What is the velocity (in ft/sec) in a rectangular concrete channel with a width of 3 feet (ft), a hydraulic grade line slope of 0.002 ft/ft, a flow depth of 1.5 ft and an assumed Manning's coefficient n = 0.014?

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the velocity in a rectangular channel with the given dimensions is 3.92 ft/sec. Solution: Use Manning's Equation and solve for V V = (K/n) R^2/3 Sf^1/2 Where: K = conversion coefficient (1.486 for English units, 1.0 for SI) n = 0.014, the Manning coefficient d = depth of flow = 1.5 ft w = width of channel = 3.0 ft Sf= channel slope = 0.002 ft/ft A = Area = d x w = 1.5 ft x 3 ft = 4.5 ft² P = wetted perimeter = w + 2d = 3 ft + 3 ft = 6 ft R = hydraulic radius = A/P = (4.5 ft² )/ (6 ft) = 0.75 ft V = (1.486/0.014) x (0.75 ft)^2/3 x (0.002 ft/ft)^1/2 = 3.92 ft/sec

What Is the Velocity (In Ft/sec) in a Rectangular Concrete