Force on 120° bend, 300 lps

Given: $d_1 = 0.3$ m, $d_2 = 0.15$ m, deflection $120^\circ$ in a vertical plane, inlet horizontal, outlet 1.5 m below the inlet, volume of fluid $= 0.09$ m³, $Q = 300$ L/s $= 0.3$ m³/s, $p_1 = 130$ kPa (gauge). Friction is neglected.

 1 --->  [ bend ]
              \ 120 deg
               \
                v 2 (1.5 m lower)

Take $x$ along the inlet flow and $y$ upward. Outlet direction: $(\cos120^\circ, -\sin120^\circ) = (-0.5, -0.866)$.

Velocities and outlet pressure

$$\begin{aligned} A_1 &= 0.070686\ \text{m}^2,\quad A_2 = 0.017671\ \text{m}^2\\ V_1 &= 4.244\ \text{m/s},\quad V_2 = 16.977\ \text{m/s} \end{aligned}$$

Bernoulli (no loss):

$$p_2 = p_1 + \gamma\left[\frac{V_1^2 - V_2^2}{2g} + 1.5\right] = 130\,000 + 9810\,(0.918 - 14.690 + 1.5) = 9620\ \text{Pa}$$

Forces

• $p_1A_1 = 130\,000\times 0.070686 = 9189\ \text{N}$
• $p_2A_2 = 9620\times 0.017671 = 170\ \text{N}$, acting on the water along $(+0.5, +0.866)$, i.e. opposite to the outlet flow
• $W = 9810\times 0.09 = 883\ \text{N}$ (downward)
• $\rho Q = 300$ kg/s

Momentum equation ($R$ = force of the bend on the water):

$$\begin{aligned} x:\ \ 9189 + 170(0.5) + R_x &= 300\,(16.977\times(-0.5) - 4.244) = -3820\\ R_x &= -13\,094\ \text{N}\\[4pt] y:\ \ 170(0.866) - 883 + R_y &= 300\,(-16.977\times 0.866) = -4411\\ R_y &= -3675\ \text{N} \end{aligned}$$

The force exerted by the water on the bend is $(+13\,094,\ +3675)$ N, i.e. $F_x = 13\,094$ N and $F_y = 3675$ N:

$$F = \sqrt{13\,094^2 + 3675^2} = 13\,600\ \text{N},\qquad \theta = \tan^{-1}\frac{3675}{13\,094} = 15.7^\circ$$

Answer: the force of the water on the bend is $\approx 13.6$ kN, acting at about $15.7^\circ$ above the horizontal in the direction of the inlet flow ($F_x = 13.09$ kN, $F_y = 3.68$ kN upward).