Question

1. A horizontal jet 0.06 mls strikes tangentially on one or another of a series of moving curved vanes, each having a deflection angle of 60 degrees. The water as it leaves the vane has an absolute velocity of 12.2 m/s and makes an angle of 30 degrees with the horizontal. Determine the horizontal force exerted on the vanes and the work done. 2. A nozzle discharges 0.05 mls horizontally, the jet stiking tangentally on one another of a series of curved vanes each having a deflection angle of 150 degrees. The velocity of the jet is 18 m/s, and the vanes have a velocity of 10 ft's in the same direction as the jet. Determine the absolute velocity of the water as it leaves the vanes and the power developed.

Answer 1

Given that Absolute velocity = 12.2 m/sec alpha = 30 degree V/sin(180 - 60 degree) = V_r/sin alpha V/sin(180 - 60 degree) = 12.2/Sin 30 degree Therefore V = 21.12 m/sec \r\n V_r = v - u Therefore u = v - v_r = > 21.13 - 12.2 Therefore u = 8.93 m/sec V_w = v = 21.13 m/sec V_r = V_r1 = 12.2 m/sec = V_1 phi = 60 degree: u_1 = u = 8.93 m/sec v_w1 = v_1 cos 60 degree = > 12.2 cos 60 degree = > 6.1 m/sec Area of Jet(a) = Q/v = > 0.06 m^3/sec/21.13 = > 2.83 times 10^-3 m^2 mass of water striking the veres per second = > wav/g = > 9810 times 2.83 times 10^-3 times 21.13/9.81 = > 59.79 kg Therefore Horizontal force (f_x) = wa/g [V_w - V_w1] = > 59.79 (21.13 - 6.1) F_z = 898.64 N work done = > f_x * u = > 898.64 times 8.93 = > 8024.88 N middot m/sec \r\n v =