--- title: "Jet on moving blade; velocity for max power" subject: "Applied Mechanics (Dynamics)" chapter: 3 key: "semester_3/applied_mechanics_dynamics:nozzle-single-blade-max-power" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/nozzle-single-blade-max-power site: BCT Engineering --- # Jet on moving blade; velocity for max power Applied Mechanics (Dynamics) · Chapter 3: System of particles - Asked in: 2072 Chaitra, 8 marks - Video: [Impact Of Liquid Jet On Series Of Flat Plate Mounted On A Wheel](https://www.youtube.com/watch?v=cpM6hF23eeQ) — BEST MECHANICAL ENGINEERING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/nozzle-single-blade-max-power - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3 ## Question A nozzle discharges a stream of water of cross-sectional area 'A' with a velocity $V_A$. The stream is deflected by single blade which moves to the right with a constant velocity $V$. Assuming that the water moves along the blade at a constant speed, determine: i) The component of forces exerted by the blade on the stream. ii) The velocity $V$ for which maximum power is developed. [Figure: stream with velocity $V_A$ from the left on a curved blade B moving right at $V$; blade exit angle $\theta$ to the horizontal.] ## Short answer Mass flow relative to blade is ρA(V_A–V). Components on stream: F_x = –ρA(V_A–V)²(1–cosθ), F_y = ρA(V_A–V)² sinθ. Power is maximized when blade speed V = V_A/3. ## Answer **Setup.** The jet has area $A$ and speed $V_A$. The single blade moves at $V$ in the jet direction. The water moves along the blade with constant speed relative to it, so its relative speed is $V_A-V$ at entry and at exit. It leaves at angle $\theta$ to the horizontal (relative to the blade). ```text V_A ---> \ \__ V_rel leaves at theta blade -> V ``` #### i) Components of the force exerted by the blade on the stream Mass of water reaching the moving blade per second: $\dot m=\rho A(V_A-V)$. Use the frame of the blade ($V$ constant), with relative speed $V_A-V$: $$ F_x=\dot m\left[(V_A-V)\cos\theta-(V_A-V)\right]=-\rho A(V_A-V)^{2}(1-\cos\theta) $$ $$ F_y=\dot m(V_A-V)\sin\theta=\rho A(V_A-V)^{2}\sin\theta $$ The negative $F_x$ means the force on the stream is opposite to the jet direction. The stream pushes the blade with $\rho A(V_A-V)^{2}(1-\cos\theta)$ forward and $\rho A(V_A-V)^{2}\sin\theta$ downward. #### ii) Velocity for maximum power Power developed on the blade (only the forward force does work): $$ P=F_{blade,x}\,V=\rho A(V_A-V)^{2}(1-\cos\theta)\,V $$ For maximum power, $dP/dV=0$: $$ \frac{d}{dV}\left[(V_A-V)^{2}V\right]=(V_A-V)^{2}-2V(V_A-V)=(V_A-V)(V_A-3V)=0 $$ $$ \boxed{V=\frac{V_A}{3}} $$ (the root $V=V_A$ gives zero power). Then $$ P_{max}=\rho A\left(\tfrac23V_A\right)^{2}(1-\cos\theta)\frac{V_A}{3}=\frac{4}{27}\rho AV_A^{3}(1-\cos\theta) $$ **Answer: (i) $F_x=-\rho A(V_A-V)^2(1-\cos\theta)$, $F_y=\rho A(V_A-V)^2\sin\theta$ on the stream; (ii) power is maximum at $V=V_A/3$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the mass flow reaching a single moving blade? A. rho A (V_A - V) B. rho A V_A C. rho A V D. rho A (V_A + V) Answer: A. Only the water arriving at relative speed V_A - V strikes the blade per second. (Key point: Mass flow on moving blade) 2. What is the horizontal force on the stream? A. rho A (V_A - V)^2 sin(theta) B. rho A V_A^2 (1 + cos theta) C. -rho A (V_A - V)^2 (1 - cos theta) D. rho A (V_A - V)(1 - cos theta) Answer: C. F_x = m dot[(V_A - V)cos theta - (V_A - V)]; the negative sign means it opposes the jet. (Key point: Horizontal force) 3. What is set to zero to find the velocity for maximum power? A. dP/dtheta B. d^2P/dV^2 C. P itself D. dP/dV, where P = rho A (V_A - V)^2 (1 - cos theta) V Answer: D. Power on the blade is force times V; setting dP/dV = 0 gives (V_A - V)(V_A - 3V) = 0. (Key point: Maximum power condition) 4. What is the optimum blade velocity? A. V = V_A/2 B. V = V_A/3 C. V = V_A D. V = 2V_A/3 Answer: B. V = V_A/3 (V = V_A gives zero power), with P_max = (4/27) rho A V_A^3 (1 - cos theta). (Key point: Optimum V) ## Related in Applied Mechanics (Dynamics) - [Water jet on moving blade, 2 in diameter](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Awater-jet-moving-blade-2in) — Applied Mechanics (Dynamics), chapter 3 - [Water jet on vane moving at 16 m/s](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Anozzle-fixed-vane-moving-4000mm2) — Applied Mechanics (Dynamics), chapter 3 - [KE of system of particles; jet on moving plate](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Akinetic-energy-system-water-plate) — Applied Mechanics (Dynamics), chapter 3 - [Water jet on fixed vane, 100 mm2](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Anozzle-fixed-vane-force-100mm2) — Applied Mechanics (Dynamics), chapter 3 ## Related in other subjects - [Normal-tangential acceleration; nozzle stream curvature](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Anormal-tangential-acceleration-nozzle) — Applied Mechanics, chapter 8 - [Nozzle stream curvature; dependent motion](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Anozzle-curvature-dependent-motion) — Applied Mechanics, chapter 8 - [Steam nozzle 4 MPa: mass flow, exit velocity, area](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteam-nozzle-4mpa-450c-to-2mpa) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [Jet striking fixed and moving curved vanes](https://www.bctengineer.com/question.md?key=semester_4%2Ffluid_mechanics_mechanical%3Ajet-on-fixed-and-moving-vane) — Fluid Mechanics, chapter 4