Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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A triangular channel with the most efficient proportion discharges water at the rate of 2 cubic meter per second. Assuming n=0.018 and S=0.0021, Calculate the normal depth
of flow in meters.
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- QUESTION 30 In the figure shown, sub-critical uniform flow occurs in a 1.2 m width rectangular channel at the depth of 0.6 m. A hump of height 10 cm is placed in the bottom of the channel. Flow rate of the channel is 0.8 m/s. Assuming no energy losses, 1. Show that the flow depth on the step is 0.453 m(Y2) 2. If the flow depth on the step were to be critical depth (Y2) calculate the minimum height of the step (AH) Y-0.6 m Y: Q=0.8 m'/s DAH = 0.1 marrow_forwardThe top width and the depth of flow in a triangular channel were measured as 5 m and 1 m respectively. The measured velocity on the centre line of flow area at the water surface, 0.2 m and 0.8 m below the surface are 0.8 m/s, 0.6 m/s and 0.5 m/s respectively. Using two point method of velocity measurement, the discharge (in m3/s) in the channel isarrow_forwardWater flows in a 8-m-wide rectangular channel that has a longitudinal slope of 0.01%. The channel has an equivalent sand roughness of 2 mm. Calculate the uniform-flow depth in m when the flow rate is 12 m3/s. Use the Darcy–Weisbach equation.arrow_forward
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