Water jet on moving blade, 2 in diameter

Assumptions (figure not shown): the blade turns the water so that, relative to the blade, it leaves at $60^\circ$ above the horizontal with unchanged relative speed (no friction). The blade moves away from the jet at $5$ ft/s. Pound-force units.

Data

$$\rho=\frac{62.4}{32.2}=1.938\ \text{slug/ft}^3,\qquad A=\frac{\pi}{4}\left(\frac{2}{12}\right)^{2}=0.02182\ \text{ft}^2$$ $$v_{rel}=25-5=20\ \text{ft/s}$$

Mass flow rate hitting the moving blade

$$\dot m=\rho A\,v_{rel}=1.938(0.02182)(20)=0.8456\ \text{slug/s}$$

Force of the blade on the water

In the frame moving with the blade, the water enters at $20$ ft/s horizontally and leaves at $20$ ft/s at $60^\circ$:

$$\begin{aligned} F_x&=\dot m(20\cos60^\circ-20)=0.8456(-10)=-8.46\ \text{lb}\\ F_y&=\dot m(20\sin60^\circ)=0.8456(17.32)=14.65\ \text{lb} \end{aligned}$$

The blade pushes on the water with $8.46$ lb opposite to the jet direction (horizontal component) and $14.65$ lb upward (vertical component); resultant $16.9$ lb.

Power on the blade

The water pushes the blade forward with $8.46$ lb at speed $5$ ft/s:

$$P=8.456(5)=42.3\ \text{ft lb/s}=\frac{42.3}{550}=0.0769\ \text{hp}$$

Answer: $F_x=-8.46$ lb, $F_y=+14.65$ lb (on the water); power developed $\approx42.3$ ft lb/s ($0.077$ hp).