Standard rate trickling filter, 1.2 MLD, NRC

Data

Flow $Q=1.2\times 10^6$ l/day $=1200$ m³/day; influent BOD (after PST) $=200$ mg/l; surface (hydraulic) loading $=1200$ l/m²/day; organic loading $=250$ g/m³/day.

Step 1: BOD load

$$W = Q\times BOD = 1200\times 200\times 10^{-3} = 240.0\ \text{kg/day}$$

Step 2: Volume (from organic loading)

$$V = \frac{W}{\text{organic loading}} = \frac{240.0}{0.25} = 960.0\ \text{m}^3$$

Step 3: Surface area (from hydraulic loading)

$$A = \frac{Q}{\text{surface loading}} = \frac{1.2\times 10^6}{1200} = 1000.0\ \text{m}^2$$

Step 4: Depth

$$\text{Depth} = \frac{V}{A} = \frac{960.0}{1000.0} = 0.96\ \text{m}$$

(Diameter of the circular filter $D=\sqrt{4A/\pi}=35.7$ m. The calculated depth is smaller than the usual 1.8-3 m of a standard-rate filter; if a depth of 1.8 m is wanted, the area is reduced to 533 m² and the hydraulic loading rises accordingly.)

Step 5: Efficiency by the NRC equation (standard rate, no recirculation, $F=1$)

$$E=\frac{100}{1+0.4432\sqrt{W/V}} = \frac{100}{1+0.4432\sqrt{240.0/960.0}} = 81.86\%$$

Effluent BOD $=200\,(1-81.86/100) = 36.3$ mg/l.

Answer: volume $=960.0$ m³, depth $=0.96$ m (area 1000.0 m²); NRC efficiency $=81.86\%$.