Tiêu đề HPGE 20 Solutions.pdf ✅

20 Hydraulics: Rotating Vessel Solutions SITUATION 1. An open cylindrical tank, 2 m in diameter, and 4 m high contains water to a depth of 3 m. It is rotated about its own vertical axis with a constant angular speed, ω. ▣ 1. If ω = 3 rad/s, how much liquid is spilled? [SOLUTION] h = ω 2 r 2 2g h = (3 rad s ) 2 (1 m) 2 2 (9.81 m s 2 ) h = 0.456 m From the figure, since h < 2 (4 m − 3 m) = 2m, then there is no spillage. Vspill = 0 m3
▣ 2. What maximum value of ω(in rpm) can be imposed without spilling any liquid? [SOLUTION] The initial height of air is 4 m − 3 m = 1 m. At start of spillage, the height of paraboloid must be 2(1 m) = 2 m. h = ω 2 r 2 2g 2 m = (ω 2π 60) 2 (1 m) 2 2 (9.81 m s 2 ) ω = 59.82 rpm ▣ 3. If ω = 8 rad s , to what depth will the water stand when brought to rest? [SOLUTION] h = (8 rad s ) 2 (1 m) 2 2 (9.81 m s 2 ) h = 3.262 m y = 4 m − h y = 0.738 m Depth of water when brought to rest: d = y + h 2 d = 0.738 m + 3.262 m 2 d = 2.369 m ▣ 4. What angular speed ω (in rpm) will just zero the depth of water at the center of the tank?
[SOLUTION] The height of the paraboloid must be the height of the cylinder. h = ω 2 r 2 2g 4 m = ω 2 (1 m) 2 2 (9.81 m s 2 ) ω = 8.86 rad s × 2π 60 ω = 84.596 rpm ▣ 5. If ω = 100 rpm, how much area at the bottom of the tank is uncovered? [SOLUTION] h = ω 2 r 2 2g h = (100 rpm × 2π 60) 2 (1 m) 2 2 (9.81 m s 2 ) h = 5.5893 m Compute the radius at the bottom
r 2 1.5893 m = (1 m) 2 5.5893 m r = 0.5332 m A = πr 2 A = π(0.5332 m) 2 A = 0.893 m2 SITUATION 2. An open cylindrical tank 0.3 m in diameter and 0.80 m high is partially filled with water to a certain depth. It is then rotated about its vertical axis at 240 rpm but no water spilled out. ▣ 6. Estimate how deep is the water in the tank, in meters. [SOLUTION] h = ω2 r 2 2g h = (240 rpm × 2π 60) 2 (0.15 m) 2 2 (9.81 m s 2 ) h = 0.2744 m hw = 0.8 m − h 2 hw = 0.4378 m ▣ 7. At what speed, in rpm, would the tank be rotated if 1.4 L of water is spilled out? [SOLUTION] Vold = πr 2h Vold = π(0.15 m) 2 (0.4378 m) Vold = 0.0309 m3 Vnew = 0.0309 m3 − 0.0014 m3 Vnew = 0.0295 m3 Compute for the height of water at rest: Vnew = πr 2hw 0.0295 m3 = π(0.15 m) 2hw hw = 0.418 m Compute for the height of air hair = 0.8 m − hw