Maximum efficiency on series of semicircular vanes

When a jet strikes a series of vanes, the whole jet discharge ($\rho aV$) is used on the moving vanes, because new vanes keep replacing the old ones. The efficiency is the work done on the vanes divided by the kinetic energy supplied by the jet. For semicircular vanes it reaches a maximum of 1 (100%) when the vane speed is $u = V/2$.

Setup

Let $V$ = jet velocity, $u$ = vane velocity (same direction), $a$ = jet area. Take the vanes as smooth, so the relative speed is unchanged.

   jet V -->   ____
 ============>/    \   vane moves
              \____/ -> u

• Relative velocity at inlet: $V_{r1} = V - u$.
• The semicircular vane turns the jet through $180^\circ$, so the relative velocity at exit is $V_{r2} = V - u$ in the opposite direction.
• Absolute exit velocity along the jet: $V_{2x} = u - (V-u) = 2u - V$.

Force and work

Mass flow striking the series of vanes = $\rho a V$.

$$\begin{aligned} F &= \rho a V\,[V - (2u - V)] = 2\rho a V (V-u) \\ \text{Work done per second} &= F u = 2\rho a V (V-u)\,u \end{aligned}$$

Efficiency

Input power (kinetic energy of the jet per second) $= \tfrac12 \rho a V \cdot V^2$.

$$\eta = \frac{2\rho a V (V-u)\,u}{\tfrac12 \rho a V^3} = \frac{4u(V-u)}{V^2}$$

Condition for maximum

$$\frac{d\eta}{du} = \frac{4}{V^2}(V - 2u) = 0 \;\Rightarrow\; u = \frac V2$$

The second derivative is $-8/V^2 < 0$, so this is a maximum.

$$\eta_{max} = \frac{4\,(V/2)(V/2)}{V^2} = 1$$

Hence the maximum efficiency is 1 (100%) at $u = V/2$. At this speed the exit absolute velocity is $2u - V = 0$, so the jet leaves with no kinetic energy and all of its energy is given to the vanes.