Two-stage trickling filter, 7665 m³/day

Data: $Q=7665$ m³/day; raw BOD$_5=190$ mg/l; 35% removed in the primary sedimentation tank, so BOD to the filters $=190\times0.65=123.5$ mg/l; final BOD $=25$ mg/l; depth $=1.85$ m; $R=0.5$ for both filters.

Data and approach

Two-stage filter (NRC equations, $W$ in kg/day, $V$ in m³):

$$E_1=\frac{100}{1+0.4432\sqrt{\dfrac{W_1}{V_1F}}},\qquad E_2=\frac{100}{1+\dfrac{0.4432}{1-E_1}\sqrt{\dfrac{W_2}{V_2F}}},\qquad F=\frac{1+R}{(1+0.1R)^2}$$

where $W_2=W_1(1-E_1)$ is the BOD load leaving the first stage ($E_1$ as a fraction in the second equation). Flow $Q=7665$ m³/day.

Step 1: Loads and efficiencies

$$\begin{aligned} BOD\ \text{applied} &= 123.5\ \text{mg/l},\quad W_1=7665\times 123.5\times10^{-3}=946.6\ \text{kg/day}\\ \text{Overall filter efficiency } E &= \frac{123.5-25}{123.5}=0.7976 \end{aligned}$$

No ratio is given, so both stages are taken to have equal efficiency $E_1=E_2=E$: $1-(1-E)^2=0.7976\Rightarrow E=1-\sqrt{1-0.7976}=0.5501$, i.e. $E_1=E_2=55.01\%$.

Step 2: Recirculation factor (same for both stages, $R=0.5$)

$$F=\frac{1+0.5}{(1+0.1\times 0.5)^2}=1.3605$$

Step 3: First stage

$$\sqrt{\frac{W_1}{V_1F}}=\frac{100/E_1-1}{0.4432}=1.8455\ \Rightarrow\ V_1=\frac{946.6}{1.3605\times 1.8455^2}=204.3\ \text{m}^3$$

Step 4: Second stage

$$\begin{aligned} W_2 &= W_1(1-E_1)=946.6\times(1-0.5501)=425.9\ \text{kg/day}\\ \sqrt{\frac{W_2}{V_2F}} &= \frac{(100/E_2-1)(1-E_1)}{0.4432}=0.8303\ \Rightarrow\ V_2=\frac{425.9}{1.3605\times 0.8303^2}=454.1\ \text{m}^3 \end{aligned}$$

Step 5: Diameters (depth $1.85$ m)

$$D_1=\sqrt{\frac{4V_1}{\pi\,d}}=\sqrt{\frac{4\times 204.3}{\pi\times 1.85}}=11.9\ \text{m},\qquad D_2=\sqrt{\frac{4\times 454.1}{\pi\times 1.85}}=17.7\ \text{m}$$

Answer: first-stage filter $D_1\approx 11.9$ m (volume 204.3 m³); second-stage filter $D_2\approx 17.7$ m (volume 454.1 m³).