Jet striking fixed and moving curved vanes

Given data

$d = 0.05\ \text{m}$, $V = 25\ \text{m/s}$, deflection angle $\theta = 165^\circ$, $u = 10\ \text{m/s}$.

Jet area: $a = \dfrac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2$

For a smooth vane with deflection $\theta$, the change of velocity component in the jet direction (relative to the vane) is $V_r(1 - \cos\theta)$.

$$1 - \cos165^\circ = 1 + 0.9659 = 1.9659$$

(a) Fixed vane

All the jet strikes the vane, $\dot m = \rho a V$ and $V_r = V$:

$$\begin{aligned} F &= \rho a V^2 (1 - \cos\theta) \\ &= 1000 \times 1.9635\times10^{-3} \times 625 \times 1.9659 \\ &= 2412.6\ \text{N} \end{aligned}$$

(b) Moving vane (single vane)

For a single moving vane the mass striking the vane per second is only $\rho a (V - u)$, because the vane is moving away from the nozzle. The relative velocity is $V - u = 15\ \text{m/s}$.

$$\begin{aligned} F &= \rho a (V - u)^2 (1 - \cos\theta) \\ &= 1000 \times 1.9635\times10^{-3} \times 15^2 \times 1.9659 \\ &= 868.5\ \text{N} \end{aligned}$$

Work done per second (power) = force $\times$ vane velocity:

$$P = F u = 868.5 \times 10 = 8685\ \text{W} = 8.685\ \text{kW}$$

Kinetic energy of the jet per second:

$$\text{KE} = \tfrac12 \rho a V^3 = 0.5 \times 1000 \times 1.9635\times10^{-3} \times 15\,625 = 15\,340\ \text{W}$$ $$\eta = \frac{8685}{15\,340} = 0.566 = 56.6\%$$

Answer: (a) $F = 2.41\ \text{kN}$; (b) $F = 868.5\ \text{N}$, $P = 8.69\ \text{kW}$, $\eta = 56.6\%$