Force of jet on flat plate, stationary and moving

Given: $d = 0.05$ m, $V = 25$ m/s, $u = 8$ m/s, $\rho = 1000$ kg/m³.

Jet area: $a = \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}$ m².

Kinetic energy of jet per second (input power): $\frac12\rho a V^3 = 15340$ W.

(a) Stationary plate

The jet loses all its momentum normal to the plate.

$$F = \rho a V^2 = 1000\times 1.9635\times10^{-3}\times 25^2 = 1227.2\ \text{N}$$

(b) Single plate moving at 8 m/s

Only the relative velocity $(V-u)$ acts, and the mass hitting the plate per second is $\rho a (V-u)$ (the plate runs away from the jet, so some water is not caught).

$$\begin{aligned} F &= \rho a (V-u)^2 = 1000\times1.9635\times10^{-3}\times 17^2 = 567.5\ \text{N}\\ W &= F\,u = 567.5\times 8 = 4540\ \text{W}\\ \eta &= \frac{F u}{\tfrac12\rho a V^3} = \frac{4540}{15340} = 29.6\% \end{aligned}$$

(c) Series of plates on a wheel

With many plates, a plate is always in front of the jet, so the whole discharge $\rho a V$ strikes the plates, but each kilogram loses only the relative velocity $(V-u)$.

$$\begin{aligned} F &= \rho a V (V-u) = 1000\times1.9635\times10^{-3}\times25\times 17 = 834.5\ \text{N}\\ W &= F u = 6676\ \text{W}\\ \eta &= \frac{2u(V-u)}{V^2} = \frac{6676}{15340} = 43.5\% \end{aligned}$$

The maximum efficiency of a series of flat plates is 50 %, at $u = V/2$.

Answer: (a) 1227.2 N; (b) 567.5 N, 4540 W, 29.6 %; (c) 834.5 N, 6676 W, 43.5 %.