1. **Problem statement:** A city needs a primary settling tank for wastewater treatment with a maximum flow of $0.570\ \text{m}^3/\text{s}$ and an overflow rate of $60.0\ \text{m/day}$. We need to find: a) The required surface area of the tank. b) The theoretical detention time if the tank depth is $3.0\ \text{m}$. c) The kilograms of BOD5 removed daily given an average flow of $0.050\ \text{m}^3/\text{s}$, BOD5 concentration of $345\ \text{mg/L}$, and 30% removal. 2. **Formulas and important notes:** - Overflow rate $O_R = \frac{Q}{A}$ where $Q$ is flow rate and $A$ is surface area. - Detention time $t = \frac{V}{Q}$ where $V$ is volume and $Q$ is flow rate. - Volume $V = A \times \text{depth}$. - To convert flow rate from $\text{m}^3/\text{s}$ to $\text{m}^3/\text{day}$, multiply by $86400$ seconds/day. - BOD5 removal mass $= \text{flow} \times \text{concentration} \times \text{removal fraction}$. - Convert mg/L to kg/m³ by noting $1\ \text{mg/L} = 0.001\ \text{kg/m}^3$. --- ### a) Required surface area Given: $$Q = 0.570\ \text{m}^3/\text{s}$$ $$O_R = 60.0\ \text{m/day}$$ Convert $Q$ to $\text{m}^3/\text{day}$: $$Q_{day} = 0.570 \times 86400 = 49248\ \text{m}^3/\text{day}$$ Use overflow rate formula: $$A = \frac{Q_{day}}{O_R} = \frac{49248}{60.0} = 820.8\ \text{m}^2$$ ### b) Theoretical detention time Tank depth $d = 3.0\ \text{m}$ Calculate volume: $$V = A \times d = 820.8 \times 3.0 = 2462.4\ \text{m}^3$$ Convert $Q$ back to $\text{m}^3/\text{s}$ for detention time: $$t = \frac{V}{Q} = \frac{2462.4}{0.570} = 4320\ \text{seconds}$$ Convert seconds to hours: $$t = \frac{4320}{3600} = 1.2\ \text{hours}$$ ### c) Kilograms of BOD5 removed daily Given: $$Q_{avg} = 0.050\ \text{m}^3/\text{s}$$ $$\text{BOD5 concentration} = 345\ \text{mg/L} = 0.345\ \text{kg/m}^3$$ $$\text{Removal fraction} = 0.30$$ Convert flow to $\text{m}^3/\text{day}$: $$Q_{day} = 0.050 \times 86400 = 4320\ \text{m}^3/\text{day}$$ Calculate mass removed: $$\text{Mass removed} = Q_{day} \times \text{concentration} \times \text{removal fraction} = 4320 \times 0.345 \times 0.30 = 446.88\ \text{kg/day}$$ **Final answers:** - a) Surface area $= 820.8\ \text{m}^2$ - b) Detention time $= 1.2\ \text{hours}$ - c) BOD5 removed $= 446.88\ \text{kg/day}$