Tiêu đề Corrections for Eval 8 Nov 2024 Solutions (1).pdf ✅

EVAL EXAM 8 SOLUTIONS SITUATION 1. Shown is the figure for the problem. ▣ 1. From Manning’s formula, Q = A n R 2/3S 1/2 First solve for the hydraulic radius R = A P R = (1.2 m)(5.4 m) 5.4 m + 2(1.2 m) R = 54 65 m Q = (1.2 m)(5.4 m) 0.013 ( 54 65 m) 2 3 (0.001) 1 2 Q = 13.93 m3 /s ▣ 2. The most efficient section for a rectangular section is when the width b and the depth d has proportions b = 2d. To design such section, the discharge must be equated. In the Manning’s equation, since n, S are not changed, then those variables will be cancelled. A1 n R1 2 3S 1 2 = A2 n R2 2 3S 1 2 A1R1 2 3 = A2R2 2 3 (1.2 m × 5.4 m) ( 54 65 m) 2 3 = (2d 2 ) ( d 2 ) 2 3 d = 1.7643 m

dm = 6d + 2d 2 6 + 4d dm = 6(0.596) + 2(0.596) 2 6 + 4(0.596) dm = 0.51148 m Solve first for the velocity of flow. v = Q A v = 10 m3 s 6(0.596 m) + 2(0.596 m) 2 v = 2.3316 m s Then solve for the Froude number FN = v √gdm FN = 2.3316 m s √(9.81 m s 2 ) (0.51148 m) FN = 1.04 ▣ 6. The critical depth of an open channel is the depth of flow at which the Froude number is exactly 1. Or as a derived equation, the critical depth is determined from: Q 2 g = A 3 B (10 m3 s ) 2 (9.81 m s 2 ) = (6dc + 2dc 2 ) 3 6 + 4dc dc = 0.6113 m ▣ 7. First consider the hydraulic radius of the channel considering that it is very wide. R = bd b + 2d = d 1 + 2 d b Now as b → ∞, the term d b approaches 0. Therefore, the converging value of the hydraulic radius is R = d. First consider the channel at the point where the depth is 1 m.