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Jet on moving curved vanes, force and work

Short answer

Jet velocity 28.52 m/s, relative velocity 16.52 m/s, exit relative 14.87 m/s (0.9×inlet). With series of vanes, Fx=1514 N, Fy=589 N; resultant force 1.62 kN at 21.3° below jet direction. Work done = Fx × u = 18.2 kW.

A 5 cm diameter jet delivering 56 liters of water per sec impinges without shock on a series of vanes moving at 12 m/s in the same direction as the jet. The vanes are curved so that they would, if stationary, deflect the jet through an angle of 135°. Fluid resistance reduces the relative velocity at exit from the vanes to 0.90 of that at entrance. Determine (a) the magnitude and direction of the resultant force on the vanes (b) The work done per second by the vanes.

For a series of vanes the whole jet discharge is used, so the force follows from the change of momentum in the direction of the jet and perpendicular to it. The resultant force is about 1.62 kN at 21.3∘21.3^\circ to the jet, and the work done is about 18.2 kW.

Given

d=5d = 5 cm, Q=0.056 m3/sQ = 0.056\ \text{m}^3/\text{s}, u=12u = 12 m/s, deflection =135∘= 135^\circ, Vr2=0.9 Vr1V_{r2} = 0.9\,V_{r1}, ρ=1000 kg/m3\rho = 1000\ \text{kg/m}^3.

Velocities

V=QA=0.056π4(0.05)2=28.52 m/sVr1=V−u=16.52 m/s,Vr2=0.9×16.52=14.87 m/s\begin{aligned} V &= \frac{Q}{A} = \frac{0.056}{\frac{\pi}{4}(0.05)^2} = 28.52\ \text{m/s} \\ V_{r1} &= V - u = 16.52\ \text{m/s},\qquad V_{r2} = 0.9 \times 16.52 = 14.87\ \text{m/s} \end{aligned} 
  Exit velocity triangle
     Vr2 at 135 deg to jet direction
   Vx2 = u + Vr2 cos135 = 12 - 10.51 = 1.49 m/s
   Vy2 =     Vr2 sin135 = 10.51 m/s

(a) Force on the vanes

Fx=ρQ (V−Vx2)=1000(0.056)(28.52−1.49)=1514 NFy=ρQ Vy2=1000(0.056)(10.51)=589 NR=15142+5892=1624 N,θ=tan⁡−15891514=21.3∘\begin{aligned} F_x &= \rho Q\,(V - V_{x2}) = 1000(0.056)(28.52 - 1.49) = 1514\ \text{N} \\ F_y &= \rho Q\,V_{y2} = 1000(0.056)(10.51) = 589\ \text{N} \\ R &= \sqrt{1514^2 + 589^2} = 1624\ \text{N},\qquad \theta = \tan^{-1}\frac{589}{1514} = 21.3^\circ \end{aligned}

The jet is turned upward, so the force on the vanes has a component FxF_x along the jet and a component FyF_y of 589 N downward.

Answer (a): R≈1.62R \approx 1.62 kN, at 21.3∘21.3^\circ below the direction of the jet.

(b) Work done per second

Work=Fx u=1514×12=18 167 J/s≈18.2 kW\text{Work} = F_x\,u = 1514 \times 12 = 18\,167\ \text{J/s} \approx 18.2\ \text{kW}

Only FxF_x does work, because the vanes move along the jet direction.

Answer (b): about 18.2 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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