Water flows through a tube assembly as shown  in the fig. Speed of flow (marked as V and 2V),  cross sectional area (A, A/2 and A/4) and the 
angles between segments has been shown in fig.  Calculate the force applied by the water flow on  the tube. Take density of water to be $\rho$ 
 

 

Let mass of water that gets divided into two streams in time interval $\mathrm{\Delta t}\text{ be }\mathrm{\Delta m}$ $\mathrm{\Delta m}=\mathrm{AV\rho \Delta t}$
Change in momentum of water in X and Y direction is
$$\begin{gathered} {\mathrm{\Delta P}}_{\mathrm{x}}=\frac{\mathrm{A}}{2}\mathrm{V\Delta t}\cdot \mathrm{\rho }\cdot \mathrm{V}\cdot \cos {60}^{\circ }+\frac{\mathrm{A}}{4}2\mathrm{V\Delta t\rho }\cdot 2\mathrm{V}\left(\frac{3}{4}\right)-\mathrm{AV\Delta t}.\mathrm{V}=\mathrm{O} \\ △P_{\mathrm{y}}=\frac{\mathrm{A}}{2}\mathrm{V}△\mathrm{t} \mathrm{\rho Vsin}{60}^{\circ }-\frac{\mathrm{A}}{4}2\mathrm{V}\cdot \mathrm{\rho }\cdot △t 2\mathrm{V}\cdot \left(\frac{\sqrt{7}}{4}\right) \\ =-\left(\frac{\sqrt{7}-\sqrt{3}}{4}\right){\mathrm{\rho AV}}^2\mathrm{\Delta t} \end{gathered}$$
$\therefore$Force on water is along negative y direction
$$\begin{gathered} \mathrm{Fy}=\frac{{\mathrm{\Delta P}}_{\mathrm{y}}}{\mathrm{\Delta t}}=-\left(\frac{\sqrt{7}-\sqrt{3}}{4}\right){\mathrm{\rho AV}}^2 \\ \therefore \text{ Force on tube, }{\mathrm{F}}_{\mathrm{y}}=\left(\frac{\sqrt{7}-\sqrt{3}}{4}\right){\mathrm{\rho AV}}^2 \end{gathered}$$