Example

Problem 1: Manning Discharge in a Rectangular Channel

A rectangular concrete channel is 3.0 m3.0\text{ m} wide and carries water 1.5 m1.5\text{ m} deep. The slope is 0.0010.001 and Manning n=0.015n=0.015. Determine discharge.

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Example

Problem 2: Normal Depth in a Rectangular Channel

A 4.0 m4.0\text{ m} wide rectangular channel carries 10 m3/s10\text{ m}^3/\text{s}. Use n=0.015n=0.015 and slope S=0.0008S=0.0008. Determine normal depth.

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Example

Problem 3: Manning Discharge in a Trapezoidal Channel

A trapezoidal channel has bottom width 2.0 m2.0\text{ m}, side slope 1.5H:1V1.5\text{H}:1\text{V}, depth 1.2 m1.2\text{ m}, slope 0.00150.0015, and n=0.025n=0.025. Determine discharge.

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Example

Problem 4: Most Efficient Rectangular Section

Determine the dimensions of the hydraulically efficient rectangular section with flow area 12 m212\text{ m}^2.

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Example

Problem 5: Most Efficient Trapezoidal Section

A hydraulically efficient trapezoidal channel must have area 20 m220\text{ m}^2 and side slopes 1H:1V1\text{H}:1\text{V}. Determine depth and bottom width.

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Example

Problem 6: Half-Full Circular Conduit

A circular conduit of diameter 1.2 m1.2\text{ m} flows half full under gravity. Use n=0.013n=0.013 and slope 0.0020.002. Determine area, hydraulic radius, and discharge.

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Example

Problem 7: Compound-Section Conveyance

A main channel subsection has A1=20 m2A_1=20\text{ m}^2, R1=1.5 mR_1=1.5\text{ m}, and n1=0.030n_1=0.030. An adjacent floodplain subsection has A2=30 m2A_2=30\text{ m}^2, R2=0.50 mR_2=0.50\text{ m}, and n2=0.060n_2=0.060. If the common energy slope is 0.0010.001, estimate total discharge by summing conveyances.

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Example

Problem 8: Required Channel Slope

A rectangular channel is 5.0 m5.0\text{ m} wide and 1.0 m1.0\text{ m} deep and must carry 12 m3/s12\text{ m}^3/\text{s}. Use n=0.018n=0.018. Determine the required uniform-flow slope.

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