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.

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