4800.0$ m³; depth $=2.40$ m; area $=2000.0$ m²; NRC efficiency $=85.35\%$.