Question 6 of 13: Force, Power and Efficiency of a Jet on a Moving Curved Plate

Question 6: Force, Power and Efficiency of a Jet on a Moving Curved Plate (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Approach. Work in the frame of the moving plate: the fluid approaches at relative velocity (V₁−U) and leaves at the same relative speed reversed (180° turn, no friction), so the linear-momentum equation gives the force, and power follows from force × plate speed; efficiency compares that power to the kinetic-energy rate of the absolute jet.

1. Part (a): force on the plate. Relative velocity V1−U = 20 m/s; only the mass flow that actually catches the moving plate matters, ρA(V1−U), and the momentum change per unit mass for a full 180° reversal is 2(V1−U): $$ F = \rho A (V_1-U)\big[2(V_1-U)\big] = 2\rho A (V_1-U)^2 $$ $$ F = 2(1000)\left(\tfrac{\pi}{4}(0.050)^2\right)(20)^2 $$ $$ \boxed{F \approx 1571\ \text{N}} $$
2. Part (b): power developed. $$ P = F\,U = (1571)(10) $$ $$ \boxed{P \approx 15\,708\ \text{W}} $$
3. Part (c): efficiency of energy transfer. Compare to the kinetic-energy rate of the absolute jet leaving the nozzle, $P_{in}=\tfrac12 \dot m V_1^2$ with ṁ=ρAV1: $$ \dot m = \rho A V_1 = (1000)(0.0019635)(30) \approx 58.9\ \text{kg/s} $$ $$ P_{in} = \tfrac12(58.9)(30)^2 \approx 26\,507\ \text{W} $$ $$ \eta = \frac{P}{P_{in}} = \frac{15\,708}{26\,507} $$ $$ \boxed{\eta \approx 59.3\%} $$
4. Part (d): plate speed for maximum efficiency. Writing power as a function of U, $P(U)=2\rho A(V_1-U)^2U$, and setting dP/dU=0 gives 3U^2-4V_1U+V_1^2=0, whose non-trivial root is: $$ \boxed{U_{opt} = \frac{V_1}{3} = \frac{30}{3} = 10\ \text{m/s}} $$ This is exactly the plate speed given in parts (a)–(c) — the problem is set up at its own optimum. The physical reason: power is the product of a shrinking relative-velocity-squared term (which falls as U rises) and a growing U term; the classic single-moving-vane result balances these two competing trends at U = V₁/3, beyond which the loss of relative momentum outweighs the gain in plate speed.