Design a sedimentation basin for a water treatment plant with the following specifications assuming it is required to supply a city with a population of 50,000 people: ● Rectangular or square basin Depth: 7-18 ft Width: 10-50 ft • Detention time: 4-8 hours • Flow through velocity: <0.55 ft/min Overflow rate: 400-1,100 gal/day-ft² Draw a figure to show calculated dimensions. You may consider to use SI units in calculations. Hint: Overflow rate is Flow/Surface Area; Detention time is Volume of tank/Flow rate; and Depth= Volume/Surface area
Design a sedimentation basin for a water treatment plant with the following specifications assuming it is required to supply a city with a population of 50,000 people: ● Rectangular or square basin Depth: 7-18 ft Width: 10-50 ft • Detention time: 4-8 hours • Flow through velocity: <0.55 ft/min Overflow rate: 400-1,100 gal/day-ft² Draw a figure to show calculated dimensions. You may consider to use SI units in calculations. Hint: Overflow rate is Flow/Surface Area; Detention time is Volume of tank/Flow rate; and Depth= Volume/Surface area
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:Design a sedimentation basin for a water treatment plant with the following specifications
assuming it is required to supply a city with a population of 50,000 people:
Rectangular or square basin
• Depth: 7-18 ft
●
.
• Width: 10-50 ft
Detention time: 4-8 hours
Flow through velocity: <0.55 ft/min
Overflow rate: 400-1,100 gal/day-ft²
Draw a figure to show calculated dimensions. You may consider to use SI units in calculations.
●
●
Hint: Overflow rate is Flow/Surface Area; Detention time is Volume of tank/Flow rate; and
Depth= Volume/Surface area
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Why did you divide the detention time by 7.48? Can you please explain? (Step 3, Line 2)
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