Jet on flat plate, force and efficiency

Notation: jet velocity $U$, area $A$, density $\rho$. The kinetic power available in the jet is $\dfrac{1}{2}\rho AU^3$.

(i) Plate at rest

$$\begin{aligned} F &= \rho AU^2\\ \text{Power developed} &= F\times 0 = 0\\ \eta &= 0 \end{aligned}$$

A stationary plate receives a force but does no work, so its efficiency is zero (the jet's energy is carried away as kinetic energy of the deflected sheet).

(ii) Plate moving with velocity u in the direction of the jet

Only the relative velocity $(U - u)$ hits the plate, and the mass flow striking the plate is $\rho A(U - u)$:

$$\begin{aligned} F &= \rho A(U - u)^2\\ P &= Fu = \rho A(U - u)^2u\\ \eta &= \frac{P}{\frac{1}{2}\rho AU^3} = \frac{2u(U - u)^2}{U^3} \end{aligned}$$

Condition for maximum efficiency:

$$\frac{d\eta}{du} = 0\;\Rightarrow\; (U - u)^2 - 2u(U - u) = 0\;\Rightarrow\; U - 3u = 0\;\Rightarrow\; u = \frac{U}{3}$$ $$\eta_{max} = \frac{2(U/3)(2U/3)^2}{U^3} = \frac{8}{27} = 29.6\ \%$$

(iii) Series of plates (e.g. wheel with many vanes)

With plates arriving in quick succession, the whole jet discharge $\rho AU$ is used, not just $\rho A(U - u)$:

$$\begin{aligned} F &= \rho AU(U - u)\\ P &= Fu = \rho AU(U - u)u\\ \eta &= \frac{2u(U - u)}{U^2} \end{aligned}$$ $$\frac{d\eta}{du} = 0\;\Rightarrow\; U - 2u = 0\;\Rightarrow\; u = \frac{U}{2},\qquad \eta_{max} = \frac{2(U/2)(U/2)}{U^2} = 0.5 = 50\ \%$$

Answer: (i) $F = \rho AU^2$, $P = 0$, $\eta = 0$; (ii) $F = \rho A(U-u)^2$, $\eta = 2u(U-u)^2/U^3$, maximum $8/27 = 29.6\%$ at $u = U/3$; (iii) $\eta_{max} = 50\%$ at $u = U/2$.