Question

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9. In a uniform flow taking place in a wide rectangular channel at a depth of 1.2 m, the velocity is found to be 1.5 m/s. If a change in the discharge causes a uniform flow at a depth of 0.88 m in this channel, the corresponding velocity of flow would be (a) 0.89 m/s (b) 1.22 m/s (c) 1.10 m/s (d) 1.50 m/s (e) 2.50 m/s 10. A triangular section is hydraulically efficient when the vertex angle is (a) 90 (b) 120 (c) 60 (d) 30 (e) None of above

Answer 1

v = 1/n R^2/3 squareroot 5 for wide rectangular channel R = y (depth of flow) v = 1/n y^2/3 squareroot s here n, squareroot s are const so v_1/v_2 = y_1^2/3/y_2^2/3 y_1 = 1.2 m v_1 = 1.5 m/s y_2 = 0.88 v_2 = ? 1.5/v_2 = (1.2)^2/3/(0.88)^2/3 = > v_2 = 1.2198 m/s for hydraulically efficient triangular channel vertex angle should be 90 degree Area = ay sin theta R = ay sin theta/2a R should be max dR/d theta = 0 = > 2a/4 cos 2 theta = 0 theta = 45 degree vertex angle = 2 theta = 90 degree