Force on a reducing pipe bend

Given data

$d_1 = 0.3\ \text{m}$, $d_2 = 0.15\ \text{m}$, $Q = 0.15\ \text{m}^3/\text{s}$, $p_1 = 200\ \text{kPa}$ (gauge), $\theta = 45^\circ$, $\rho = 1000\ \text{kg/m}^3$. Take $x$ along the inlet flow and $y$ perpendicular to it, towards the side to which the flow turns.

Areas and velocities

$$\begin{aligned} A_1 &= \frac{\pi}{4}(0.3)^2 = 0.07069\ \text{m}^2, & A_2 &= \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2 \\ V_1 &= \frac{0.15}{0.07069} = 2.122\ \text{m/s}, & V_2 &= \frac{0.15}{0.01767} = 8.488\ \text{m/s} \end{aligned}$$

Outlet pressure (Bernoulli, horizontal, no loss)

$$p_2 = p_1 + \frac{\rho}{2}(V_1^2 - V_2^2) = 200\,000 + 500(4.503 - 72.05) = 166\,226\ \text{Pa}$$

Pressure forces

$$p_1A_1 = 200\,000 \times 0.07069 = 14\,137\ \text{N}, \qquad p_2A_2 = 166\,226 \times 0.01767 = 2937\ \text{N}$$

Momentum equation (forces on the fluid)

Let $R_x$ and $R_y$ be the forces exerted by the bend on the fluid. $\rho Q = 150\ \text{kg/s}$.

In the $x$-direction:

$$p_1A_1 - p_2A_2\cos45^\circ + R_x = \rho Q (V_2\cos45^\circ - V_1)$$ $$\begin{aligned} 14\,137 - 2077 + R_x &= 150(6.002 - 2.122) = 582 \\ R_x &= -11\,478\ \text{N} \end{aligned}$$

In the $y$-direction:

$$\begin{aligned} -p_2A_2\sin45^\circ + R_y &= \rho Q V_2\sin45^\circ \\ R_y &= 2077 + 150 \times 6.002 = 2977\ \text{N} \end{aligned}$$

Force of water on the bend

By Newton's third law, the force on the bend is equal and opposite:

$$F_x = 11\,478\ \text{N}, \qquad F_y = -2977\ \text{N}$$ $$\begin{aligned} F &= \sqrt{11\,478^2 + 2977^2} = 11\,858\ \text{N} \\ \alpha &= \tan^{-1}\frac{2977}{11\,478} = 14.5^\circ \end{aligned}$$

$F_x$ acts in the direction of the incoming flow, and $F_y$ acts away from the side to which the flow turns.

Answer: $F = 11.86\ \text{kN}$ at $14.5^\circ$ to the inlet flow direction, tilted away from the side to which the water turns ($F_x = 11.48\ \text{kN}$, $F_y = 2.98\ \text{kN}$)