Water jet on fixed vane, 100 mm2

Linear and angular momentum of a system of particles

• Linear momentum: $\vec L=\sum m_i\vec v_i=m\vec{\bar v}$, where $m$ is the total mass and $\vec{\bar v}$ the velocity of the mass centre. Newton's law: $\sum\vec F=\dot{\vec L}$.
• Angular momentum about O: $\vec H_O=\sum\vec r_i\times m_i\vec v_i$, with $\sum\vec M_O=\dot{\vec H}_O$.

Force of a fixed vane on a steady stream

Assumption: the stream arrives horizontally (to the right) and leaves at C at $60^\circ$ above the horizontal, in the forward direction. Since the vane is fixed and friction is neglected, the speed is unchanged: $v_B=v_C=60$ m/s.

   V_B --->   ______
              \      \  V_C (60 deg above horizontal)
               \______\
                 vane

Mass flow rate:

$$\dot m=\rho Av=1000(100\times10^{-6})(60)=6.0\ \text{kg/s}$$

Momentum equation (force on the stream by the vane):

$$\vec F=\dot m(\vec v_C-\vec v_B)$$ $$\begin{aligned} F_x&=6\left(60\cos60^\circ-60\right)=-180\ \text{N}\\ F_y&=6\left(60\sin60^\circ-0\right)=311.8\ \text{N}\\ F&=\sqrt{180^{2}+311.8^{2}}=360\ \text{N} \end{aligned}$$

The direction is $\tan^{-1}(311.8/180)=60^\circ$ above the negative x-axis (that is $120^\circ$ from +x). Check: $F=2\dot mv\sin(\tfrac{\alpha}{2})=2(6)(60)\sin30^\circ=360$ N, where $\alpha=60^\circ$ is the deflection angle.

Answer: the vane exerts $\vec F=(-180\hat i+311.8\hat j)$ N on the stream, magnitude $360$ N at $120^\circ$ to the original flow (the stream pushes on the vane with an equal and opposite force).