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A triangular roadside channel is poorly lined with riprap. The channel has side slopes of 2:1 (H:V) and longitudinal slope of 2.5 percent. Determine the flow rate (cubic meters per second) in the channel if flow is uniform and the top width of the flowing channel is 5 meters.

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jolis1796

Answer:

Q = 14.578 m³/s

Explanation:

Given

We use the Manning Equation as follows

Q = (1/n)*A*(∛R²)*(√S)

where

  • Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)
  • A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)
  • S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025
  • n = Manning roughness coefficient (empirical constant), dimensionless = 0.023
  • R = hydraulic radius = A/P (m for S.I.) where :
  • A = cross-sectional area of flow as defined above,
  • P = wetted perimeter of cross-sectional flow area (m for S.I.)

we get A as follows

A = (B*h)/2

where

B = 5 m (the top width of the flowing channel)

h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m   (the deep)

A = (5 m*1.25 m/2) = 3.125 m²

then we find P

P = 2*√((B/2)²+h²)   ⇒  P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m

⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m

Substituting values into the Manning equation gives:

Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)

⇒ Q = 14.578 m³/s

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