Certainly! Here is the content from the provided image, **rephrased in a more humanized manner** without using two equal signs on the same line: --- # Problem 3.8.1: Municipal Wastewater Plant Design A municipal wastewater plant must treat up to 60,000 m³/day. The target particle for settling has these properties: - Particle diameter: \( D_p = 200 \times 10^{-6} \) m - Shape factor: \( k = 0.05 \) - Drag coefficient: \( f = 0.025 \) - Particle density: \( \rho_p = 1.25 \times 10^3 \) kg/m³ For the rectangular settling basin: - Length to width ratio should be more than 6 - Overflow rate should be at least four times the settling velocity - Horizontal velocity should not exceed one-third of the scour velocity **Tasks**: a) Find the dimensions of the rectangular tank b) Determine the detention time --- ## Solution ### 1. Settling Velocity Calculation The **horizontal velocity** is one-third of the scour velocity: \[ \text{Horizontal velocity} = \frac{V_h}{3} = 0.02951 \text{ m/s} \] The **settling velocity** (\( u_t \)) can be calculated as: \[ u_t = \frac{g(\rho_p - \rho_f) D_p^2}{18 \mu_f} \] Plugging in the values gives: \[ u_t = 5.44 \times 10^{-3} \text{ m/s} \] The **Reynolds number** is: \[ Re = \frac{\rho_f u_t D_p}{\mu_f} = 1.088 \] The **overflow rate** will be three times the settling velocity: \[ \text{Overflow rate} = 3 \times u_t = 21.76 \times 10^{-3} \text{ m/s} \] --- ### 2. Basin Dimensions Let \( W \) be the width, \( L \) the length, and \( H \) the height of the tank. - The cross-sectional area required (width × height): \[ W \times H = \frac{\text{Flow rate}}{\text{Horizontal velocity}} \] Plug in the numbers: \[ W \times H = \frac{60,000}{24 \times 60 \times 60} \div 0.02951 \] The result is approximately \( 23.54 \) m². - The plan area (length × width): \[ L \times W = \frac{\text{Flow rate}}{\text{Overflow rate}} \] \[ L \times W = \frac{60,000}{24 \times 60 \times 60} \div 21.76 \times 10^{-3} \] This gives about \( 31.905 \) m². - Given the ratio \( \frac{L}{W} > 6 \): \[ \frac{L}{W} = \frac{31.905}{W^2} \] Set \( \frac{L}{W} = 6 \): \[ 6 = \frac{31.905}{W^2} \] Solving for \( W \): \[ W^2 = \frac{31.905}{6} \] \[ W \approx 2.305 \text{ m} \] Now, \( L = 6 \times W \approx 13.83 \) m - Find \( H \): \[ H = \frac{23.54}{2.305} \approx 10.21 \text{ m} \] --- ### 3. Volume and Detention Time - **Tank volume**: \[ \text{Volume} = L \times W \times H = 13.83 \times 2.305 \times 10.21 \approx 325.47 \text{ m}^3 \] - **Detention time**: \[ t = \frac{\text{Volume} \times 24 \times 60}{\text{Flow rate}} \] \[ t = \frac{325.47 \times 24 \times 60}{60,000} \] \[ t \approx 7.81 \text{ min} \] --- ## **Summary of Results** - **Width (\( W \))**: 2.305 m - **Length (\( L \))**: 13.83 m - **Height (\( H \))**: 10.21 m - **Detention time**: 7.81 minutes --- Let me know if you need any further clarifications or calculations!