Standard rate filter, 3 MLD, 175 mg/l, NRC

Data: $Q=3$ MLD $=3000$ m³/day, BOD$_5$ $=175$ mg/l, organic loading $=150$ g/m³/day (the "surface loading" given in g/m³/day is an organic loading). The recirculation ratio 1 : 2 is read as $R=Q_R/Q=0.5$. The hydraulic loading is not given, so a depth is assumed.

Step 1: BOD load

$$W=Q\times BOD=3000\times 175\times10^{-3}=525.0\ \text{kg/day}$$

Step 2: Volume (organic loading 150 g/m³/day)

$$V=\frac{W}{\text{organic loading}}=\frac{525.0}{0.150}=3500.0\ \text{m}^3$$

Step 3: Depth and area

No hydraulic loading is given, so adopt a depth of 2.00 m (standard-rate filters are 1.8-3 m deep):

$$A=\frac{V}{2.00}=1750.0\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=47.2\ \text{m},\qquad \text{hydraulic loading}=\frac{3000}{1750.0}=1.71\ \text{m}^3/\text{m}^2\text{/day}$$

Step 4: Efficiency by the NRC equation

With the given recirculation ratio $R=0.5$: $F=\dfrac{1+R}{(1+0.1R)^2}=1.361$

$$E=\frac{100}{1+0.4432\sqrt{W/(VF)}}=\frac{100}{1+0.4432\sqrt{525.0/(3500.0\times 1.361)}}=87.17\%$$

Effluent BOD $=175\,(1-87.17/100)=22.4$ mg/l.

(If the ratio is read as $R=2$, then $F=2.083$ and $E\approx 89.4\%$.)

Answer: volume $=3500.0$ m³; depth $=2.00$ m; area $=1750.0$ m²; NRC efficiency $=87.17\%$.