Question

Water flows out of a big tank along a tube bent at right angles; the inside radius of the tube is equal to $r=0.50cm$ (figure shown above). The length of the horizontal section of the tube is equal to $l=22cm$. The water flow rate is $Q=0.50$ litres per second. Find the moment of reaction forces of flowing water, acting on the tube's walls, relative to the point $O.$

Answer 1

Let the velocity of water flowing through the tube at a certain instant
of time be $u$, then $u=\frac{Q}{\pi r^2}$, where
$Q$ is the rate of flow of water and $\pi r^2$ is the cross section
area of the tube.
From impulse momentum theorem, for the stream of water striking the tube corner, in x-direction in the time interval $dt$,
$F_{x}dt=-\rho Qudt$ or $F_x=-\rho Qu$
and similarly, $F_y=\rho Qu$
Therefore, the force exerted on the water stream by the tube,
$\overrightarrow{F}=-\rho Qu\overrightarrow{i}+\rho Qu\overrightarrow{j}$
According to third law, the reaction force on the tube's wall by the stream equals $(-\overrightarrow{F})$
$=\rho Qu\overrightarrow{i}-\rho Qu\overrightarrow{j}$.
Hence, the sought moment of force about $0$ becomes
$\overrightarrow{N}=l(-\overrightarrow{i})\times (\rho Qu\overrightarrow{i}-\rho Qu\overrightarrow{j})=\rho Qul\overrightarrow{k}=\frac{\rho Q^2}{\pi r^2}l\overrightarrow{k}$
and $|\overrightarrow{N}|=\frac{\rho Q^{2}l}{\pi r^2}=0.70Nm=70Ncm$