HR trickling filter, BOD 280 mg/l

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

$$\begin{aligned} Q &= 5000\ \text{m}^3/\text{day}\\ \text{BOD after primary} &= 280\times(1-0.25) = 210.0\ \text{mg/l}\\ W &= Q\times BOD = 5000\times 210.0\times 10^{-3} = 1050.0\ \text{kg/day}\\ E &= \frac{210.0-25}{210.0}\times 100 = 88.10\% \end{aligned}$$

Step 2: Recirculation factor

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

Step 3: Volume

$$88.10 = \frac{100}{1+0.4432\sqrt{W/(VF)}}\Rightarrow \sqrt{\frac{W}{VF}} = 0.3049 \Rightarrow VF = \frac{1050.0}{0.3049^2} = 11294.1$$ $$V = \frac{VF}{F} = \frac{11294.1}{1.890} = 5974.6\ \text{m}^3$$

Step 4: Diameter

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

$$A = \frac{V}{d} = 2987.3\ \text{m}^2,\qquad D = \sqrt{\frac{4A}{\pi}} = 61.67\ \text{m (adopt 61.7 m)}$$

Step 5: Checks

$$\begin{aligned} \text{Organic loading} &= \frac{W}{V} = 0.18\ \text{kg/m}^3\text{/day}\ (\text{high rate: 0.32-1.0})\\ \text{Hydraulic loading} &= \frac{Q(1+R)}{A} = 4.2\ \text{m}^3/\text{m}^2\text{/day}\ (\text{high rate: 10-40}) \end{aligned}$$

Both loadings are below the usual high-rate range. This is because a single stage must give the high efficiency of 88%, so the NRC equation requires a large volume. A diameter of 61.7 m is also more than the usual maximum of about 60 m for a rotary distributor. In practice, provide two filters in parallel, each of area $2987.3/2 = 1493.7$ m² and diameter $\sqrt{4\times1493.7/\pi} = 43.6$ m (depth 2.0 m), or use a two-stage filter.

Answer: single-stage high-rate filter of volume about 5974.6 m³, depth 2.0 m, diameter 61.7 m (in practice 2 filters of 43.6 m diameter each).