Tiêu đề HPGE 30 Solutions.pdf ✅

30 Hydraulics: Non-uniform Flow Solutions SITUATION 1. Water flows in a rectangular channel at a depth of 30 cm with a velocity of 16 m/s, as shown. If a downstream sill (not shown in the figure) forces a hydraulic jump, ▣ 1. What will be the depth of the water downstream of the jump? [SOLUTION] y2 = y1 2 [−1 + √1 + 8Fr1 2 ] Compute for the Froude number at the start Fr1 = v √gy1 Fr1 = 16 m s √(9.81 m s 2 ) (0.3 m) Fr1 = 9.327 Solve for y2 y2 = 0.3 m 2 [−1 + √1 + 8(9.327) 2] y2 = 3.81 m ▣ 2. What is the velocity downstream of the jump? [SOLUTION] Q1 = Q2
A1V1 = A2V2 v1y1 = v2y2 (16 m s ) (0.3 m) = v2 (3.81 m) v2 = 1.26 m s ▣ 3. What is the Froude number downstream of the jump? [SOLUTION] Fr2 = v2 √gy2 Fr2 = 1.26 m s √(9.81 m s 2 ) (3.81 m) Fr2 = 0.206 ▣ 4. What is the headloss due to the jump? [SOLUTION] For a rectangular channel, hL = (y2 − y1 ) 3 4y1y2 hL = (3.81 m − 0.3 m) 3 4(0.3 m)(3.81 m) hL = 9.457 m SITUATION 2. Water flows in a triangular channel (side slops 1H:1V) at a rate of 5 m3 s . It is then forced to undergo a jump at a depth of 0.8 m. ▣ 5. What is the flow depth downstream of the jump? [SOLUTION]
From conservation of momentum, ∑F = ρQ(v2 − v1 ) F1 − F2 = ρQ(v2 − v1) Solve for the velocities v1 = Q A1 v1 = 5 m3 s (0.8 m) 2 v1 = 7.8125 m s v2 = 5 m3 s y2 Compute for the forces F1 = γhA1 F1 = (9810 N m3 ) ( 0.8 m 3 ) (0.8 m) 2 F1 = 1674.24 N F2 = γhA2 F2 = (9810 N m3 ) ( y2 3 ) (y2 2 ) F2 = 3270y2 2 Substitute to the equation