HR trickling filter, 5 MLD, RR 1.5

Data and approach

High-rate filter design uses the NRC (National Research Council) equation; $W$ is in kg/day of BOD applied to the filter, $V$ in m³:

$$E = \frac{100}{1+0.4432\sqrt{\dfrac{W}{VF}}},\qquad F=\frac{1+R}{(1+0.1R)^2}$$

Step 1: BOD applied to the filter and required efficiency

$$\begin{aligned} Q &= 5000\ \text{m}^3/\text{day}\\ \text{BOD after primary} &= 250\times(1-0.3) = 175.0\ \text{mg/l}\\ W &= Q\times BOD = 5000\times 175.0\times 10^{-3} = 875.0\ \text{kg/day}\\ E &= \frac{175.0-30}{175.0}\times 100 = 82.86\% \end{aligned}$$

Step 2: Recirculation factor

$$F = \frac{1+1.5}{(1+0.1\times 1.5)^2} = 1.890$$

Step 3: Volume

$$82.86 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow \sqrt{\frac{W}{VF}} = 0.4668 \Rightarrow VF = \frac{875.0}{0.4668^2} = 4015.1$$ $$V = \frac{VF}{F} = \frac{4015.1}{1.890} = 2124.0\ \text{m}^3$$

Step 4: Diameter

Adopt depth $d = 2.0$ m (high-rate filters are 1.8-3 m deep).

$$A = \frac{V}{d} = 1062.0\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 36.77\ \text{m (adopt 36.8 m)}$$

Step 5: Checks

$$\begin{aligned} \text{Organic loading} &= \frac{W}{V} = 0.41\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.8-1.6})\\ \text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 11.8\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40}) \end{aligned}$$

The organic loading comes out below the usual high-rate range because a single stage has to give a high efficiency (about 83%); the hydraulic loading is within the high-rate range.

Answer: single-stage high-rate filter of volume about 2124.0 m³, depth 2.0 m, diameter 36.8 m.