Resultant force on bifurcating bend

The resultant force is found by applying the Bernoulli equation to get the inlet pressure and then the momentum equation in the $x$ and $y$ directions. The result is about 3.71 kN, acting at about $47^\circ$ to the inlet axis.

Assumptions

The figure does not give the inlet diameter, so I take $d_1 = 20$ cm. The inlet pipe lies along the $x$-axis, nozzle 2 ($d_2 = 12$ cm) continues straight, and nozzle 3 ($d_3 = 10$ cm) leaves at $60^\circ$ to the axis. The plane is horizontal, so weight does not matter. Both nozzles discharge to the atmosphere, so $p_2 = p_3 = 0$ (gauge). $\rho = 1000\ \text{kg/m}^3$.

                    / --> V3 (d3, 60 deg)
   V1, p1  ------->(
                    \ --> V2 (d2, straight)

Discharges and inlet velocity

$$\begin{aligned} A_1 &= 0.03142\ \text{m}^2,\; A_2 = 0.011310\ \text{m}^2,\; A_3 = 0.007854\ \text{m}^2 \\ Q_2 &= A_2V_2 = 0.2262\ \text{m}^3/\text{s},\quad Q_3 = A_3V_3 = 0.1571\ \text{m}^3/\text{s} \\ Q_1 &= Q_2 + Q_3 = 0.3833\ \text{m}^3/\text{s},\quad V_1 = Q_1/A_1 = 12.20\ \text{m/s} \end{aligned}$$

Inlet pressure (Bernoulli, no losses)

$$p_1 = \tfrac12\rho\,(V_2^2 - V_1^2) = \tfrac12(1000)(20^2 - 12.2^2) = 125\,580\ \text{Pa} = 125.6\ \text{kPa}$$

Momentum equation (force of bend on fluid, $F_x$, $F_y$)

$$\begin{aligned} F_x + p_1A_1 &= \rho\,(Q_2V_2 + Q_3V_3\cos 60^\circ - Q_1V_1) \\ F_x &= 1000\,(4.524 + 1.571 - 4.676) - 3945 = -2526\ \text{N} \\ F_y &= \rho Q_3V_3\sin 60^\circ = 1000(0.1571)(20)(0.866) = 2721\ \text{N} \end{aligned}$$

The force exerted on the bend by the water is equal and opposite:

$$R_x = 2526\ \text{N},\quad R_y = -2721\ \text{N}$$ $$R = \sqrt{2526^2 + 2721^2} = 3713\ \text{N},\qquad \theta = \tan^{-1}\frac{2721}{2526} = 47.1^\circ$$

Answer: $R \approx 3.71$ kN, directed $47.1^\circ$ below the inlet axis (away from the branch side), pushing the bend along the flow direction.