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Give step-by-step solution with explanation and final answer: Design a flocculation basin by determine the basin volume, tank dimensions, required input power and impeller diameter using the following parameters and the manufacture's data. Design flow rate = 1 1 . 5 x 1 0 3 m / d Flocculation time = 3 0 minutes Three flocculator compartments with G = 7 0 , 5 0 , 3 0 s - 1 Water temperature = 5 0 C Place impeller at one - third the water depth. From the manufacture's data the following impellers are available

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Give step-by-step solution with explanation and final answer: Design a flocculation basin by determine the basin volume, tank dimensions, required input power and impeller diameter using the following parameters and the manufacture's data. Design flow rate = 1 1 . 5 x 1 0 3 m / d Flocculation time = 3 0 minutes Three flocculator compartments with G = 7 0 , 5 0 , 3 0 s - 1 Water temperature = 5 0 C Place impeller at one - third the water depth. From the manufacture's data the following impellers are available

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Asked by: varrisanju5@gmail.com
Created at: 2025-10-31 06:44:04

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# Flocculation Basin Design Step-by-Step ## **Given Data** - **Design flow rate**: \( 1.5times 10^3 \) m³/d - **Flocculation time (total)**: 30 minutes - **Number of compartments**: 3 -Mixing intensity (G)** per compartment: - Compartment 1: \( G_1 = 70\, \text{s}^{-1} \) - Compartment 2: \( G_2 = 50\, \text{s}^{-1} \) - Compartment 3: \( G_3 = 30\, \text{s}^{-1} \) - **Water temperature**: \( 5\,^\circ\text{C} \) - **Impeller placed at one-third water depth** - **Available impeller data**: (Not specified—assume will choose suitable diameter in process) --- ## **1. Convert Flow Rate to m³/s** \[ Q = \frac{1.5 \times 10^3\, \text{m}^3}{24 \times 60 \times 60\, \text{s}} = \frac{1.5 \times 10^3}{86400} \approx .01736\, \text{m}^3/\text{s} \] --- ## **2. Calculate Total Basin Volume** Flocculation time (\( t \)) = 30 min = 180 s \[ V = Q \times t = .01736\, \text{m}^3/\text{s} \times 180\, \text{s} = 31.25\, \text{m}^3 \] --- ## **3. Compartment Volumes** Assume **equal detention time** in each compartment (unless stated otherwise): \[ t_\text{compartment} = \frac{30\, \text{min}}{3} = 10\, \text{min} = 600\, \text{s} \] Each compartment volume: \[ V_\text{compartment} = Q \times t_\text{compartment} = .01736\, \text{m}^3/\text{s} \times 600\, \text{s} = 10.416\, \text{m}^3 \] --- ## **4. Tank Dimensions** Assume **rectangular tanks** and a typical **water depth** of 3 m (common for flocculation): \[ \text{Surface Area per compartment} = \frac{V_\text{compartment}}{\text{depth}} = \frac{10.416\, \text{m}^3}{3\, \text{m}} \approx 3.47\, \text{m}^2 \] Assume **Length : Width = 3:1** (common for mixing tanks): Let \( W = w \) and \( L = 3w \): \[ L \times W = 3.47 \implies 3w \times w = 3.47 \implies w^2 = 1.156 \implies w \approx 1.075\, \text{m} \] \[ L = 3 \times 1.075 \approx 3.225\, \text{m} \] **Each compartment:** - Length: 3.2 m - Width: 1.1 m - Depth: 3. m --- ## **5. Water Viscosity at 5°C** At 5°C, dynamic viscosity (\( \mu \)) ≈ \( 1.52 \times 10^{-3} \) N·s/m². --- ## **6. Power Input for Each Compartment** The **mixing intensity** is given by: \[ G = \sqrt{\frac{P}{\mu V}} \] Where: - \( G \) = velocity gradient (s⁻¹) - \( P \) = power input (W) - \( \mu \) = dynamic viscosity (N·s/m²) - \( V \) = compartment volume (m³) **Rearrange to solve for P:** \[ P = G^2 \cdot \mu \cdot V \] ### **Compartment 1:** \[ P_1 = (70)^2 \times 1.52 \times 10^{-3} \times 10.416 \] \[ = 490 \times 1.52 \times 10^{-3} \times 10.416 \] \[ = 490 \times .01584432 \] \[ = 77.637\, \text{W} \] ### **Compartment 2:** \[ P_2 = (50)^2 \times 1.52 \times 10^{-3} \times 10.416 \] \[ = 250 \times .01584432 = 39.6108\, \text{W} \] ### **Compartment 3:** \[ P_3 = (30)^2 \times 1.52 \times 10^{-3} \times 10.416 \] \[ = 900 \times .01584432 = 14.2599\, \text{W} \] --- ## **7. Impeller Diameter Selection** ### **Impeller Placement** - Placed at **1/3 water depth**: Depth = 3. m → Position = 1 m below water surface ### **Estimate Impeller Diameter** A common guideline: **Impeller diameter = .3 to .5 of tank width** \[ \text{Tank width} \approx 1.1\, \text{m} \] \[ D_\text{impeller} = .4 \times 1.1 \approx .44\, \text{m} \] If manufacturer's impeller sizes include .4 m and .5 m, **select .4 m diameter**. --- ## **8. Summary Table** | Compartment | G (s⁻¹) | Volume (m³) | Power (W) | Tank (L × W × D, m) | Impeller diam. (m) | |-------------|---------|-------------|-----------|---------------------|--------------------| | 1 | 70 | 10.4 | 77.6 | 3.2 × 1.1 × 3. | .4 | | 2 | 50 | 10.4 | 39.6 | 3.2 × 1.1 × 3. | .4 | | 3 | 30 | 10.4 | 14.3 | 3.2 × 1.1 × 3. | .4 | --- ## **Final Answers** ### **Basin Volume:** \[ \boxed{31.25\, \text{m}^3} \] ### **Compartment Dimensions (each):** \[ \boxed{3.2\, \text{m} \times 1.1\, \text{m} \times 3.\, \text{m}} \] ### **Input Power per Compartment:** - Compartment 1: \( \boxed{77.6\, \text{W}} \) - Compartment 2: \( \boxed{39.6\, \text{W}} \) - Compartment 3: \( \boxed{14.3\, \text{W}} \) ### **Impeller Diameter:** \[ \boxed{.4\, \text{m}} \] --- ## **Image Example** ![Schematic of a 3-compartment flocculation tank with impellers at 1/3 depth](https://i.imgur.com/ImpelFloc.png) *Alt text: Schematic of a rectangular flocculator with three compartments, each with an impeller at one-third water depth.* --- **Note:** If more specific manufacturer's impeller data is provided, the exact diameter may be adjusted accordingly.

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