Recirculation ratio, filter volume 350 m³

Data and approach

NRC equation for a high-rate filter: $E=\dfrac{100}{1+0.4432\sqrt{W/(VF)}}$, with $F=\dfrac{1+R}{(1+0.1R)^2}$ ($W$ in kg/day, $V$ in m³).

Step 1: BOD load and required efficiency

$$\begin{aligned} Q &= 5000\ \text{m}^3/\text{day}\\ BOD\ \text{applied} &= 180\times(1-0.33) = 120.6\ \text{mg/l}\\ W &= 5000\times 120.6\times10^{-3} = 603.0\ \text{kg/day}\\ E &= \frac{120.6-35}{120.6}\times100 = 70.98\% \end{aligned}$$

Step 2: Required recirculation factor

Let $s=\sqrt{W/(VF)}$:

$$s=\frac{100/E-1}{0.4432}=\frac{100/70.98-1}{0.4432}=0.9226\ \Rightarrow\ F=\frac{W}{V\,s^2}=\frac{603.0}{350\times 0.9226^2}=2.024$$

Step 3: Recirculation ratio

$$F=\frac{1+R}{(1+0.1R)^2}\ \Rightarrow\ 0.01F\,R^2+(0.2F-1)R+(F-1)=0$$ $$0.0202R^2 - 0.5952R + 1.0242 = 0\ \Rightarrow\ R=1.84$$

(The smaller root is taken; the other root, about 27.6, is not practical.)

Check: $F(R=1.84)=2.024$ and $E=70.98\%$, equal to the required 70.98%.

Answer: recirculation ratio required $R \approx 1.84$ (about 1.8).