# Step-by-Step Solution ## **Given Data- **Initial solids concentration (\(C_\))**: 240 mg/L - **Underflow concentration (\(C_u\))**: 12,000 mg/L - **Inflow rate (\(Q\))**: 1,416 m³/day - **Cylinder diameter**: .4 m (radius = .2 m) - **From the graph**: Need to determine **critical settling velocity**. --- ## **Step 1: Find the Height Corresponding to Critical Concentration** At the critical concentration, the solids concentration increases from \(C_\) to \(C_u\). Mass balance gives: \[ C_ \times h_ = C_u \times h_u \] where: - \(h_\) = initial interface height (from the graph, at time = ) - \(h_u\) = interface height at critical concentration From the graph: - \(h_ \approx .38\, \text{m}\) - \(C_ = 240\, \text{mg/L}\), \(C_u = 12,000\, \text{mg/L}\) \[ h_u = \frac{C_}{C_u} \times h_ = \frac{240}{12000} \times .38 = .076\, \text{m} \] --- ## **Step 2: Find the Corresponding Time and Slope (Settling Velocity)** From the graph, find the time at which the interface reaches **.076 m**. - At \(h = .076\) m, **time ≈ 53 min** (from the graph, interpolate if necessary). Now, calculate the **settling velocity (\(v_s\))**: \[ v_s = \frac{h_ - h_u}{t} \] where \(t\) is the time to reach \(h_u\): \[ v_s = \frac{.38\, \text{m} - .076\, \text{m}}{53\, \text{min}} = \frac{.304}{53} = .00574\, \text{m/min} \] Convert to m/day: \[ .00574\, \text{m/min} \times 144\, \text{min/day} = 8.26\, \text{m/day} \] --- ## **Step 3: Calculate Surface Area (\(A\)) Required** \[ A = \frac{Q}{v_s} \] Where: - \(Q = 1,416\, \text{m}^3/\text{day}\) - \(v_s = 8.26\, \text{m/day}\) \[ A = \frac{1416}{8.26} = 171.5\, \text{m}^2 \] --- ## **Final Answer** ### **The required surface area is \( \boxed{171.5\ \text{m}^2} \).** --- #### **Summary Table** | Parameter | Value | |--------------------------|---------------------| | Initial Height, \(h_\) | .38 m | | Underflow Height, \(h_u\)| .076 m | | Time to \(h_u\) | 53 min | | Settling Velocity, \(v_s\)| 8.26 m/day | | Inflow Rate, \(Q\) | 1,416 m³/day | | Surface Area, \(A\) | **171.5 m²** |