Jet on moving blade; velocity for max power

Setup. The jet has area $A$ and speed $V_A$. The single blade moves at $V$ in the jet direction. The water moves along the blade with constant speed relative to it, so its relative speed is $V_A-V$ at entry and at exit. It leaves at angle $\theta$ to the horizontal (relative to the blade).

  V_A --->  \
             \__  V_rel leaves at theta
   blade ->  V

i) Components of the force exerted by the blade on the stream

Mass of water reaching the moving blade per second: $\dot m=\rho A(V_A-V)$.

Use the frame of the blade ($V$ constant), with relative speed $V_A-V$:

$$F_x=\dot m\left[(V_A-V)\cos\theta-(V_A-V)\right]=-\rho A(V_A-V)^{2}(1-\cos\theta)$$ $$F_y=\dot m(V_A-V)\sin\theta=\rho A(V_A-V)^{2}\sin\theta$$

The negative $F_x$ means the force on the stream is opposite to the jet direction. The stream pushes the blade with $\rho A(V_A-V)^{2}(1-\cos\theta)$ forward and $\rho A(V_A-V)^{2}\sin\theta$ downward.

ii) Velocity for maximum power

Power developed on the blade (only the forward force does work):

$$P=F_{blade,x}\,V=\rho A(V_A-V)^{2}(1-\cos\theta)\,V$$

For maximum power, $dP/dV=0$:

$$\frac{d}{dV}\left[(V_A-V)^{2}V\right]=(V_A-V)^{2}-2V(V_A-V)=(V_A-V)(V_A-3V)=0$$ $$\boxed{V=\frac{V_A}{3}}$$

(the root $V=V_A$ gives zero power). Then

$$P_{max}=\rho A\left(\tfrac23V_A\right)^{2}(1-\cos\theta)\frac{V_A}{3}=\frac{4}{27}\rho AV_A^{3}(1-\cos\theta)$$

Answer: (i) $F_x=-\rho A(V_A-V)^2(1-\cos\theta)$, $F_y=\rho A(V_A-V)^2\sin\theta$ on the stream; (ii) power is maximum at $V=V_A/3$.