1. **State the problem:** We need to find the diameter of a single-stage rock media filter that reduces BOD5 from 125 mg/L to 25 mg/L with a flow rate of 0.14 m³/s, recirculation ratio of 12, filter depth of 1.83 m, and temperature 20°C. 2. **Relevant formula:** The NRC equation for BOD removal in a filter is given by: $$C = C_0 \exp\left(-\frac{k n z}{1 + r}\right)$$ where: - $C_0$ = influent BOD5 concentration (125 mg/L) - $C$ = effluent BOD5 concentration (25 mg/L) - $k$ = reaction rate constant at 20°C - $n$ = porosity of the filter media (assumed typical value 0.4) - $z$ = filter depth (1.83 m) - $r$ = recirculation ratio (12) 3. **Determine $k$:** At 20°C, typical $k$ for BOD removal in rock media filters is about 0.23 1/day. Convert to 1/s: $$k = \frac{0.23}{24 \times 3600} = 2.66 \times 10^{-6} \text{ s}^{-1}$$ 4. **Calculate the exponential term:** $$\frac{k n z}{1 + r} = \frac{2.66 \times 10^{-6} \times 0.4 \times 1.83}{1 + 12} = \frac{1.95 \times 10^{-6}}{13} = 1.5 \times 10^{-7}$$ 5. **Calculate expected effluent concentration:** $$C = 125 \times \exp(-1.5 \times 10^{-7}) \approx 125 \times (1 - 1.5 \times 10^{-7}) = 125 \text{ mg/L}$$ This is not close to 25 mg/L, so the filter depth or flow assumptions need adjustment. 6. **Calculate required filter volume $V$ using flow and reaction rate:** The BOD removal rate is: $$Q (C_0 - C) = k V C_{avg}$$ Assuming $C_{avg} \approx \frac{C_0 + C}{2} = \frac{125 + 25}{2} = 75 \text{ mg/L}$ Rearranged for $V$: $$V = \frac{Q (C_0 - C)}{k C_{avg}}$$ Convert concentrations to kg/m³: $$125 \text{ mg/L} = 0.125 \text{ kg/m}^3, \quad 25 \text{ mg/L} = 0.025 \text{ kg/m}^3, \quad 75 \text{ mg/L} = 0.075 \text{ kg/m}^3$$ Calculate $V$: $$V = \frac{0.14 (0.125 - 0.025)}{2.66 \times 10^{-6} \times 0.075} = \frac{0.14 \times 0.1}{2.0 \times 10^{-7}} = \frac{0.014}{2.0 \times 10^{-7}} = 70000 \text{ m}^3$$ 7. **Calculate filter cross-sectional area $A$:** $$A = \frac{V}{z} = \frac{70000}{1.83} = 38251 \text{ m}^2$$ 8. **Calculate diameter $D$ of the filter:** $$A = \pi \frac{D^2}{4} \Rightarrow D = 2 \sqrt{\frac{A}{\pi}} = 2 \sqrt{\frac{38251}{3.1416}} = 2 \times 110.3 = 220.6 \text{ m}$$ **Final answer:** The diameter of the rock media filter is approximately **221 meters**. This large diameter suggests either the reaction rate constant or assumptions may need review for practical design.