Tiêu đề HPGE 31 Solutions.pdf ✅

31 Hydraulics: Hydrodynamics Solutions SITUATION 1. Find the force exerted by a jet of water of diameter 75 mm moving at 20 m/s on a flat plate. ▣ 1. When the jet strikes a stationary plate normally. [SOLUTION] F = ρAv 2 F = (1000 kg m3 ) [ π 4 (0.075 m) 2 ] (20 m s ) 2 F = 1767.15 N ▣ 2. When the jet strikes a plate moving away from the jet at a speed of 15 m/s. [SOLUTION] F = ρA(vf − vp) 2 F = (1000 kg m3 ) [ π 4 (0.075 m) 2 ] (20 m s − 15 m s ) 2 F = 110.45 N ▣ 3. When the jet strikes a plate moving towards the jet at a speed of 15 m/s. [SOLUTION] F = ρA(vf − vp) 2 F = (1000 kg/m3 [ π 4 (0.075 m) 2 ][20 m s − (−15 m s )] 2 F = 5411.88 N SITUATION 2. The jets from a garden sprinkler are 25 mm in diameter and are normal to the 0.6 -m radius. Cv = 0.85 and Cc = 1.0. If the pressure at the base of the nozzle is 414 kPa.

SITUATION 3. A reducer is used to connect two horizontal water pipe lines of diameters 600 mm and 300 mm as shown in the figure. At the inlet end of the reducer, the pressure is 300 kPa. At the outlet end of the reducer, the velocity is 6 m/s. ▣ 7. What is the velocity in the larger pipe? [SOLUTION] A1v1 = A2v2 π 4 (0.6 m) 2v1 = π 4 (0.3 m) 2 (6 m s ) v1 = 1.5 m s ▣ 8. Determine the pressure at the outlet end of the reducer. [SOLUTION] From energy equation, P1 γ + v1 2 2g = P2 γ + v2 2 2g 300 kPa 9.81 kN m3 + (1.5 m s ) 2 2 (9.81 m s 2 ) = P2 9.81 kN m3 + (6 m s ) 2 2 (9.81 m s 2 ) P2 = 283.125 kPa ▣ 9. Determine the force on the reducer. [SOLUTION] Conservation of momentum