Let's break down the problem step by step: ### **Given Data** - **Feed rate:** \( 1000 \) ton/day = \( 1,000,000 \) kg/day - **Feed concentration (\( C_f \)):** \( 200 \) kg/m³ - **Underflow concentration (\( C_u \)):** \( 1500 \) kg/m³ - **Solid density (\( \rho_s \)):** \( 4440 \) kg/m³ #### **Batch Sedimentation Data** | Time (hr) | 0 | 0.1 | 0.25 | 0.5 | 1.0 | 2.0 | 4.0 | |-----------|-----|-----|------|-----|-----|-----|-----| | Height (m)| 0.91| 0.61| 0.43 | 0.27| 0.15| 0.08| 0.03| --- ## **Step 1: Calculate the Volumetric Flow Rates** ### **Feed Volumetric Flow Rate** \[ Q_f = \frac{\text{Feed rate}}{\text{Feed concentration}} = \frac{1,000,000 \ \mathrm{kg/day}}{200 \ \mathrm{kg/m^3}} = 5000 \ \mathrm{m^3/day} \] \[ Q_f = \frac{5000}{24 \times 3600} = 0.05787 \ \mathrm{m^3/s} \] ### **Underflow Volumetric Flow Rate** \[ Q_u = \frac{\text{Feed rate}}{\text{Underflow concentration}} = \frac{1,000,000 \ \mathrm{kg/day}}{1500 \ \mathrm{kg/m^3}} = 666.67 \ \mathrm{m^3/day} \] \[ Q_u = \frac{666.67}{24 \times 3600} = 0.00772 \ \mathrm{m^3/s} \] --- ## **Step 2: Calculate the Minimum Area Using Batch Data** ### **Use the Talmage and Fitch Method (Graphical/Tabular Approach)** First, calculate the critical solids flux: - Find the point where the settling curve has the least slope (this is usually at the inflection point or where interface velocity is minimum). #### **Calculate Initial Pulp Volume** \[ \text{Initial volume, } V_0 = \frac{\text{Feed mass}}{\text{Feed concentration}} \] Let's assume we work with **1 m² area** (for ease), initial height \( H_0 = 0.91 \) m: \[ V_0 = 1 \ \mathrm{m^2} \times 0.91 \ \mathrm{m} = 0.91 \ \mathrm{m^3} \] \[ \text{Initial mass of solids} = V_0 \times C_f = 0.91 \times 200 = 182 \ \mathrm{kg} \] \[ \text{Final height for 1500 kg/m³ (underflow)}: \] \[ V_u = \frac{\text{Same mass}}{C_u} = \frac{182}{1500} = 0.121 \ \mathrm{m^3} \] \[ \text{Final underflow height} = \frac{0.121}{1 \ \mathrm{m^2}} = 0.121 \ \mathrm{m} \] --- ### **Calculate the Minimum Settling Velocity** Find the slope (change in height/change in time) at the critical point (i.e., at the interface where the concentration equals the underflow concentration). From the table, when the height reaches **0.15 m at 1.0 hr**, it's close to the desired underflow concentration (0.121 m). Take the slope between 0.27 m at 0.5 hr and 0.15 m at 1.0 hr: \[ \text{Slope} = \frac{0.27 - 0.15}{1.0 - 0.5} = \frac{0.12}{0.5} = 0.24 \ \mathrm{m/hr} \] But, let's interpolate for 0.121 m: Between 0.15 m (1 hr) and 0.08 m (2 hr): \[ \text{Rate} = \frac{0.15 - 0.08}{2.0 - 1.0} = 0.07 \ \mathrm{m/hr} \] At underflow height (0.121 m), approximate time: \[ \text{Fraction to reach 0.121} = \frac{0.15 - 0.121}{0.15 - 0.08} = \frac{0.029}{0.07} = 0.414 \] So, time = 1 hr + 0.414 × 1 hr ≈ 1.41 hr Settling velocity near underflow height: \[ \text{Settling velocity} = \frac{0.15 - 0.121}{1.41 - 1.0} = \frac{0.029}{0.41} = 0.071 \ \mathrm{m/hr} \] --- ## **Step 3: Calculate Required Area** The area required, \( A \), is given by: \[ A = \frac{Q_u}{\text{Settling velocity}} \] Where \( Q_u \) = underflow volumetric rate \( = 666.67 \ \mathrm{m^3/day} \). Convert settling velocity to m/day: \[ 0.071 \ \mathrm{m/hr} \times 24 = 1.704 \ \mathrm{m/day} \] \[ A = \frac{666.67}{1.704} = 391.4 \ \mathrm{m^2} \] --- ## **Step 4: Calculate the Diameter** \[ A = \pi D^2 / 4 \] \[ D = \sqrt{\frac{4A}{\pi}} \] \[ D = \sqrt{\frac{4 \times 391.4}{3.1416}} = \sqrt{498.37} = 22.34 \ \mathrm{m} \] --- ## **Final Answer** **Minimum required diameter of the thickener is:** \[ \boxed{22.3 \ \text{m}} \] --- If you need more detailed steps or a plot for the sedimentation data, let me know!