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Alchen [17]
3 years ago
8

g A 30-m-diameter sedimentation basin has an average water depth of 3.0 m. It is treating 0.3 m3/s wastewater flow. Compute over

flow rate and detention time.
Engineering
1 answer:
stepladder [879]3 years ago
6 0

Answer:

The overflow rate is 4.24×10^-4 m/s.

The detention time is 7069.5 s

Explanation:

Overflow rate is given as volumetric flow rate ÷ area

volumetric flow rate = 0.3 m^3/s

area = πd^2/4 = 3.142×30^2/4 = 706.95 m^2

Overflow rate = 0.3 m^3/s ÷ 706.95 m^2 = 4.24×10^-4 m/s

Detention time = volume ÷ volumetric flow rate

volume = area × depth = 706.95 m^2 × 3 m = 2120.85 m^3

Detention time = 2120.85 m^3 ÷ 0.3 m^3/s = 7069.5 s

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algol [13]

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5 0
2 years ago
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madreJ [45]

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