A sedimentation basin with a settling zone surface area of 2400 m accepts a flow of 46,000 m'/day. For a particle-settling velocity of 0.2 mm/s, what is the approximate removal efficiency of the sedimentation basin?
A sedimentation basin with a settling zone surface area of 2400 m accepts a flow of 46,000 m'/day. For a particle-settling velocity of 0.2 mm/s, what is the approximate removal efficiency of the sedimentation basin?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![**Sedimentation Basin Analysis**
This problem involves a sedimentation basin with specific parameters.
- **Surface Area of Settling Zone**: 2400 m²
- **Flow Rate**: 46,000 m³/day
- **Particle-Settling Velocity**: 0.2 mm/s
**Objective**: Calculate the approximate removal efficiency of the sedimentation basin.
**Concepts**:
The removal efficiency of a sedimentation basin can be determined using the overflow rate (also known as the surface loading rate) and the particle-settling velocity. The formula for removal efficiency (\(E\)) is typically represented as:
\[ E = \left( \frac{V_s}{V_o} \right) \times 100 \]
Where:
- \(V_s\) is the particle-settling velocity
- \(V_o\) is the overflow rate, calculated as \( \frac{\text{Flow rate}}{\text{Surface area}} \)
**Steps**:
1. **Convert Particle-Settling Velocity**:
- Convert from mm/s to m/day.
- \( 0.2 \, \text{mm/s} = 0.2 \times 86,400 \, \text{m/day} = 17,280 \, \text{m/day} \)
2. **Calculate Overflow Rate**:
- \( V_o = \frac{46,000 \, \text{m}^3/\text{day}}{2400 \, \text{m}^2} = 19.17 \, \text{m/day} \)
3. **Calculate Removal Efficiency**:
- \( E = \left( \frac{17,280}{19.17} \right) \times 100 \approx 90.21\% \)
This simplification provides a framework to assess the efficiency of the sedimentation process based on the given parameters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fded4d143-e6bc-43a7-af80-2ef19d02a5d5%2F2bdfa70d-b528-4b0f-8bda-206839015bda%2F63polzr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Sedimentation Basin Analysis**
This problem involves a sedimentation basin with specific parameters.
- **Surface Area of Settling Zone**: 2400 m²
- **Flow Rate**: 46,000 m³/day
- **Particle-Settling Velocity**: 0.2 mm/s
**Objective**: Calculate the approximate removal efficiency of the sedimentation basin.
**Concepts**:
The removal efficiency of a sedimentation basin can be determined using the overflow rate (also known as the surface loading rate) and the particle-settling velocity. The formula for removal efficiency (\(E\)) is typically represented as:
\[ E = \left( \frac{V_s}{V_o} \right) \times 100 \]
Where:
- \(V_s\) is the particle-settling velocity
- \(V_o\) is the overflow rate, calculated as \( \frac{\text{Flow rate}}{\text{Surface area}} \)
**Steps**:
1. **Convert Particle-Settling Velocity**:
- Convert from mm/s to m/day.
- \( 0.2 \, \text{mm/s} = 0.2 \times 86,400 \, \text{m/day} = 17,280 \, \text{m/day} \)
2. **Calculate Overflow Rate**:
- \( V_o = \frac{46,000 \, \text{m}^3/\text{day}}{2400 \, \text{m}^2} = 19.17 \, \text{m/day} \)
3. **Calculate Removal Efficiency**:
- \( E = \left( \frac{17,280}{19.17} \right) \times 100 \approx 90.21\% \)
This simplification provides a framework to assess the efficiency of the sedimentation process based on the given parameters.
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