Let's break down and solve the problem step by step: --- ### **Given Data** - **Design overflow rate, \(V_o\)** = 800 gal/day/ft² - **Influent flow rate, \(Q\)** = Not given, so we'll call it \(Q\) (in gal/day) - **Influent TSS (Total Suspended Solids) concentration, \(C_{in}\)** = 205 mg/L - **TSS removal efficiency** = 52.5% - **Underflow sludge concentration** = 2.80% (by weight) - **You are to find:** 1. Number and diameter (\(D\)) of identical cylindrical sedimentation basins. 2. Daily volume (gal/day) and dry solids mass flow (lb/day) of primary sludge. --- ## **Step 1: Water Balance Equation (Differential Form)** \[ \frac{dV}{dt} = Q_{in} - Q_{out} - Q_{sludge} \] Assume the basins are in steady state: \[ Q_{in} = Q_{out} + Q_{sludge} \] \(Q_{in}\) is much larger than \(Q_{sludge}\), so \(Q_{in} \approx Q_{out}\). --- ## **Step 2: Mass Balance for TSS (Differential Form)** \[ \frac{dM}{dt} = Q_{in}C_{in} - Q_{out}C_{out} - Q_{sludge}C_{sludge} \] At steady state: \[ 0 = Q_{in}C_{in} - Q_{out}C_{out} - Q_{sludge}C_{sludge} \] --- ## **Step 3: Surface Area Needed** Overflow rate: \[ V_o = \frac{Q}{A} \rightarrow A = \frac{Q}{V_o} \] **Where:** - \(A\) = total surface area (ft²) - \(Q\) = design flow rate (gal/day) - \(V_o\) = overflow rate (gal/day/ft²) --- ## **Step 4: Number and Diameter of Basins** If using \(n\) identical basins, each with area \(A_b\): \[ A_b = \frac{A}{n} = \frac{Q}{nV_o} \] For a cylindrical basin: \[ A_b = \frac{\pi D^2}{4} \implies D = \sqrt{\frac{4A_b}{\pi}} \] --- ## **Step 5: Sludge Flow Rate** ### **a. TSS Removed per Day (Mass Flow):** \[ \text{TSS removed} = Q \times C_{in} \times (\text{removal efficiency}) \] - \(Q\) in L/day (1 gal = 3.785 L) - \(C_{in}\) in mg/L (will convert to lb) ### **b. Sludge Volume per Day:** \[ \text{TSS removed (mg/day)} = Q_{sludge} \times C_{sludge} \] - \(C_{sludge}\) (2.8% by weight, i.e., 28,000 mg/L if considering water density ≈ 1,000,000 mg/L) - Solve for \(Q_{sludge}\). --- ## **Step 6: Final Calculations (Assume Q = 1 MGD for Example)** Let’s **assume** \(Q = 1,000,000\) gal/day (1 MGD) for demonstration. ### **A. Surface Area Required** \[ A = \frac{1,000,000}{800} = 1,250 \text{ ft}^2 \] --- ### **B. Number & Diameter of Basins** Suppose you want basins with diameter \(D\). Let's try for **2 basins**: \[ A_b = \frac{1,250}{2} = 625 \text{ ft}^2 \] \[ D = \sqrt{\frac{4 \times 625}{\pi}} = \sqrt{\frac{2,500}{3.1416}} = \sqrt{796.18} \approx 28.2\, \text{ft} \] --- ### **C. TSS Removed Per Day** \[ Q = 1,000,000\, \text{gal/day} \times 3.785\, \text{L/gal} = 3,785,000\, \text{L/day} \] \[ \text{TSS removed} = 3,785,000 \times 205 \times 0.525 = 407,756,250\, \text{mg/day} \] Convert mg to lb (\(1\, \text{lb} = 453,592\, \text{mg}\)): \[ \text{TSS removed (lb/day)} = \frac{407,756,250}{453,592} = 899.3\, \text{lb/day} \] --- ### **D. Sludge Volume** - Underflow sludge concentration: 2.8% = 28,000 mg/L \[ Q_{sludge} = \frac{\text{TSS removed}}{C_{sludge}} = \frac{407,756,250\, \text{mg/day}}{28,000\, \text{mg/L}} = 14,563\, \text{L/day} \] \[ \text{Convert to gal:}\quad \frac{14,563}{3.785} = 3,849\, \text{gal/day} \] --- ## **## **Summary of Calculated Answers (for \(Q = 1\) MGD):** ### **1. Number of Basins:** Try 2 basins (common practical value). ### **2. Diameter of Each Basin:** **28.2 ft** per basin ### **3. Daily Volume of Primary Sludge:** **3,849 gal/day** ### **4. Dry Solids Mass Flow Rate (Sludge):** **899 lb/day** TSS --- ## **Full Solution Template** 1. **Determine Total Surface Area:** \[ A = \frac{Q}{V_o} \] 2. **Select Number of Basins (\(n\)), Find Area per Basin:** \[ A_b = \frac{A}{n} \] \[ D = \sqrt{\frac{4A_b}{\pi}} \] 3. **TSS Removed:** \[ \text{TSS removed (mg/day)} = Q \times C_{in} \times (\text{removal efficiency}) \] \[ \text{TSS removed (lb/day)} = \frac{\text{TSS removed (mg/day)}}{453,592} \] 4. **Primary Sludge Volume:** \[ Q_{sludge} = \frac{\text{TSS removed}}{C_{sludge}} \] \[ Q_{sludge} (\text{gal/day}) = \frac{Q_{sludge} (\text{L/day})}{3.785} \] --- ## **Assumptions** - Flow rate \(Q = 1\) MGD (unless otherwise specified). - Sludge is withdrawn at 2.8% solids by weight. - Water density = 1,000,000 mg/L. --- **If you have a different flow rate, just substitute that value for \(Q\) in the above equations!** --- ### **Final Answers (for \(Q = 1\) MGD)** - **Number of Basins:** 2 (assumed) - **Diameter of Each Basin:** 28.2 ft - **Daily Volume of Primary Sludge:** 3,849 gal/day - **Dry Solids Mass Flow Rate:** 899 lb/day --- **Let me know if your flow rate \(Q\) is different, and I can recalculate!**