Title HPGE 29 Solutions.pdf ✅

29 Hydraulics: Critical Flow Solutions SITUATION 1. A 3-m wide rectangular channel having a specific energy of 2 m has a discharge of 5 m3 s . ▣ 1. Determine the two depths having the given specific energy. [SOLUTION] E = y + v 2 2g 2 m = y + [ 5 m3 s (3 m)y ] 2 2 (9.81 m s 2 ) y = 1.963 m, 0.288 m ▣ 2. Determine the critical depth of flow. [SOLUTION] yc = √ q 2 g 3 yc = √ ( 5 m3 s 3 m ) 2 9.81 m s 2 3 yc = 0.6567 m ▣ 3. Determine the critical velocity. [SOLUTION]
v = Q A v = 5 m3 s (3 m)(0.6567 m) v = 2.538 m s ▣ 4. Determine the minimum specific energy. [SOLUTION] Emin = 3 2 yc Emin = 3 2 (0.6567 m) Emin = 0.985 m SITUATION 2. Water flows at a rate of 12 m3/s in a triangular channel that has a vertex angle of 70°. If the flow depth in the channel is 1 m, ▣ 5. Determine the type of flow in the channel. [SOLUTION] Compute for B: B = 2 × (1m) tan 70° 2 B = 1.4 m

▣ 8. Determine the alternate depth corresponding to the actual flow depth. [SOLUTION] Compute the area in terms of depth. A = 1 2 (2y tan 70° 2 ) y A = y 2 tan 35° Compute for the alternate depth E = y2 + v 2 2g 15.98 m = y2 + ( 12 m3 s y2 2 tan 35°) 2 2 (9.81 m s 2 ) y2 = 15.978 m ▣ 9. Determine the critical depth. [SOLUTION] For the critical depth, Q 2 g = A 3 B (12 m3 s ) 2 9.81 m s 2 = (yc 2 tan 35°) 2 2yc tan 35° yc = 2.267 m ▣ 10. A triangular channel with side slopes of 3V:2H and bed slope of 0.002 is designed to carry a discharge of 8 m3/s. The channel is made of clean, excavated earth materials with n = 0.022. Calculate the specific energy of water. [SOLUTION]