Air jet on curved vane with pitot

Given: nozzle $d = 2$ cm, standard air $\rho_a = 1.225$ kg/m³, water manometer reading $h = 7$ cm (water $\rho_w = 1000$ kg/m³). The jet is deflected by the fixed vane through $30^\circ$ and leaves at the same speed (no friction).

Speed of the air from the pitot reading

A stagnation pitot tube in the nozzle exit plane (where static pressure is atmospheric) reads the dynamic pressure:

$$p_0 - p = \frac{1}{2}\rho_aV^2 = \rho_wgh$$ $$V = \sqrt{\frac{2\rho_wgh}{\rho_a}} = \sqrt{\frac{2(1000)(9.81)(0.07)}{1.225}} = \sqrt{1121.1} = 33.5\ \text{m/s}$$

Horizontal force on the vane

$$A = \frac{\pi}{4}(0.02)^2 = 3.1416\times 10^{-4}\ \text{m}^2$$ $$\dot m = \rho_aAV = 1.225\times 3.1416\times 10^{-4}\times 33.48 = 0.01288\ \text{kg/s}$$

The jet enters horizontally with velocity $V$ and leaves at $30^\circ$ to the horizontal with the same speed (pressure atmospheric everywhere, so only momentum flux matters). Taking $x$ along the incoming jet:

$$\begin{aligned} -R_x &= \dot m\,(V\cos30^\circ - V)\\ R_x &= \dot mV(1 - \cos30^\circ) = 0.01288\times 33.48\times 0.13397\\ &= 0.0578\ \text{N} \end{aligned}$$

where $R_x$ is the force on the vane in the direction of the incoming jet.

Answer: speed of air $V \approx 33.5$ m/s; horizontal force on the vane $\approx 0.058$ N in the direction of the jet. (Equivalently $R_x = \rho_aAV^2(1-\cos30^\circ) = 0.4315 \times 0.134$.)