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OverLord2011 [107]
3 years ago
7

A watershed is divided into two sub-basins A and B. The surface runoff from sub-basin A enters a channel at point 1 and flows to

point 2 where the runoff from sub-basin B is added to the hydrograph and the combined flow is then routed through a reservoir. a. Determine the discharge hydrograph from the reservoir assuming that the reservoir is initially empty (i.e. no baseflow). (Only plot up to hour 12). Also, provide ERH and SHG plots for each sub-basinb. What are the areas of sub-basins A and B in square miles?
Engineering
1 answer:
Dimas [21]3 years ago
8 0

Answer:

I want to answer this question but the information provided is incomplete to plot from which the areas of sub basins A and B can be calculated.

Explanation:

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g A 12,000 m3/day treatment plant has a rectangular sedimentation basin with dimensions 12 meters wide, 3 meters deep, and 25 me
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Answer:

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

Explanation:

Given;

volumetric flow rate of the treatment, Q₀ = 12,000 m³/day

length of the rectangular tank, L = 25 m

width of the tank, W = 12 m

height of the tank, H = 3 m

settling velocity of the particles, V_s = 6 x 10⁻³ m/s

The overflow rate of the sediments are calculated as follows;

V_o = \frac{Q_o}{A_s}

where;

As is the surface area of the tank, m²

Q₀ is the flow rate, m³/s

As = 2LW + 2LH + 2WH

As = (2 x 25 x 12) + (2 x 25 x 3) + (2 x 12 x 3)

As = 822 m²

Q_0 (m^3/s)= \frac{12,000 \ m^3}{day} \times \frac{1 \ day}{24 \ hr} \times \frac{1 \ hour}{60 \ \min} \times \frac{1 \ \min}{60 \ s} = \frac{12,000}{24 \times 60 \times 60}  (m^3/s)= \frac{12,000}{86,400} \ m^3/s\\\\Q_o = 0.139 \ m^3/s

The overflow rate;

V_o = \frac{Q_0}{A_s} = \frac{0.139}{822} = 1.69 \times 10^{-4} \ m/s

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

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Answer:

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Explanation:

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