--- title: "Jet on moving curved vanes, force and work" subject: "Fluid Mechanics" chapter: 7 key: "semester_3/fluid_mechanics:jet-on-moving-curved-vanes" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-on-moving-curved-vanes site: BCT Engineering --- # Jet on moving curved vanes, force and work Fluid Mechanics · Chapter 7: Momentum principle and flow analysis - Asked in: 2073 Shrawan, 5+3 marks - Video: [Force Exerted by Jet on An Unsymmetrical Moving Curved Plate (I)](https://www.youtube.com/watch?v=eiyWErGvrMg) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-on-moving-curved-vanes - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7 ## Question 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. ## 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. ## Answer 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 $21.3^\circ$ to the jet, and the work done is about **18.2 kW**. #### Given $d = 5$ cm, $Q = 0.056\ \text{m}^3/\text{s}$, $u = 12$ m/s, deflection $= 135^\circ$, $V_{r2} = 0.9\,V_{r1}$, $\rho = 1000\ \text{kg/m}^3$. #### Velocities $$ \begin{aligned} V &= \frac{Q}{A} = \frac{0.056}{\frac{\pi}{4}(0.05)^2} = 28.52\ \text{m/s} \\ V_{r1} &= V - u = 16.52\ \text{m/s},\qquad V_{r2} = 0.9 \times 16.52 = 14.87\ \text{m/s} \end{aligned} $$ ```text 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 $$ \begin{aligned} F_x &= \rho Q\,(V - V_{x2}) = 1000(0.056)(28.52 - 1.49) = 1514\ \text{N} \\ F_y &= \rho Q\,V_{y2} = 1000(0.056)(10.51) = 589\ \text{N} \\ R &= \sqrt{1514^2 + 589^2} = 1624\ \text{N},\qquad \theta = \tan^{-1}\frac{589}{1514} = 21.3^\circ \end{aligned} $$ The jet is turned upward, so the force on the vanes has a component $F_x$ along the jet and a component $F_y$ of 589 N downward. **Answer (a): $R \approx 1.62$ kN, at $21.3^\circ$ below the direction of the jet.** #### (b) Work done per second $$ \text{Work} = F_x\,u = 1514 \times 12 = 18\,167\ \text{J/s} \approx 18.2\ \text{kW} $$ Only $F_x$ does work, because the vanes move along the jet direction. **Answer (b): about 18.2 kW.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the whole discharge Q used in the momentum equation here? A. Because a single vane moves at the jet speed B. The vanes form a series, so all the jet water is deflected C. Because the vane is stationary D. Because the relative velocity is zero Answer: B. For a series of vanes the full jet mass flow is intercepted. (Key point: Series of vanes) 2. What is the relative velocity at inlet and exit? A. Vr1 = V - u = 16.52 m/s and Vr2 = 0.9 Vr1 = 14.87 m/s B. Vr1 = V + u and Vr2 = Vr1 C. Vr1 = V = 28.52 m/s and Vr2 = 0.9 V D. Vr1 = u = 12 m/s and Vr2 = 0.9 u Answer: A. The vane moves with the jet, so relative inlet speed is V - u; friction cuts the exit value to 0.9 of it. (Key point: Relative velocity) 3. Which expression gives the force on the vanes along the jet? A. Fx = rho Q (V + Vx2) B. Fx = rho Q Vr2 C. Fx = rho Q u D. Fx = rho Q (V - Vx2) with Vx2 = u + Vr2 cos135 Answer: D. Change of momentum along the jet is inlet V minus exit absolute x-velocity. (Key point: Force components) 4. Why is the work done per second equal to Fx times u only? A. Fy is zero B. Fy acts but is cancelled by Fx C. The vanes move along the jet direction, so only Fx does work D. Work always uses the resultant force Answer: C. Work needs displacement along the force; vanes move in the x direction only. 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