Two-stage trickling filter diameters, 4 MLD

Data: $Q=4$ MLD $=4000$ m³/day; BOD fed to the first filter $=200$ mg/l; final BOD $\le 30$ mg/l; $E_1=2E_2$; depth $=1.2$ m; $R=2$ 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 $Q=4000$ m³/day.

Step 1: Loads and efficiencies

$$\begin{aligned} BOD\ \text{applied} &= 200\ \text{mg/l},\quad W_1=4000\times 200\times10^{-3}=800.0\ \text{kg/day}\\ \text{Overall filter efficiency } E &= \frac{200-30}{200}=0.8500 \end{aligned}$$

Let $E_2=x$, so $E_1=2x$ (as given). Overall efficiency $E=E_1+E_2(1-E_1)$, i.e. $1-(1-2x)(1-x)=0.8500$.

$$2x^2-3x+0.8500=0\ \Rightarrow\ x=0.3792$$

So $E_2=37.92\%$ and $E_1=75.84\%$.

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

$$F=\frac{1+2}{(1+0.1\times 2)^2}=2.0833$$

Step 3: First stage

$$\sqrt{\frac{W_1}{V_1F}}=\frac{100/E_1-1}{0.4432}=0.7189\ \Rightarrow\ V_1=\frac{800.0}{2.0833\times 0.7189^2}=743.1\ \text{m}^3$$

Step 4: Second stage

$$\begin{aligned} W_2 &= W_1(1-E_1)=800.0\times(1-0.7584)=193.3\ \text{kg/day}\\ \sqrt{\frac{W_2}{V_2F}} &= \frac{(100/E_2-1)(1-E_1)}{0.4432}=0.8926\ \Rightarrow\ V_2=\frac{193.3}{2.0833\times 0.8926^2}=116.5\ \text{m}^3 \end{aligned}$$

Step 5: Diameters (depth $1.2$ m)

$$D_1=\sqrt{\frac{4V_1}{\pi\,d}}=\sqrt{\frac{4\times 743.1}{\pi\times 1.2}}=28.1\ \text{m},\qquad D_2=\sqrt{\frac{4\times 116.5}{\pi\times 1.2}}=11.1\ \text{m}$$

Answer: first-stage filter $D_1\approx 28.1$ m (volume 743.1 m³); second-stage filter $D_2\approx 11.1$ m (volume 116.5 m³).