Two-stage trickling filter diameters, 5 MLD

Data: $Q=5$ MLD $=5000$ m³/day; BOD fed to the first filter $=180$ mg/l; final BOD $\le30$ 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 $Q=5000$ m³/day.

Step 1: Loads and efficiencies

$$\begin{aligned} BOD\ \text{applied} &= 180\ \text{mg/l},\quad W_1=5000\times 180\times10^{-3}=900.0\ \text{kg/day}\\ \text{Overall filter efficiency } E &= \frac{180-30}{180}=0.8333 \end{aligned}$$

Let $E_2=x$, so $E_1=2x$ (as given). Overall efficiency $E=E_1+E_2(1-E_1)$, i.e. $1-(1-2x)(1-x)=0.8333$.

$$2x^2-3x+0.8333=0\ \Rightarrow\ x=0.3681$$

So $E_2=36.81\%$ and $E_1=73.62\%$.

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

$$\sqrt{\frac{W_1}{V_1F}}=\frac{100/E_1-1}{0.4432}=0.8083\ \Rightarrow\ V_1=\frac{900.0}{2.0833\times 0.8083^2}=661.1\ \text{m}^3$$

Step 4: Second stage

$$\begin{aligned} W_2 &= W_1(1-E_1)=900.0\times(1-0.7362)=237.4\ \text{kg/day}\\ \sqrt{\frac{W_2}{V_2F}} &= \frac{(100/E_2-1)(1-E_1)}{0.4432}=1.0216\ \Rightarrow\ V_2=\frac{237.4}{2.0833\times 1.0216^2}=109.2\ \text{m}^3 \end{aligned}$$

Step 5: Diameters (depth $1.2$ m)

$$D_1=\sqrt{\frac{4V_1}{\pi\,d}}=\sqrt{\frac{4\times 661.1}{\pi\times 1.2}}=26.5\ \text{m},\qquad D_2=\sqrt{\frac{4\times 109.2}{\pi\times 1.2}}=10.8\ \text{m}$$

Answer: first-stage filter $D_1\approx 26.5$ m (volume 661.1 m³); second-stage filter $D_2\approx 10.8$ m (volume 109.2 m³).