Question 5 of 13: Jet on a Moving Curved Plate (180° Turn)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examinations, December 2014. Three (3) hours, closed book. Section A: Calculative (9 questions, do 7); Section B: Analytical/Graphical (4 questions, do 3). Ten questions constitute a complete paper (50 marks). Every printed question is solved below, including the two "extra" questions in each Section beyond the minimum required.
Find. (a) U for maximum efficiency; (b) force F on the plate; (c) power P developed; (d) efficiency η.
Fig. Q5 — jet of speed V1 strikes a curved (bucket-like) plate moving at U in the same direction; the flow leaves reversed (180° turn) relative to the plate.
Approach. Work in the frame moving with the plate: the relevant relative velocity is Vr = V1 − U, and only the relative mass flow (ρA·Vr) participates in momentum exchange. Maximising the power-efficiency ratio over U gives the classic single-vane result.
(a) Plate speed for maximum efficiency. The efficiency η(U) = 4U(V1−U)/V12 is maximised where dη/dU = 0, i.e. at the midpoint of the parabola:
$$U = \frac{V_1}{2} = \frac{30}{2} = \boxed{15.0\text{ m/s}}$$
(b) Force on the plate. Relative velocity Vr = 30−15 = 15 m/s; jet area A = π(0.050)²/4 = 1.9635×10−3 m²; relative mass flow ṁrel = ρAVr = 29.45 kg/s. A full 180° turn reverses the relative velocity, so ΔVrel = 2Vr:
$$F = \dot m_{rel}(2V_r) = 29.45\times 2(15) = \boxed{883.6\text{ N}}$$
(c) Power developed.
$$P = FU = 883.6\times 15 = \boxed{13.25\text{ kW}}$$
(d) Efficiency of energy transfer. Compared against the full kinetic power carried by the jet itself:
$$P_{jet} = \tfrac{1}{2}\rho A V_1^3 = \tfrac{1}{2}(1000)(1.9635\times10^{-3})(30)^3 = 26.51\text{ kW}$$
$$\eta = \frac{P}{P_{jet}} = \frac{13.25}{26.51} = \boxed{50.0\%}$$
