Force on 120° pipe bend, 0.23 m³/s

Given: $d_1 = 0.3$ m, $d_2 = 0.15$ m, deflection $120^\circ$, $Q = 0.23$ m³/s, outlet 1.4 m below inlet, volume of fluid in bend $= 0.085$ m³, $p_1 = 140$ kPa (gauge), $h_L = 0.25V_1^2/2g$.

 inlet 1  --->  [ bend ]
 (horizontal)        \ 120 deg
                      \
                       v  outlet 2 (1.4 m lower)

Take $x$ along the inlet flow and $y$ upward. The outlet velocity direction is $120^\circ$ from the inlet direction, pointing downward: unit vector $(\cos120^\circ, -\sin120^\circ) = (-0.5, -0.866)$.

Velocities and outlet pressure

$$\begin{aligned} A_1 &= 0.070686\ \text{m}^2,\quad A_2 = 0.017671\ \text{m}^2\\ V_1 &= \frac{0.23}{0.070686} = 3.254\ \text{m/s},\quad V_2 = \frac{0.23}{0.017671} = 13.015\ \text{m/s}\\ h_L &= 0.25\times\frac{3.254^2}{19.62} = 0.135\ \text{m} \end{aligned}$$ $$\frac{p_2}{\gamma} = \frac{p_1}{\gamma} + \frac{V_1^2 - V_2^2}{2g} + 1.4 - h_L$$ $$p_2 = 140\,000 + 9810\left(0.5396 - 8.6338 + 1.4 - 0.1349\right) = 73\,005\ \text{Pa}$$

Forces

• $p_1A_1 = 140\,000\times 0.070686 = 9896\ \text{N}$ (in $+x$)
• $p_2A_2 = 73\,005\times 0.017671 = 1290\ \text{N}$ (acts opposite to the outlet flow direction, i.e. along $(+0.5, +0.866)$)
• Weight of fluid $W = 9810\times 0.085 = 834\ \text{N}$ (downward)
• $\rho Q = 1000\times 0.23 = 230$ kg/s

Momentum equation (forces on fluid, $R$ = force of the bend on fluid):

$$\begin{aligned} x:\ \ p_1A_1 + p_2A_2\,(0.5) + R_x &= \rho Q\,(V_2\cos120^\circ - V_1)\\ 9896 + 645 + R_x &= 230(-6.508 - 3.254) = -2245\\ R_x &= -12\,786\ \text{N}\\[4pt] y:\ \ p_2A_2\,(0.866) - W + R_y &= \rho Q\,(V_2\sin(-120^\circ))\\ 1117 - 834 + R_y &= 230(-11.272) = -2593\\ R_y &= -2876\ \text{N} \end{aligned}$$

The force exerted by the water on the bend is the opposite of $R$:

$$F_x = +12\,786\ \text{N},\qquad F_y = +2876\ \text{N}$$ $$F = \sqrt{12\,786^2 + 2876^2} = 13\,106\ \text{N},\qquad \theta = \tan^{-1}\frac{2876}{12\,786} = 12.7^\circ$$

Answer: net force on the bend $\approx 13.1$ kN, acting at $12.7^\circ$ above the horizontal (in the direction of the inlet flow), with $F_x = 12.79$ kN and $F_y = 2.88$ kN upward.