Arectangular concrete (n=0.013) channel 20 feet wide, on a 2.5% slope, is discharging 400 ft3 /sec into a stilling basin. the stilling basin is also 20 feet wide and has a water depth of 8 ft determined from the downstream channel condition. what is the length of the stilling basin? what is the height of the endsill?

Length of stilling basin = 32.9 feet

Height of end sill = 6.58 feet

Explanation:

Discharge = Q = 400 ft^3 /sec

Slope = 2.5 ft

Width = 20 feet

n = 0.013

we will assume the depth of flow as "d"

Q = 1/n (R)^2/3 (slope)^1/2 A   ( here R is the hydraulic Radius)

by substituting the given data in above formula we get:

400 = 1/0.013 * (R)^2/3 * sqrt (2.5/100) * 20d

R = A/P

here, A is the flow area and P is the wetted perimeter

400*0.013 = (20d/(20+20d))^2/3 * sqrt(2.5/100) * 20d

d = 1.42 feet

Depth of stilling channel before the jump will be = d1 =  8 feet

Depth of stilling channel after the jump will be = d2 = 1.42 feet

Length of stilling basin = 5(d2 - d1)

                                      = 5( 8 - 1.42)

Length of stilling basin = 32.9 feet

Now calculating the height of end sill:

Jump height = (8 - 1.42)

Height of end sill = 6.58 feet

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answer: 500 wph.

explanation: