- Practice · 6 marks
Force of jet on symmetrical curved vane
Short answer
(a) Fixed vane: F = ρaV²(1+cos20°) = 4936 N. (b) Moving vane: relative velocity (V-u)=20 m/s, mass flow ρa(V-u), F = ρa(V-u)²(1+cos20°) = 2194 N, power = Fu = 21.94 kW.
A jet of water 60 mm in diameter and moving at 30 m/s strikes at the centre of a symmetrical curved vane which deflects the jet through 160 degrees (each half leaves at 20 degrees to the original jet direction, back towards the source). Neglecting friction, find (a) the force exerted on the vane in the direction of the jet when it is fixed, and (b) the force and power developed when the vane moves at 10 m/s in the direction of the jet.
Given: m, m/s, deflection , so the exit angle with the original jet direction is and the exit velocity component along the jet is (opposite to the jet).
Jet area m².
(a) Fixed vane
The vane is symmetrical and the jet splits into two equal halves, which leave with velocity (no friction, no change in speed). Change of velocity along the jet per kg: .
(b) Vane moving at 10 m/s in the jet direction
A single vane moving away receives mass flow per second, and the relative speed m/s is unchanged along the vane (no friction). Relative to the vane the jet is turned in the same way, so
Answer: (a) 4936 N (fixed vane); (b) 2194 N and 21.94 kW (moving vane).
✓ Answer checked by Claude Sonnet 5.5
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