Culverts are designed to carry a specified discharge and are based on estimates

Question

Culverts are designed to carry a specified discharge and are based on estimates of runoff. The runoff estimates, in turn, are often based on the magnitude of precipitation events (e.g., the 50-year or 100-year flood): basin characteristics (e.g., drainage area, topography): and land use. With the estimates of runoff, engineers can determine an appropriate culvert size. At a certain region, the estimated maximum discharge following a rainfall event is 12 ft^3/s. What should the diameter of a culvert be to handle such a discharge, if the culvert is to be made of iron (n = 0.015), and constructed at a slope of 0.01? For pipes or culverts, the cross-sectional area (A) is equal to the area of a circle (pi r^2), while the wetted perimeter is equal to the circumference (2 pi r). Therefore, the hydraulic radius R = r/2.

Explanation / Answer

Manning's equation: Q = VA = (1.49/n) AR(2/3) S0.5 = 12 ft3 /s n = 0.015 R = r/2 S = 0.01 = (1.49/0.015) x pi x r x r x (r/2)2/3 x 0.01 0.5 = ( 311.906 x r2 x r2/3 x 0.010.5 ) / 22/3 = 31.1906 / 22/3 x r4/3 =31.1906/1.58 x r4/3 = 19.74 x r4/3 12 = 19.74 x r4/3 r4/3 = 12/19.74 r = 0.6073/4 r = 0.688 ft Diameter = 0.688 x 2 = 1.376 The diameter of the culvert to handle the discharge of 12 ft3 /s is 1.376 ft.

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